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4.9: Problem-Solving Strategies

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Learning Objectives

By the end of this section, you will be able to:

  • Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure (a). Then, as in Figure (b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

APPLYING NEWTON'S SECOND LAW

Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: \(F_{net} = ma\)

For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions: \[ F_{net \, x} = ma \]

\[ F_{net \, y} = 0 \]

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

  • To solve problems involving Newton’s laws of motion, follow the procedure described:
  • Draw a sketch of the problem.
  • Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
  • Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the \(x\)-direction) then \(F_{net \, x} = 0 \). If the object does accelerate in that direction, \(F_{net \, x} = ma \).
  • Check your answer. Is the answer reasonable? Are the units correct?

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High School Physics : Calculating Force

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : net force.

problem solving about force

Plug these into the equation to solve for acceleration.

problem solving about force

Example Question #2 : Calculating Force

problem solving about force

Plug in the values given to us and solve for the force.

problem solving about force

Example Question #1 : Calculating Force

problem solving about force

Plug in the given values to solve for the mass.

problem solving about force

(Assume the only two forces acting on the object are friction and Derek).

problem solving about force

Plug in the information we've been given so far to find the force of friction.

problem solving about force

Friction will be negative because it acts in the direction opposite to the force of Derek.

problem solving about force

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

problem solving about force

Example Question #6 : Calculating Force

problem solving about force

We can find the net force by adding the individual force together.

problem solving about force

If the object has a constant velocity, that means that the net acceleration must be zero.

problem solving about force

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

problem solving about force

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

problem solving about force

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

problem solving about force

We are given the mass, but we will need to calculate the acceleration to use in the formula.

problem solving about force

Plug in our given values and solve for acceleration.

problem solving about force

Now we know both the acceleration and the mass, allowing us to solve for the force.

problem solving about force

Example Question #9 : Calculating Force

problem solving about force

We can calculate the gravitational force using the mass.

problem solving about force

Example Question #10 : Calculating Force

problem solving about force

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4.6 Problem-Solving Strategies

Learning objectives.

By the end of this section, you will be able to:

  • Understand and apply a problem-solving procedure to solve problems using Newton's laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure 4.20 (a). Then, as in Figure 4.20 (b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 4.20 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well. A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 4.20 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure 4.20 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying Newton’s Second Law

Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: F net = ma F net = ma .

For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions:

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

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StickMan Physics

StickMan Physics

Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

F=ma Equations

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

1

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

2

                a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

3

               a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

4

            a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

5

             a= 4.47 m/s 2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

different masses

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

                F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

force at an angle

            v f = 8.58 m/s

  • Back to the Newtons Second Law Lesson
  • Continue to Mass and Weight
  • Back to the Main Forces Page
  • Back to the Stickman Physics Home Page
  • Equation Sheet

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problem solving about force

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problem solving about force

Simplifying a Difficult Problem

Consider the situation below in which a force is directed at an angle to the horizontal. In such a situation, the applied force could be resolved into two components. These two components can be considered to replace the applied force at an angle. By doing so, the situation simplifies into a familiar situation in which all the forces are directed horizontally and vertically.

Once the situation has been simplified, the problem can be solved like any other problem. The task of determining the acceleration involves first determining the net force by adding up all the forces as vectors and then dividing the net force by the mass to determine the acceleration. In the above situation, the vertical forces are balanced (i.e., F grav , F y , and F norm add up to 0 N), and the horizontal forces add up to 29.3 N, right (i.e., 69.3 N, right + 40 N, left = 29.3 N, right). The net force is 29.3 N, right and the mass is 10 kg (m = F grav /g); therefore, the acceleration is 2.93 m/s/s, right.

Your Turn to Practice

To test your understanding, analyze the two situations below to determine the net force and the acceleration. When finished, click the button to view the answers.

The net force is 69.9 N, right and the acceleration is 3.5 m/s/s, right .

Note that the vertical forces balance but the horizontal forces do not. The net force is

F net = 129.9 N, right - 60 N, left = 69.9 N, right

The mass is

m = (F grav / g) = 20 kg

So the acceleration is

a = (69.9 N) / (20 kg) =3.50 m/s/s.

The net force is 30.7 N, right and the acceleration is 1.23 m/s/s, right .

F net = 70.7 N, right - 40 N, left = 30.7 N, right
m = (F grav / g) = 25 kg
a = (30.7 N) / (25 kg) =1.23 m/s/s.

What's Up with the Normal Force?

There is one peculiarity about these types of problems that you need to be aware of. The normal force (F norm ) is not necessarily equal to the gravitational force (F grav ) as it has been in problems that we have previously seen. The principle is that the vertical forces must balance if there is no vertical acceleration. If an object is being dragged across a horizontal surface, then there is no vertical acceleration. For this reason, the normal force (F norm ) plus the vertical component (F y ) of the applied force must balance the gravitational force (F grav ). A quick review of these problems shows that this is the case. If there is an acceleration for an object being pulled across a floor, then it is a horizontal acceleration; and thus the only imbalance of force would be in the horizontal direction .

Now consider the following situation in which a force analysis must be conducted to fill in all the blanks and to determine the net force and acceleration. In a case such as this, a thorough understanding of the relationships between the various quantities must be fully understood. Make an effort to solve this problem. When finished, click the button to view the answers. (When you run into difficulties, consult the help from a previous unit .)

 The F grav is

F grav = m • g = (10 kg) • (9.8 m/s/s) = 98 N

Using the sine function,

F y = (60 N) • sine (30 degrees) = 30 N

Since vertical forces are balanced, F norm = 68 N.

Now F frict can be found

F frict = mu • F norm = ( 0.3) • (68 N) = 20.4 N

Using the cosine function,

F x = (60 N) • cosine (30 degrees) = 52.0 N

Now since all the individual force values are known, the F net can be found:

F net = 52.0 N,right + 20.4 N, left = 31.6 N, right.

The acceleration is

a = 3.16 m/s/ s, right.

In conclusion, a situation involving a force at an angle can be simplified by using trigonometric relations to resolve that force into two components. Such a situation can be analyzed like any other situation involving individual forces. The net force can be determined by adding all the forces as vectors and the acceleration can be determined as the ratio of Fnet/mass.

   

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Check your understanding.

The following problems provide plenty of practice with F net = m • a problems involving forces at angles. Try each problem and then click the button to view the answers.

Glen Brook and Warren Peace are incorrect. Warren Peace perhaps believes that the Fnorm = Fgrav; but this is only the case when there are only two vertical forces and no vertical acceleration; sorry Warren - there is a second vertical force in this problem (F app ).

Glen Brook perhaps thinks that the F app force is 50 N upwards and thus the F norm must be 50 N upwards to balance the Fgrav. Sorry Glen - the F app is only 25 N upwards (50 N) • sine 30 degrees).

"Datagal Olive!"

2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

First use the mass to determine the force of gravity.

F grav = m • g = (20 kg) • (9.8 m/s/s) = 196 N

Now find the vertical component of the applied force using a trigonometric function.

F y = (80 N) • sine (45 degrees) = 56.7 N

Thus, F norm =139.3 N in order for the vertical forces to balance.

The horizontal component of the applied force can be found as

F x = (80 N) • cosine (45 degrees) = 56.7 N

Since the speed is constant, the horizontal forces must also balance; and so F frict = 56.7 N.

The value of "mu" can be found using the equation

3. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Answer: "mu" = 0.25

 The F grav can be calculated from the mass of the object.

The vertical component of the applied force can be calculated using a trigonometric function:

F y = (80 N) • sine (30 degrees) = 40 N

In order for the vertical forces to balance, F norm + F y = F grav . Thus,

F norm = F grav - F y = = 196 N - 40 N = 156 N

The horizontal component of the applied force can be calculated using a trigonometric function:

F x = (80 N) • cosine (30 degrees) = 69.2 N

The net force is the sum of all the forces when added as vectors. Thus,

F net = (69.2 N, right) + (40 N,left) = 29.2 N, right
a = F net / m = (29.2 N, right) / (20 kg) = 1.46 m/s/s, right.

The value of "mu" can be found using the equation "mu" = F frict / F norm .

4. The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The F grav can be calculated from the mass of the object.

F grav = m • g = (5 kg) • (9.8 m/s/s) = 49 N

The vertical component of the applied force can be calculated using a trigonometric function.

F y = (15 N) • sine (45 degrees) =10.6 N
F norm = F grav - F y = = 49 N - 10.6 N = 38.4 N
F x = (15 N) • cosine (45 degrees) = 10.6 N

Since the speed is constant, the horizontal forces must balance. Therefore, F frict = 10.6 N.

The value of "mu" can be found using the equation "mu"= F frict / F norm :

"mu" = 0.276

5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Since the velocity is constant, the acceleration and the net force are 0 m/s/s and 0 N respectively. The F grav can be calculated from the mass of the object.

The object moves at constant speed; thus, the horizontal forces must balance. For this reason, F x = 10 N.

The applied force can now be found using a trigonometric function and the horizontal component:

cosine (60 degrees) = (10 N) / (F app )

Proper algebra yields

F app = (10 N) / [cosine (60 degrees) ] = 20 N
F y = (20 N) • sine (60 degrees) = 17.3 N
F norm = F grav - F y = 49 N - 17.3 N = 31.7 N

6. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +2 m/s/s.

The net force can be calculated as:

F net = m • a = (10 kg) • (2 m/s/s) = 20 N, right.
F x = (70 N) • cosine (45 degrees) = 49.5 N

The net force is the vector sum of all the forces. Thus, F net = F x + F frict . That is,

20 N, right = 49.5 N, right + F frict

Therefore, F frict must be 29.5 N, left.

F y = (70 N) • sine (45 degrees) = 49.5 N
F norm = F grav - F y = = 98 N - 49.5 N = 48.5 N

The value of "mu" can be found using the equation "mu" = F frict / F norm :

7. Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force for 2.0 seconds.

F y = (50 N) • sine (45 degrees) = 35.4 N

In order for the vertical forces to balance, F norm + F y = F grav . Algebraic rearrangement leads to:

F norm = F grav - F y = 196 N - 35.4 N = 160.6 N
F x = (50 N) • cosine (45 degrees) = 35.4 N

The F net is 35.4 N, right since the only force which is not balanced is F x .

The acceleration is:

a = F net / m = (35.4 N) / (20 kg) = 1.77 m/s/s

The velocity after 2.0 seconds can be calculated using a kinematic equation:

v f = v i + a • t = 0 m/s + (1.77 m/s/s) • (2.0 s) v f = 3.54 m/s

8. A student pulls a 2-kg backpack across the ice (assume friction-free) by pulling at a 30-degree angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force.

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +0.125 m/s/s.

F net = m • a = (2 kg) • (0.125 m/s/s) = 0.250 N, right

Since the acceleration is horizontal, the vertical forces balance each other. The horizontal component of the applied force (F x ) supplies the horizontal force required for the acceleration. Thus, the horizontal component of the applied force is 0.250 N.

Using trigonometry, the applied force (F app ) can be calculated:

cosine (30 degrees) = ( 0.250 N) / (F app )

Algebraic rearrangement of this equation leads to:

F app = 0.289 N

9. The following object is moving to the right and encountering the following forces. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F grav = m • g = (10 kg) • (10 m/s/s) = 100 N
F norm = F grav - F y = 100 N - 35.4 N = 64.6 N

This x-component of the applied force (F x ) is directed leftward. This horizontal component of force is not counteracted by a rightward force. For this reason, the net force is 35.4 N.  Knowing F net , allows us to determine the acceleration of the object:

a = F net / m = (35.4 N) / (10 kg) = ~3.5 m/s/s

The acceleration of an object is the velocity change per time. For an acceleration of 3.5 m/s/s, the velocity change should be 3.5 m/s for each second of time change. In the velocity-time table, the velocity is decreasing by 3.5 m/s each second. Thus, the values should read 17.5 m/s, 14.0 m/s, 10.5 m/s, 7.0 m/s, 3.5 m/s, 0 m/s.

10. The 10-kg object is being pulled to the left at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = F grav - F norm = 98 N - 80 N = 18 N

The applied force can be determined using a trigonometric function:

sine 30 (degrees) = (18 N) / F app

Algebraic rearrangement leads to:

F app = (18 N) / [ sine (30 degrees) ] = 36 N

Similar trigonometry allows one to determine the x-component of the applied force:

F x = (36 N) • cosine(30 degrees) = 31.2 N

Since the speed is constant, the horizontal forces must also balance. Thus the force of friction is equal to the F x value. F frict = 31.2 N

The value of "mu" can be found using the equation "mu" = F frict / F norm

11. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = (100 N) • sine (45 degrees) = 70.7 N
F norm = F grav - F y = 98 N - 70.7 N = 27.3 N
F x = (100 N) • cosine (45 degrees) = 70. N

The net force is the vector sum of all the forces.

F net = 50 N, right + 70.7 N, left = 20.7 N, left

The acceleration is can be found from a = F net / m :

  • › Force and Motion
  • › This article

Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide

This step-by-step guide is meant to show you how to approach problems where you have to deal with moving objects subject to friction and other forces, and you need to apply Newton's Laws. We will go through many problems, so you can have a clear idea of the process involved in solving them.

The problems we will examine include objects that

  • are pushed/pulled horizontally with an angle
  • move up or down an incline
  • hang from ropes attached to the ceiling
  • hang from ropes that run over pulleys
  • move connected by a string
  • are pushed in contact with each other (Coming soon!)
  • Box pulled at an angle over a horizontal surface
  • Block pushed over the floor with a downward and forward force
  • Object moving at constant velocity over a horizontal surface
  • Block pushed up a frictionless ramp
  • Mass pulled up an incline with friction
  • A mass hanging from two ropes
  • Two hanging objects connected by a rope
  • Two masses on a pulley
  • Two blocks connected by a string are pulled horizontally

Centripetal Force

Practice problem 1.

  • at the bottom (and rising)
  • halfway to the top
  • 45° past the top
  • Determine the minimum coefficient of static friction needed to complete the stunt as planned.

The only forces present in this problem are weight (which always points down), normal (which always points toward the center of the loop), and friction (which is always tangential to the loop). The weight of the motorcycle never changes. The normal varies from a maximum at the bottom of the loop to a minimum at the top. As the motorcycle drives up the loop, the friction force acts along the direction of motion to keep gravity from slowing it down. As the motorcycle drives down the loop, the friction force acts opposite the direction of motion to keep gravity from speeding it up. The bike isn't speeding up or slowing down, but it is changing direction. This means the net force always points toward the center of motion.

Weight points down and normal points up, so the net force is their difference. The normal force points toward the center, so it should be given the positive value. The net force is the centripetal force.

Friction isn't necessary since the motorcycle isn't accelerating horizontally.

f i  =  0 N

The normal force points horizontally, toward the center of the loop. There isn't anything to balance this force at this position. This is our net force — our centripetal force.

Weight points down and normal points inward (toward the center of the loop). They aren't opposite anymore, so they can't cancel. The motorcycle isn't accelerating vertically at this moment, so there must be something to balance the weight. That's where friction steps in. Friction and weight are equal.

Both weight and normal point down, so the net force is their sum.

Once again, friction isn't necessary since the motorcycle isn't accelerating horizontally.

f iii  =  0 N

The normal and friction forces are at right angles to each other. This makes them "good" vectors. Normal points toward the center and contributes to the centripetal force. Friction is tangential to the circle and contributes nothing to the centripetal force. Weight is the "bad" vector. It needs to be broken up into components. The component pointing toward the center contributes to the centripetal force.

The component of the weight tangent to the track is balanced by the friction (so that the speed stays constant).

As I was solving this problem, I realized there is a conceptually easier way to solve it. What stays constant in this problem? Weight and centripetal force. What changes? Direction. Compute weight and centripetal force once, then combine them as appropriate to the location.

Weight points down and normal points up. The net force (their difference) is the centripetal force.

Normal points toward the center, which makes it the centripetal force.

Friction counteracts weight to keep the motorcycle moving with constant speed around the loop.

Both weight and normal point down toward the center of the loop. The net force (their sum) is the centripetal force.

Again, friction isn't necessary since the motorcycle isn't accelerating horizontally.

The position for this part of the problem was rigged so the two components would have the same magnitude and there'd be fewer things to calculate.

The normal force and a component of the weight point toward the center of the loop (the component perpendicular to the loop), so they get to be the centripetal force.

The other component of the weight (the component parallel to the loop) is balanced by friction so the net force tangential to the loop is zero and the speed does not change.

Static friction needs to be greatest when the motorcycle is vertical. The tires need to grip the loop to balance out the weight and keep the bike from accelerating tangential to the loop. At these two points (one where the bike is going up and the other where the bike is going down) friction equals weight and normal provides the centripetal force. I'll solve this both ways — once using the fancy, formal method…

and again using the conceptually easier method…

practice problem 2

Book cover

  • meters per second
  • Earth days per rotation
  • rotations per Earth year

Use the centripetal acceleration equation and solve for speed. Substitute values for the acceleration due to gravity on Earth and the radius of the Earth's orbit (also known as an astronomical unit).

Sounds fast, but things in space tend to move fast anyway. How does this compare to the yearly motion of the Earth around the Sun? That's the goal of the next parts of the question.

We'll solve this practice problem two ways. First we'll use the definition of speed and substitute the value calculated above and the distance traveled in one rotation (the circumference).

The fancier way is to start from the centripetal acceleration equation, replace speed with circumference over period, and simplify. This gives us an equation that some people like so much they memorize it.

Solve for period and substitute.

That's a quick "year" — if we use the astronomical definition of the year as the period of one trip around the Sun. How does it compare to a regular calendar year?

This problem is best solved by dimensional analysis.

We will return to this problem in the section on power .

practice problem 3

The curve radii of modern high-speed systems result in dependence on the speed and the maximum possible superelevation of the guideway to compensate for the centrifugal forces occurring. The Transrapid's guideway can have a maximum superelevation of 12 degree (up to 16 degree in special cases) which allows smaller radii at higher speeds than in the case of conventional wheel-on-rail systems. Minimal radius: 350 m 200 km/h: 0 705 m 400 km/h: 2825 m 500 km/h: 4415 m ThyssenKrupp Transrapid GmbH, 2002
  • the maximum centripetal acceleration (in m/s 2 and g) implied by these specifications
  • the speed limit (in m/s and km/h) on a curved section of track with the minimal radius

Once you get past reading the awkward translation from German to English, this is a conceptually easy question. Set up a table like the one below and complete the missing parts. Use a graphing calculator as it eliminates the drudgery of repeated calculation.

Start by converting the maximum speeds from km/h to m/s.

Then apply the equation for centripetal acceleration.

The resulting values will be in m/s 2 . To convert to g, divide by the standard value for the acceleration due to gravity.

a c [g]  =  a c [m/s 2 ] ÷   9.80665 m/s 2

This procedure gives a set of numbers that are reasonably close to one another: three values in m/s 2 and three in g. Find the mean of both of these triplets. A partially completed table is provided below.

For the second part of this question, follow the logic of the first part in reverse order. Assume that the maximum permissible centripetal acceleration is the same for all curves, regardless of size. Use the mean value we just calculated to determine the speed limit on a curve with a 350 m radius.

Then convert from m/s to km/h.

Add these results to the table.

A more advanced technique is to solve the problem graphically.

Starting from the equation for centripetal acceleration.

Make v 2 the subject and compare to the equation for a straight line.

The equation is telling us we should put speed squared on the y  axis and radius on the x  axis.

This makes the centripetal acceleration equal the slope of the line of best fit.

The intercept value of 7.28 m 2 /s 2 is effectively equal to zero. Since the speed squared values are all on the order of several thousand, an intercept that's less than ten is "small" in comparison.

Use the coefficients from the line of best fit to find the speed limit on the minimum radius curve. Finish by converting the speed back to km/h from m/s.

practice problem 4

  • Complete the first two columns using astronomical data from a reliable source. Be sure to specify the units used for each entry.
  • Complete the last two columns using a calculator. Be sure to state your answers in SI units.

The data for the moon and planets came from a page in this book. The data for the Sun came from a page in The Physics Factbook. The distance from the Earth to the Sun is called an astronomical unit (au). I plan on discussing the au in the section of this book on miscellaneous units (1 au = 1.496 × 10 11  m). I know you know the period of Earth's orbit. You may just have to pause and think for a second. The first two columns of the table are done.

Speed is the rate of change of distance with time, which in circular motion means circumference divided by period.

We are going to use this equation over and over again. Please watch those units. Convert to meters and seconds as appropriate.

Now that we have all the speeds, we can compute the centripetal accelerations.

Again we'll do it over and over again…

The table is now ready for completion.

Dynamics: Force and Newton’s Laws of Motion

Problem-solving strategies, learning objective.

By the end of this section, you will be able to:

  • Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure 1(a). Then, as in Figure 1(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Figure 1. (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, F T is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. F T is no longer shown, because it is not a force acting on the system of interest; rather, F T  acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = –w , if Tarzan is stationary.

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 1(c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 1(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure 1(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying Newton’s Second Law

image

F net x  = ma ,

F net y = 0.

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

Section Summary

To solve problems involving Newton’s laws of motion, follow the procedure described:

  • Draw a sketch of the problem.
  • Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
  • Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then  F net x  = 0 . If the object does accelerate in that direction,  F net x  = ma .
  • Check your answer. Is the answer reasonable? Are the units correct?

Problems & Exercises

1. A 5.00 × 10 5 -kg rocket is accelerating straight up. Its engines produce 1.250 × 10 7  of thrust, and air resistance is 4.50 × 10 6 N. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

2. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s 2 , what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.

3. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

4. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

5. A freight train consists of two 8.00 × 10 4   engines and 45 cars with average masses of 5.50 × 10 4 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10 -2  if the force of friction is 7.50 × 10 5 , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

6. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75 × 10 5  backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s 2 , what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

7. A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s 2 ? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

8. (a) Find the magnitudes of the forces F 1 and F 2  that add to give the total force F tot  shown in Figure 4. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of   F 1 and F 2 . (c) Find the direction and magnitude of some other pair of vectors that add to give F tot . Draw these to scale on the same drawing used in part (b) or a similar picture.

A right triangle is shown made up of three vectors. The first vector, F sub one, is along the triangle’s base toward the right; the second vector, F sub two, is along the perpendicular side pointing upward; and the third vector, F sub tot, is along the hypotenuse pointing up the incline. The magnitude of F sub tot is twenty newtons. In a free-body diagram, F sub one is shown by an arrow pointing right and F sub two is shown by an arrow acting vertically upward.

9. Two children pull a third child on a snow saucer sled exerting forces  F 1 and F 2 as shown from above in Figure 4 . Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of  F 1 and F 2 .

An overhead view of a child sitting on a snow saucer sled. Two forces, F sub one equal to ten newtons and F sub two equal to eight newtons, are acting toward the right. F sub one makes an angle of forty-five degrees from the x axis and F sub two makes an angle of thirty degrees from the x axis in a clockwise direction. A friction force f is equal to seven point five newtons, shown by a vector pointing in negative x direction. In the free-body diagram, F sub one and F sub two are shown by arrows toward the right, making a forty-five degree angle above the horizontal and a thirty-degree angle below the horizontal respectively. The friction force f is shown by an arrow along the negative x axis.

10. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 6 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?

Figure of car stuck in the mud and a rope connected to a tree trunk in an attempt to pull out the car.

11. What force is exerted on the tooth in Figure 7 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.

Cross-section of jaw with sixteen teeth is shown. Braces are along the outside of the teeth. Three forces are acting on the protruding tooth. The applied force, F sub app, is shown by an arrow vertically downward; a second force, T, is shown by an arrow making an angle of fifteen degrees below the positive x axis; and a third force, T, is shown by an arrow making an angle of fifteen degrees below the negative x axis.

Figure 7. Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, F app , points straight toward the back of the mouth.

12. Figure 9 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.

Two caped superheroes hang on a rope suspended vertically from a bar.

Figure 9. Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?

13. A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0º below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?

14. Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.

15. Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.

16. Unreasonable Results (a) Repeat Exercise 7, but assume an acceleration of 1.20 m/s 2  is produced. (b) What is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable?

17. Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50 × 10 6  at takeoff, the engines of which produce a thrust of 2.00 × 10 6 ? Do not neglect gravity. (b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)

Selected Solutions to Problems & Exercises

1. Using the free-body diagram:

An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

  • [latex]{F}_{\text{net}}=T-f-mg=\text{ma}\\[/latex] ,

[latex]a=\frac{T-f-\text{mg}}{m}=\frac{1\text{.}\text{250}\times {\text{10}}^{7}\text{N}-4.50\times {\text{10}}^{\text{6}}N-\left(5.00\times {\text{10}}^{5}\text{kg}\right)\left(9.{\text{80 m/s}}^{2}\right)}{5.00\times {\text{10}}^{5}\text{kg}}=\text{6.20}{\text{m/s}}^{2}\\[/latex]

3. Use Newton’s laws of motion.

Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.

[latex]F=\left(\text{70.0 kg}\right)\left[\left(\text{39}\text{.}{\text{2 m/s}}^{2}\right)+\left(9\text{.}{\text{80 m/s}}^{2}\right)\right]\\[/latex] [latex]=3.\text{43}\times {\text{10}}^{3}\text{N}\\[/latex].  The force exerted by the high-jumper is actually down on the ground, but F is up from the ground and makes him jump.

  • This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 10 3 N.

5. (a) 4.41 × 10 5 N (b) 1.50 × 10 5 N

7. (a) 910 N (b) 1.11 × 10 3

9. (a) a = 0.139 m/s, θ = 12.4º

11. Use Newton’s laws since we are looking for forces.

  • Draw a free-body diagram:

A horizontal dotted line with two vectors extending downward from the mid-point of the dotted line, both at angles of fifteen degrees. A third vector points straight downward from the intersection of the first two angles, bisecting them; it is perpendicular to the dotted line.

  • The tension is given as T = 25.0 N. Find F app . Using Newton’s laws gives:[latex]\sigma{F}_{y}=0\\[/latex], so that applied force is due to the y -components of the two tensions: F app = 2 T  sin  θ  = 2(25.0 N) sin(15º) = 12.9 N The x -components of the tension cancel. [latex]\sum{F}_{x}=0\\[/latex].
  • This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.
  • College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

4.2: Moment of a Force: Problem Solving

Chapter 0: physics basics, chapter 1: an introduction to statics, chapter 2: force vectors, chapter 3: equilibrium of a particle, chapter 4: force system resultants, chapter 5: equilibrium of a rigid body, chapter 6: structural analysis, chapter 7: internal forces, chapter 8: friction, chapter 9: center of gravity and centroid, chapter 10: moment of inertia, chapter 11: virtual work, chapter 12: kinematics of a particle, chapter 13: kinetics of a particle: force and acceleration, chapter 14: kinetics of a particle: impulse and momentum, chapter 15: planar kinematics of a rigid body, chapter 16: 3-dimensional kinetics of a rigid body, chapter 17: concept of stress, chapter 18: stress and strain - axial loading, chapter 19: torsion, chapter 20: bending, chapter 21: analysis and design of beams for bending, chapter 22: shearing stresses in beams and thin-walled members, chapter 23: transformations of stress and strain, chapter 24: principal stresses under a given loading, chapter 25: deflection of beams, chapter 26: columns, chapter 27: energy methods.

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problem solving about force

Consider a man pulling a nail from a wooden wall using a crowbar. The crowbar contacts the wall at point A . The man applies a vertical force F 1 on the grip of the crowbar, whereas it takes a force F 2 at the claw to pull out the nail. Determine the minimum force the man requires to pull out the nail.

At point A , the moment due to force F 1 can be expressed as the sum of the product of the magnitude of the force and the perpendicular distances. By substituting the values, the moment at A due to F 1 can be determined.

Similarly, resolving the force F 2 at the claw into its components, the moment due to F 2 at point A can be determined using the sine component of F 2 .

Since the moment of force, F 1 is smaller than F 2 , force F 1 is insufficient to remove the nail.

By comparing both the moment of force equations and substituting the value, the minimum force required to pull out the nail can be determined.

Understanding the scalar formulation of the moment of a force and applying it correctly through problem-solving is crucial in designing and analyzing mechanical systems. Here are the steps for problem-solving with the moment of a force:

  • Draw a free-body diagram (FBD) of the system . The FBD is a basic problem-solving step. It is a diagrammatic representation of all the forces acting on the system. Every force acting on the system must be identified and included in the FBD.
  • Identify the axis or point about which the moment is calculated. In some cases, the axis or point about which the net moment needs to be calculated is provided. However, in other cases, the problem requires selecting an appropriate axis or point.
  • Calculate the perpendicular distance between the force and the axis or point. The perpendicular distance is the shortest distance between the force and the chosen axis or point.
  • Calculate the moment of each force. In this step, the magnitudes of the individual forces and their distances to the chosen axis or point are multiplied to calculate their respective moments. Here, the moment of a force is a vector quantity, and its direction is perpendicular to the plane of the force and the chosen axis or point.
  • Sum the moments. Once the moments of all the forces have been calculated, they must be algebraically summed to obtain the net moment acting on the system.
  • Determine the effect of the net moment. If the net moment is zero, there is no rotational motion. If the net moment is negative, the object will rotate clockwise, and if the net moment is positive, the object will rotate counterclockwise.
  • Solve the problem. The final step of problem-solving is to use the calculated net moment to solve the given problem. For instance, if the angular acceleration of an object needs to be calculated, the net moment acting on the object is divided by the moment of inertia of the object.
  • Russell C. Hibbeler. (2016). Engineering Mechanics: Statics & Dynamics . Pearson. Page no - 121-124.

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Net force word problems 

Find here in this lesson some easy and challenging net force word problems.  

Problem #1:

What is the net force on the airplane in the figure below?

The airplane is moving with a force of 800 N. However, there are two forces moving in the opposite direction on the airplane.

Just add these two forces: 40 N + 60 N = 100 N

Subtract to get the net force: 800 N  - 100 N  = 700 N

The net force is 700 N.

The airplane will move with a force of 700 N as a result of air friction and wind. 

Net force on an airplane

Problem #2 : You and your brother are pushing a car with a dead battery with forces of 20 N and 25 N in the same direction. What is the net force applied on the car?

Solution: 

Since you are pushing the car in the same direction, the forces will be added together.

Net force = 20 N + 25 N

Net force  = 45 N.

Problem #3 : A brother is pulling a toy from his sister with a force of 6  N. The sister is pulling back with a force of 8 N. 

Who gets the toy?

What is the net force? 

The sister gets the toy of course since she is pulling with a stronger force.

Net force = 8 N - 6 N

Net force  = 2 N.

More challenging net force word problems

Problem #4 :

4 people are playing a tug of war. Two are pulling on the right side. Two are pulling on the left side.  On the right side, one is pulling with a force of 60 N and the other with a force of 70 N. On the left side, one is pulling with a force of 30 N. How much force should the second person on the left apply to keep the rope in equilibrium? 

The rope will be in equilibrium is the net force is 0.

The forces on the right is equal to 60 N + 70 N = 130 N

Let x be the force that must be applied by the second person on the left.

30 N + x = 130 N

Since 30 N + 100 N = 130 N,  x = 100 N

The other person should pull with a force of 100 N to keep the rope in equilibrium.

Equilibrium

Problem #5 :

Your friend is pulling upward on an object with force of 3 N. You are pulling to the right with a force of 4 N.

Find the net force and the direction the object moves.

net force

Just build the rectangle and find the resultant. The red arrow shows the direction the object will move. 

Net force

This net force word problem is a little challenging. To find the net force, we need use the Pythagorean Theorem .

What is a net force?

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Mechanics (Essentials) - Class 11th

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HBR On Strategy podcast series

A Better Framework for Solving Tough Problems

Start with trust and end with speed.

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When it comes to solving complicated problems, the default for many organizational leaders is to take their time to work through the issues at hand. Unfortunately, that often leads to patchwork solutions or problems not truly getting resolved.

But Anne Morriss offers a different framework. In this episode, she outlines a five-step process for solving any problem and explains why starting with trust and ending with speed is so important for effective change leadership. As she says, “Let’s get into dialogue with the people who are also impacted by the problem before we start running down the path of solving it.”

Morriss is an entrepreneur and leadership coach. She’s also the coauthor of the book, Move Fast and Fix Things: The Trusted Leader’s Guide to Solving Hard Problems .

Key episode topics include: strategy, decision making and problem solving, strategy execution, managing people, collaboration and teams, trustworthiness, organizational culture, change leadership, problem solving, leadership.

HBR On Strategy curates the best case studies and conversations with the world’s top business and management experts, to help you unlock new ways of doing business. New episodes every week.

  • Listen to the full HBR IdeaCast episode: How to Solve Tough Problems Better and Faster (2023)
  • Find more episodes of HBR IdeaCast
  • Discover 100 years of Harvard Business Review articles, case studies, podcasts, and more at HBR.org .

HANNAH BATES: Welcome to HBR On Strategy , case studies and conversations with the world’s top business and management experts, hand-selected to help you unlock new ways of doing business.

When it comes to solving complicated problems, many leaders only focus on the most apparent issues. Unfortunately that often leads to patchwork or partial solutions. But Anne Morriss offers a different framework that aims to truly tackle big problems by first leaning into trust and then focusing on speed.

Morriss is an entrepreneur and leadership coach. She’s also the co-author of the book, Move Fast and Fix Things: The Trusted Leader’s Guide to Solving Hard Problems . In this episode, she outlines a five-step process for solving any problem. Some, she says, can be solved in a week, while others take much longer. She also explains why starting with trust and ending with speed is so important for effective change leadership.

This episode originally aired on HBR IdeaCast in October 2023. Here it is.

CURT NICKISCH: Welcome to the HBR IdeaCast from Harvard Business Review. I’m Curt Nickisch.

Problems can be intimidating. Sure, some problems are fun to dig into. You roll up your sleeves, you just take care of them; but others, well, they’re complicated. Sometimes it’s hard to wrap your brain around a problem, much less fix it.

And that’s especially true for leaders in organizations where problems are often layered and complex. They sometimes demand technical, financial, or interpersonal knowledge to fix. And whether it’s avoidance on the leaders’ part or just the perception that a problem is systemic or even intractable, problems find a way to endure, to keep going, to keep being a problem that everyone tries to work around or just puts up with.

But today’s guest says that just compounds it and makes the problem harder to fix. Instead, she says, speed and momentum are key to overcoming a problem.

Anne Morriss is an entrepreneur, leadership coach and founder of the Leadership Consortium and with Harvard Business School Professor Francis Frei, she wrote the new book, Move Fast and Fix Things: The Trusted Leaders Guide to Solving Hard Problems . Anne, welcome back to the show.

ANNE MORRISS: Curt, thank you so much for having me.

CURT NICKISCH: So, to generate momentum at an organization, you say that you really need speed and trust. We’ll get into those essential ingredients some more, but why are those two essential?

ANNE MORRISS: Yeah. Well, the essential pattern that we observed was that the most effective change leaders out there were building trust and speed, and it didn’t seem to be a well-known observation. We all know the phrase, “Move fast and break things,” but the people who were really getting it right were moving fast and fixing things, and that was really our jumping off point. So when we dug into the pattern, what we observed was they were building trust first and then speed. This foundation of trust was what allowed them to fix more things and break fewer.

CURT NICKISCH: Trust sounds like a slow thing, right? If you talk about building trust, that is something that takes interactions, it takes communication, it takes experiences. Does that run counter to the speed idea?

ANNE MORRISS: Yeah. Well, this issue of trust is something we’ve been looking at for over a decade. One of the headlines in our research is it’s actually something we’re building and rebuilding and breaking all the time. And so instead of being this precious, almost farbege egg, it’s this thing that is constantly in motion and this thing that we can really impact when we’re deliberate about our choices and have some self-awareness around where it’s breaking down and how it’s breaking down.

CURT NICKISCH: You said break trust in there, which is intriguing, right? That you may have to break trust to build trust. Can you explain that a little?

ANNE MORRISS:  Yeah, well, I’ll clarify. It’s not that you have to break it in order to build it. It’s just that we all do it some of the time. Most of us are trusted most of the time. Most of your listeners I imagine are trusted most of the time, but all of us have a pattern where we break trust or where we don’t build as much as could be possible.

CURT NICKISCH: I want to talk about speed, this other essential ingredient that’s so intriguing, right? Because you think about solving hard problems as something that just takes a lot of time and thinking and coordination and planning and designing. Explain what you mean by it? And also, just  how we maybe approach problems wrong by taking them on too slowly?

ANNE MORRISS: Well, Curt, no one has ever said to us, “I wish I had taken longer and done less.” We hear the opposite all the time, by the way. So what we really set out to do was to create a playbook that anyone can use to take less time to do more of the things that are going to make your teams and organizations stronger.

And the way we set up the book is okay, it’s really a five step process. Speed is the last step. It’s the payoff for the hard work you’re going to do to figure out your problem, build or rebuild trust, expand the team in thoughtful and strategic ways, and then tell a real and compelling story about the change you’re leading.

Only then do you get to go fast, but that’s an essential part of the process, and we find that either people under emphasize it or speed has gotten a bad name in this world of moving fast and breaking things. And part of our mission for sure was to rehabilitate speed’s reputation because it is an essential part of the change leader’s equation. It can be the difference between good intentions and getting anything done at all.

CURT NICKISCH: You know, the fact that nobody ever tells you, “I wish we had done less and taken more time.” I think we all feel that, right? Sometimes we do something and then realize, “Oh, that wasn’t that hard and why did it take me so long to do it? And I wish I’d done this a long time ago.” Is it ever possible to solve a problem too quickly?

ANNE MORRISS: Absolutely. And we see that all the time too. What we push people to do in those scenarios is really take a look at the underlying issue because in most cases, the solution is not to take your foot off the accelerator per se and slow down. The solution is to get into the underlying problem. So if it’s burnout or a strategic disconnect between what you’re building and the marketplace you’re serving, what we find is the anxiety that people attach to speed or the frustration people attach to speed is often misplaced.

CURT NICKISCH: What is a good timeline to think about solving a problem then? Because if we by default take too long or else jump ahead and we don’t fix it right, what’s a good target time to have in your mind for how long solving a problem should take?

ANNE MORRISS: Yeah. Well, we’re playful in the book and talking about the idea that many problems can be solved in a week. We set the book up five chapters. They’re titled Monday, Tuesday, Wednesday, Thursday, Friday, and we’re definitely having fun with that. And yet, if you count the hours in a week, there are a lot of them. Many of our problems, if you were to spend a focused 40 hours of effort on a problem, you’re going to get pretty far.

But our main message is, listen, of course it’s going to depend on the nature of the problem, and you’re going to take weeks and maybe even some cases months to get to the other side. What we don’t want you to do is take years, which tends to be our default timeline for solving hard problems.

CURT NICKISCH: So you say to start with identifying the problem that’s holding you back, seems kind of obvious. But where do companies go right and wrong with this first step of just identifying the problem that’s holding you back?

ANNE MORRISS: And our goal is that all of these are going to feel obvious in retrospect. The problem is we skip over a lot of these steps and this is why we wanted to underline them. So this one is really rooted in our observation and I think the pattern of our species that we tend to be overconfident in the quality of our thoughts, particularly when it comes to diagnosing problems.

And so we want to invite you to start in a very humble and curious place, which tends not to be our default mode when we’re showing up for work. We convince ourselves that we’re being paid for our judgment. That’s exactly what gets reinforced everywhere. And so we tend to counterintuitively, given what we just talked about, we tend to move too quickly through the diagnostic phase.

CURT NICKISCH: “I know what to do, that’s why you hired me.”

ANNE MORRISS: Exactly. “I know what to do. That’s why you hired me. I’ve seen this before. I have a plan. Follow me.” We get rewarded for the expression of confidence and clarity. And so what we’re inviting people to do here is actually pause and really lean into what are the root causes of the problem you’re seeing? What are some alternative explanations? Let’s get into dialogue with the people who are also impacted by the problem before we start running down the path of solving it.

CURT NICKISCH: So what do you recommend for this step, for getting to the root of the problem? What are questions you should ask? What’s the right thought process? What do you do on Monday of the week?

ANNE MORRISS: In our experience of doing this work, people tend to undervalue the power of conversation, particularly with other people in the organization. So we will often advocate putting together a team of problem solvers, make it a temporary team, really pull in people who have a particular perspective on the problem and create the space, make it as psychologically safe as you can for people to really, as Chris Argyris so beautifully articulated, discuss the undiscussable.

And so the conditions for that are going to look different in every organization depending on the problem, but if you can get a space where smart people who have direct experience of a problem are in a room and talking honestly with each other, you can make an extraordinary amount of progress, certainly in a day.

CURT NICKISCH: Yeah, that gets back to the trust piece.

ANNE MORRISS: Definitely.

CURT NICKISCH: How do you like to start that meeting, or how do you like to talk about it? I’m just curious what somebody on that team might hear in that meeting, just to get the sense that it’s psychologically safe, you can discuss the undiscussable and you’re also focusing on the identification part. What’s key to communicate there?

ANNE MORRISS: Yeah. Well, we sometimes encourage people to do a little bit of data gathering before those conversations. So the power of a quick anonymous survey around whatever problem you’re solving, but also be really thoughtful about the questions you’re going to ask in the moment. So a little bit of preparation can go a long way and a little bit of thoughtfulness about the power dynamic. So who’s going to walk in there with license to speak and who’s going to hold back? So being thoughtful about the agenda, about the questions you’re asking about the room, about the facilitation, and then courage is a very infectious emotion.

So if you can early on create the conditions for people to show up bravely in that conversation, then the chance that you’re going to get good information and that you’re going to walk out of that room with new insight in the problem that you didn’t have when you walked in is extraordinarily high.

CURT NICKISCH: Now, in those discussions, you may have people who have different perspectives on what the problem really is. They also bear different costs of addressing the problem or solving it. You talked about the power dynamic, but there’s also an unfairness dynamic of who’s going to actually have to do the work to take care of it, and I wonder how you create a culture in that meeting where it’s the most productive?

ANNE MORRISS: For sure, the burden of work is not going to be equitably distributed around the room. But I would say, Curt, the dynamic that we see most often is that people are deeply relieved that hard problems are being addressed. So it really can create, and more often than not in our experience, it does create this beautiful flywheel of action, creativity, optimism. Often when problems haven’t been addressed, there is a fair amount of anxiety in the organization, frustration, stagnation. And so credible movement towards action and progress is often the best antidote. So even if the plan isn’t super clear yet, if it’s credible, given who’s in the room and their decision rights and mandate, if there’s real momentum coming out of that to make progress, then that tends to be deeply energizing to people.

CURT NICKISCH: I wonder if there’s an organization that you’ve worked with that you could talk about how this rolled out and how this took shape?

ANNE MORRISS: When we started working with Uber, that was wrestling with some very public issues of culture and trust with a range of stakeholders internally, the organization, also external, that work really started with a campaign of listening and really trying to understand where trust was breaking down from the perspective of these stakeholders?

So whether it was female employees or regulators or riders who had safety concerns getting into the car with a stranger. This work, it starts with an honest internal dialogue, but often the problem has threads that go external. And so bringing that same commitment to curiosity and humility and dialogue to anyone who’s impacted by the problem is the fastest way to surface what’s really going on.

CURT NICKISCH: There’s a step in this process that you lay out and that’s communicating powerfully as a leader. So we’ve heard about listening and trust building, but now you’re talking about powerful communication. How do you do this and why is it maybe this step in the process rather than the first thing you do or the last thing you do?

ANNE MORRISS: So in our process, again, it’s the days of the week. On Monday you figured out the problem. Tuesday you really got into the sandbox in figuring out what a good enough plan is for building trust. Wednesday, step three, you made it better. You created an even better plan, bringing in new perspectives. Thursday, this fourth step is the day we’re saying you got to go get buy-in. You got to bring other people along. And again, this is a step where we see people often underinvest in the power and payoff of really executing it well.

CURT NICKISCH: How does that go wrong?

ANNE MORRISS: Yeah, people don’t know the why. Human behavior and the change in human behavior really depends on a strong why. It’s not just a selfish, “What’s in it for me?” Although that’s helpful, but where are we going? I may be invested in a status quo and I need to understand, okay, if you’re going to ask me to change, if you’re going to invite me into this uncomfortable place of doing things differently, why am I here? Help me understand it and articulate the way forward and language that not only I can understand, but also that’s going to be motivating to me.

CURT NICKISCH: And who on my team was part of this process and all that kind of stuff?

ANNE MORRISS: Oh, yeah. I may have some really important questions that may be in the way of my buy-in and commitment to this plan. So certainly creating a space where those questions can be addressed is essential. But what we found is that there is an architecture of a great change story, and it starts with honoring the past, honoring the starting place. Sometimes we’re so excited about the change and animated about the change that what has happened before or what is even happening in the present tense is low on our list of priorities.

Or we want to label it bad, because that’s the way we’ve thought about the change, but really pausing and honoring what came before you and all the reasonable decisions that led up to it, I think can be really helpful to getting people emotionally where you want them to be willing to be guided by you. Going back to Uber, when Dara Khosrowshahi came in.

CURT NICKISCH: This is the new CEO.

ANNE MORRISS: The new CEO.

CURT NICKISCH: Replaced Travis Kalanick, the founder and first CEO, yeah.

ANNE MORRISS: Yeah, and had his first all-hands meeting. One of his key messages, and this is a quote, was that he was going to retain the edge that had made Uber, “A force of nature.” And in that meeting, the crowd went wild because this is also a company that had been beaten up publicly for months and months and months, and it was a really powerful choice. And his predecessor, Travis was in the room, and he also honored Travis’ incredible work and investment in bringing the company to the place where it was.

And I would use words like grace to also describe those choices, but there’s also an incredible strategic value to naming the starting place for everybody in the room because in most cases, most people in that room played a role in getting to that starting place, and you’re acknowledging that.

CURT NICKISCH: You can call it grace. Somebody else might call it diplomatic or strategic. But yeah, I guess like it or not, it’s helpful to call out and honor the complexity of the way things have been done and also the change that’s happening.

ANNE MORRISS: Yeah, and the value. Sometimes honoring the past is also owning what didn’t work or what wasn’t working for stakeholders or segments of the employee team, and we see that around culture change. Sometimes you’ve got to acknowledge that it was not an equitable environment, but whatever the worker, everyone in that room is bringing that pass with them. So again, making it discussable and using it as the jumping off place is where we advise people to start.

Then you’ve earned the right to talk about the change mandate, which we suggest using clear and compelling language about the why. “This is what happened, this is where we are, this is the good and the bad of it, and here’s the case for change.”

And then the last part, which is to describe a rigorous and optimistic way forward. It’s a simple past, present, future arc, which will be familiar to human beings. We love stories as human beings. It’s among the most powerful currency we have to make sense of the world.

CURT NICKISCH: Yeah. Chronological is a pretty powerful order.

ANNE MORRISS: Right. But again, the change leaders we see really get it right, are investing an incredible amount of time into the storytelling part of their job. Ursula Burns, the Head of Xerox is famous for the months and years she spent on the road just telling the story of Xerox’s change, its pivot into services to everyone who would listen, and that was a huge part of her success.

CURT NICKISCH: So Friday or your fifth step, you end with empowering teams and removing roadblocks. That seems obvious, but it’s critical. Can you dig into that a little bit?

ANNE MORRISS: Yeah. Friday is the fun day. Friday’s the release of energy into the system. Again, you’ve now earned the right to go fast. You have a plan, you’re pretty confident it’s going to work. You’ve told the story of change the organization, and now you get to sprint. So this is about really executing with urgency, and it’s about a lot of the tactics of speed is where we focus in the book. So the tactics of empowerment, making tough strategic trade-offs so that your priorities are clear and clearly communicated, creating mechanisms to fast-track progress. At Etsy, CEO Josh Silverman, he labeled these projects ambulances. It’s an unfortunate metaphor, but it’s super memorable. These are the products that get to speed out in front of the other ones because the stakes are high and the clock is sticking.

CURT NICKISCH: You pull over and let it go by.

ANNE MORRISS: Yeah, exactly. And so we have to agree as an organization on how to do something like that. And so we see lots of great examples both in young organizations and big complex biotech companies with lots of regulatory guardrails have still found ways to do this gracefully.

And I think we end with this idea of conflict debt, which is a term we really love. Leanne Davey, who’s a team scholar and researcher, and anyone in a tech company will recognize the idea of tech debt, which is this weight the organization drags around until they resolve it. Conflict debt is a beautiful metaphor because it is this weight that we drag around and slows us down until we decide to clean it up and fix it. The organizations that are really getting speed right have figured out either formally or informally, how to create an environment where conflict and disagreements can be gracefully resolved.

CURT NICKISCH: Well, let’s talk about this speed more, right? Because I think this is one of those places that maybe people go wrong or take too long, and then you lose the awareness of the problem, you lose that urgency. And then that also just makes it less effective, right? It’s not just about getting the problem solved as quickly as possible. It’s also just speed in some ways helps solve the problem.

ANNE MORRISS: Oh, yeah. It really is the difference between imagining the change you want to lead and really being able to bring it to life. Speed is the thing that unlocks your ability to lead change. It needs a foundation, and that’s what Monday through Thursday is all about, steps one through four, but the finish line is executing with urgency, and it’s that urgency that releases the system’s energy, that communicates your priorities, that creates the conditions for your team to make progress.

CURT NICKISCH: Moving fast is something that entrepreneurs and tech companies certainly understand, but there’s also this awareness that with big companies, the bigger the organization, the harder it is to turn the aircraft carrier around, right? Is speed relative when you get at those levels, or do you think this is something that any company should be able to apply equally?

ANNE MORRISS: We think this applies to any company. The culture really lives at the level of team. So we believe you can make a tremendous amount of progress even within your circle of control as a team leader. I want to bring some humility to this and careful of words like universal, but we do think there’s some universal truths here around the value of speed, and then some of the byproducts like keeping fantastic people. Your best people want to solve problems, they want to execute, they want to make progress and speed, and the ability to do that is going to be a variable in their own equation of whether they stay or they go somewhere else where they can have an impact.

CURT NICKISCH: Right. They want to accomplish something before they go or before they retire or finish something out. And if you’re able to just bring more things on the horizon and have it not feel like it’s going to be another two years to do something meaningful.

ANNE MORRISS: People – I mean, they want to make stuff happen and they want to be around the energy and the vitality of making things happen, which again, is also a super infectious phenomenon. One of the most important jobs of a leader, we believe, is to set the metabolic pace of their teams and organizations. And so what we really dig into on Friday is, well, what does that look like to speed something up? What are the tactics of that?

CURT NICKISCH: I wonder if that universal truth, that a body in motion stays in motion applies to organizations, right? If an organization in motion stays in motion, there is something to that.

ANNE MORRISS: Absolutely.

CURT NICKISCH: Do you have a favorite client story to share, just where you saw speed just become a bit of a flywheel or just a positive reinforcement loop for more positive change at the organization?

ANNE MORRISS: Yeah. We work with a fair number of organizations that are on fire. We do a fair amount of firefighting, but we also less dramatically do a lot of fire prevention. So we’re brought into organizations that are working well and want to get better, looking out on the horizon. That work is super gratifying, and there is always a component of, well, how do we speed this up?

What I love about that work is there’s often already a high foundation of trust, and so it’s, well, how do we maintain that foundation but move this flywheel, as you said, even faster? And it’s really energizing because often there’s a lot of pent-up energy that… There’s a lot of loyalty to the organization, but often it’s also frustration and pent-up energy. And so when that gets released, when good people get the opportunity to sprint for the first time in a little while, it’s incredibly energizing, not just for us, but for the whole organization.

CURT NICKISCH: Anne, this is great. I think finding a way to solve problems better but also faster is going to be really helpful. So thanks for coming on the show to talk about it.

ANNE MORRISS:  Oh, Curt, it was such a pleasure. This is my favorite conversation. I’m delighted to have it anytime.

HANNAH BATES: That was entrepreneur, leadership coach, and author Anne Morriss – in conversation with Curt Nickisch on HBR IdeaCast.

We’ll be back next Wednesday with another hand-picked conversation about business strategy from Harvard Business Review. If you found this episode helpful, share it with your friends and colleagues, and follow our show on Apple Podcasts, Spotify, or wherever you get your podcasts. While you’re there, be sure to leave us a review.

When you’re ready for more podcasts, articles, case studies, books, and videos with the world’s top business and management experts, you’ll find it all at HBR.org.

This episode was produced by Mary Dooe, Anne Saini, and me, Hannah Bates. Ian Fox is our editor. Special thanks to Rob Eckhardt, Maureen Hoch, Erica Truxler, Ramsey Khabbaz, Nicole Smith, Anne Bartholomew, and you – our listener. See you next week.

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Partner Center

Problem solving in the Middle East

23 May 2024

If there’s an issue on the airfield, Air Force Corporal John Chapman can get it done. 

The ground support equipment technician, deployed to Australia’s main operating base in the Middle East on Operation Accordion, has become the person coalition forces turn to in a bind. 

That’s what happened on April 1 when the mechanic from Muswellbrook, NSW, received a late-night call for help. 

A large Egyptian military cargo plane was stuck on the runway and needed an air start before it could take off. 

For more than two decades, international partner nations have been guests on the Emirati base. Corporal Chapman, the only Air Force technician deployed on Operation Accordion, was the only person on the base with the right tools and experience to get the job done. 

The right tools in this case was an air-start cart – a turbine compressor on wheels that pumps pressurised air into a jet engine to ‘jump start’ the turbines. 

It was well after midnight and the phone was ringing off the hook. 

“Our partners requested help, I was pretty sure I could make it happen but I wasn’t 100 per cent sure it was going to work,” Corporal Chapman said. 

Luckily, the Egyptian aeroplane was similar to the Australian C-130 Hercules and the equipment could be attached to the Antonov without issue. 

“[The Egyptians] were all smiles when we got the prop started and they got on their way,” he said. 

It wasn’t the first time the Australian has been called on to help partner militaries. 

Over his eight-month deployment to the Middle East, he has helped United States generals and become mates with the British. 

When a US general visited the base, they needed equipment to power their plane when it was on the runway. 

Then there was the time the British blew a tire and needed compressed nitrogen. 

Luckily, the Australian workshop was able to help. 

“That’s how it is between the international countries here,” Corporal Chapman said. 

“We just call each other up if we need a hand – there’s a lot of favours for favours.” 

With less than a month until he returns to Australia, Corporal Chapman will look back on his deployment as a career highlight. 

“I’m operating by myself and I’ve used every workshop and airside skill I’ve been taught,” he said. 

“I love getting asked, ‘how do I solve this?’ Mechanics love to solve problems.” 

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By Angela Fritz, Elise Hammond and Chris Lau, CNN

Incredible lighthouse picture from Maine

From CNN's Chris Lau

A long-exposure photo shows the aurora borealis over Portland, Maine, on May 10.

Among a flurry of surreal images capturing the dazzling auroras is one taken by Benjamin Williamson of a lighthouse in Portland, Maine.

"It's one of the most incredible things I've ever seen, the awe and wonder," Williamson told CNN.

He said he used a long-exposure technique to snap the shot, but did not edit it.

Watch the full interview with Williamson here .

Things could be about to ramp up

If you still haven't seen the aurora, hold on for another 30 minutes to an hour, according to CNN meteorologist Chad Myers.

The next wave of coronal mass ejections, or CMEs, which cause the aurora, is about to arrive, he said.

"Just wait a minute because things are going to start to ramp up here," he said, adding that the increase could arrive "anytime now." "When it comes, get outside, get ready, put your coat on."

For those who are too busy to witness the phenomenon tonight, Myers said the aurora is expected to last three nights.

Why does the aurora last for a weekend?

By CNN's Chris Lau

The northern lights can be seen from Eaton Rapids, Michigan, on May 10.

Generally, it takes just eight minutes for light to travel 93 million miles to the Earth from the sun, but astrophysicist Janna Levin said the energized particles causing the current wave of aurora travel a lot slower, causing the phenomenon to last for the weekend.

"Some of these mass ejections are trillions of kilograms," she said. "They're slower. So they're taking longer, but still hours, maybe tens of hours."

Here's how the solar storm looks in the South and on the East Coast

The aurora was visible across the East Coast and in the South Friday.

Here's how it looked in Chester, South Carolina.

Down in Florida, waves of color swam through the sky.

Up north in New Jersey, a purple-ish haze could be seen in the sky.

Will solar storms get more intense and risky in the future?

The answer is probably not in the short term, according to astrophysicist Hakeem Oluseyi.

He said scientists study what is constantly happening on the surface of the sun and have found a pattern.

“Geological data shows us that in the past the sun was way more active than it is today. It has cycles where it goes very quiet ... and you have events that show that the solar activity was much, much greater,” he told CNN. “So there's no evidence that we're going to see those big maxima this cycle." 

But the astrophysicist also spoke of a caveat - the limitations of modern science.

“Even though it's predictable in the short term, we still don't quite understand what creates the magnetic fields in the sun,” he said, adding: “That's why NASA has so many satellites looking at the sun.”

In Pictures: Auroras light the sky during rare solar storm

From CNN Digital's Photo Team

The northern lights glow in the night sky in Brandenburg, Germany, on May 10.

A series of solar flares and coronal mass ejections from the sun are creating dazzling auroras across the globe .

The rare solar storm may also disrupt communications. The last time a solar storm of this magnitude reached Earth was in October 2003, according to the National Oceanic and Atmospheric Administration's Space Weather Prediction Center.

See more photos of the aurora from tonight.

Behind dazzling aurora could lie “real danger,” Bill Nye the Science Guy says

Bill Nye the Science Guy speaks to CNN on Friday, May 10.

The massive solar storm could present “a real danger,” especially with the modern world relying so much on electricity, according to Bill Nye the Science Guy , a science educator and engineer.

Scientists are warning an increase in solar flares and coronal mass ejections from the sun have the potential to disrupt communication on Earth into the weekend. Solar flares can affect communications and GPS almost immediately because they disrupt Earth’s ionosphere, or part of the upper atmosphere. Energetic particles released by the sun can also disrupt electronics on spacecraft and affect astronauts without proper protection within 20 minutes to several hours.

In comparison to tonight's event, Nye drew comparisons with another incident in 1859, known as the Carrington Event, when telegraph communications were severely affected.

“The other thing, everybody, that is a real danger to our technological society, different from 1859, is how much we depend on electricity and our electronics and so on,” Nye said. "None of us really in the developed world could go very long without electricity."

He noted that there are systems in place to minimize the impact, but “stuff might go wrong,” stressing that not all transformers are equipped to withstand such a solar event.

“It depends on the strength of the event and it depends on how much of our infrastructures are prepared for this the sort of thing,” he said.

Bill Nye breaks down significance of the solar storm | CNN

Bill Nye breaks down significance of the solar storm | CNN

This post has been updated with more details on solar flares' impact on electronics.

Here's where clouds will block the view of the northern lights in the US

From CNN's Angela Fritz

An infrared satellite image taken around 10:30 p.m. ET.

After an incredibly stormy week, most of the Lower 48 has clear skies to see the northern lights. But there are some areas where clouds and rainy weather are spoiling the view.

A deck of clouds is blocking the sky in the Northeast, from parts of Virginia into Maine, as an area of low pressure spins off the East Coast.

In the Midwest, the aurora will be hard to see through thick clouds in parts of Wisconsin, Michigan — including the Upper Peninsula — and Illinois.

A stripe of clouds is tracking across Texas, including Dallas-Forth Worth, and into Louisiana.

And in the Southwest, patchy clouds across the the Four Corners region could make the northern lights difficult to spot.

Aurora seen at least as far south as Georgia

Barely visible to the naked eye, the aurora can be seen in Atlanta in the 10 p.m. ET hour. 

It is easier to see through photographs using a long exposure. The photos below, taken by CNN's Eric Zerkel and Emily Smith, used 3- and 10-second exposures.

Aurora seen in Atlanta around 10:15 p.m. ET.

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  1. Learn How to Solve Force Problems

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  2. Guide to Solving Force Problems

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  3. Introduction to Forces

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  4. Problem solving help force level 4 example16

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  5. Dynamics: Force, Friction, FBD, & Problem Solving

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COMMENTS

  1. Forces in Physics, tutorials and Problems with Solutions

    The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions.

  2. Vectors and Forces Problem Sets

    Problem 2: Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk. Hector yanks on the chain with a 67.0 N force at an angle of 30.0° above the horizontal. Determine the horizontal and vertical components of the tension force.

  3. 6.1 Solving Problems with Newton's Laws

    A classic problem in physics, similar to the one we just solved, is that of the Atwood machine, which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure 6.7, m 1 = 2.00 kg and m 2 = 4.00 kg.

  4. 6.2: Solving Problems with Newton's Laws (Part 1)

    Problem-Solving Strategy: Applying Newton's Laws of Motion. Identify the physical principles involved by listing the givens and the quantities to be calculated. Sketch the situation, using arrows to represent all forces. Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.

  5. Force Problems

    Hints And Answers For Force Problems. Hint and answer for Problem # 7. The force of gravity pulling down on the block is F1 = Mg sin θ. The maximum friction force opposing the sliding is F2 = Mg cos θμs. At some angle θ the block will be on the verge of sliding. This is the maximum angle θ and occurs when F1 = F2.

  6. 4.9: Problem-Solving Strategies

    Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure ...

  7. Newton's Law Problem Sets

    Problem 22: Brandon is the catcher for the Varsity baseball team. He exerts a forward force on the .145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m. Determine the acceleration of the ball and the force which is applied to it by Brandon.

  8. Forces and Newton's laws of motion

    Magnetic forces, magnetic fields, and Faraday's law. Unit 14. Electromagnetic waves and interference. Unit 15. Geometric optics. Unit 16. Special relativity . Unit 17. Quantum Physics. ... Masses on incline system problem (Opens a modal) Up next for you: Unit test. Level up on all the skills in this unit and collect up to 300 Mastery points!

  9. Calculating Force

    In this problem there will be two forces acting upon the airplane: the weight of the plane (force of gravity) and the lifting force. Since we are looking for the minimum force to lift the plane, we can set the two forces equal to each other: . We can calculate the gravitational force using the mass.

  10. 4.6 Problem-Solving Strategies

    Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation.

  11. Finding force

    Okay, so F is gonna be equal to m, which is 60 kilograms times the acceleration, which we just calculated, four meters per second squared. And we can now figure out what that force is. We just have to multiply. Six times four is 24 and there's a zero. And the units become kilogram meters per second squared.

  12. F=ma Practice Problems

    F=ma Problem Set . Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations. In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems. 1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

  13. Centripetal force problem solving (video)

    The secret to solving centripetal force problems is that you solve them the same way you solve any force problem. In other words, first, you draw a quality force diagram. And then you use Newton's second law for one of the directions at a time. And if the direction you chose to analyze Newton's second law for didn't get you to where you needed ...

  14. Net Force Problems Revisited

    The following problems provide plenty of practice with F net = m • a problems involving forces at angles. Try each problem and then click the button to view the answers. 1. A 50-N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook, Olive N. Glenveau, and Warren Peace are ...

  15. Solving problems which involve forces, friction, and Newton's ...

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  16. Centripetal Force

    practice problem 1. A 250 kg motorcycle is driven around a 12 meter tall vertical circular track at a constant speed of 11 m/s. Determine the normal and friction forces at the four points labeled in the diagram below. Determine the minimum coefficient of static friction needed to complete the stunt as planned.

  17. Dynamics (Force or Newtons 2nd Law) Problems

    How to Solve Force Problems. 1.Identify the Problem. Any problem that asks you to relate force and motion is a Newton's Second Law problem, no matter what was given or requested in the problem.In some cases, Newton's Second Law is easy to identify—for example, a problem might ask you for the value of a particular force.

  18. Problem-Solving Strategies

    Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure ...

  19. Moment of a Force: Problem Solving (Video)

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  20. Net Force Word Problems

    Problem #5: Your friend is pulling upward on an object with force of 3 N. You are pulling to the right with a force of 4 N. Find the net force and the direction the object moves. Just build the rectangle and find the resultant. The red arrow shows the direction the object will move. This net force word problem is a little challenging.

  21. Newton's second law: Solving for force, mass, and acceleration

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  22. Solving Force Problems in Physics by Using Free-Body Diagrams

    Whether it's to pass that big test, qualify for that big promotion or even master that cooking technique; people who rely on dummies, rely on it to learn the critical skills and relevant information necessary for success. In physics, force problems typically ask you to predict what will happen when you apply force to an object, and usually ...

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  26. Brute-Force Integer Max Flow Algorithm for Networks Under Reachability

    An exact algorithm for solving integer maximum flow problem under reachability restrictions is proposed. A computational complexity of brute-force method was obtained. Presented examples show main issues arising while solving network flow problems under reachability restrictions. Enhancements for reducing operations number and redundant computations are provided. Tree-structured networks are ...

  27. Aurora lights up the sky in geomagnetic storm

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