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2.4: Problem Solving

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• Page ID 18390

• Tom Weideman
• University of California, Davis

Next we are going to focus on elements of problem-solving. We have all the tools we need, so this will not involve any new physics, but the idea is to introduce you to some common themes that come up in physics mechanics problems.

One of the favorite devices for physics problems is the pulley. As was stated in the description of the tension force, to start out we use the simplest model, which means we will assume that pulleys are massless and frictionless. Pulleys get especially interesting in situations like the following example, where at least one of the pulleys is able to move. The two blocks remain at rest in the system of ropes and pulleys shown in the diagram. Given this information, can you conclude how the two masses compare?

Figure 2.4.1 – Blocks Hanging from Multiple Pulleys

By now we know that when it comes to analyzing the forces present in a system, there is no better tool than the FBD. We begin there:

Figure 2.4.2 – FBD of the Block and Pulley

[We have taken the liberty of defining coordinate systems in our FBDs – up is the \(+y\)-direction for both – which we will need shortly.]

One might ask why there are two tension force vectors drawn for the pulley. The simplest answer is to consider what you would feel if you cut the rope on both sides of the pulley and held one end in each hand. Clearly you would feel both ends of the rope pulling down. Therefore by Newton's third law, both ends of the rope are pulling up on the pulley. With the pulley massless and frictionless, these two tension forces must also be equal, which explains why they are labeled the same. Note that the tension vector on the block is also labeled with the same variable name. This is because it is the same rope , and our assumption of massless, frictionless pulleys ensures that everywhere that we measure the tension for a single piece of rope, it will be the same.

If we were to draw the tension force vector pulling up on the right pulley, we would not be able to label it the same. Not all tension vectors in a single physical system are equal, just the magnitudes of all tension vectors derived from the same rope.

Another curious aspect of this FBD is the weight label of the left pulley. Technically, that force is acting on the block, and the block is pulling on the pulley. The pull on the pulley by the block happens to equal the weight of the block in this case, and the pulley has no weight of its own, so we are justified in taking this little shortcut. Another way to justify it is to treat the block + pulley as a single system, and the gravity force on the system is the force vector shown.

The next step in our analysis is to sum the forces for each object and apply Newton's second law, which in this case involves zero acceleration. In taking the sum of forces, we have to take care to correctly use our coordinate system:

\[ \left. \begin{array}{l} 0 = a_1 = \dfrac{F_{net \; 1}}{m_1} = \dfrac{2T - m_1 g}{m_1} \;\;\; \Rightarrow \;\;\; T = \dfrac{m_1 g}{2} \\ 0 = a_2 = \dfrac{F_{net \; 2}}{m_2} = \dfrac{T - m_2 g}{m_2} \;\;\;\; \Rightarrow \;\;\; T = m_2 g \\ \end{array} \right\} \;\;\; \Rightarrow \;\;\; m_1 = 2 m_2\]

Notice that the light weight \(m_2\) holds up the heavier one because the placement of the pulley allows us to use the tension from the same rope twice on the heavier mass. This trick can actually be repeated as many times as we like (the pulley can have multiple tracks in it), and this enables us to lift very heavy weights with very little force. This invention is called a block and tackle . They are used for sailing ships (the heavy sails and boom can be pulled tighter), lifting engine blocks, and many other applications.

Constraints

Next we take on a tricky concept known as constraint s. A constraint is a condition that exists for a physical system that restricts how it can behave. What makes the concept "tricky" is that these ultimately play a mathematical role in the solution of the problem, but this role is often difficult to extract from the statement of the problem. Put simply, constraints relate variables in the problem to each other, providing additional equations (beyond Newton's second law) to work with. We have already seen an example of a constraint. It is the relationship between the friction force and the normal force. For kinetic friction this provides an equation that relates these two forces, while for static friction it sets an upper limit on the magnitude of friction for a given normal force.

One of the most common examples of a constraint is related to ropes moving through pulleys. This constraint relates the motion of one object to that of another when they are connected through a system of pulleys. Let us return to the system shown in Figure 2.4.1 and ask the following question: If the block \(m_2\) drops a distance \(\Delta y \), what happens to the block \(m_1\)?

First of all, it should be clear that \(m_1\) rises as \(m_2\) drops, so the only question is, how far? This may not be apparent at first, but think of it this way: When the pulley holding \(m_1\) moves up 1 unit, both segments of string going up from the pulley get shorter by 1 unit. These two units of string don't simply vanish, and in fact they are taken-up by the free end of the string, which is attached to \(m_2\). This means that as \(m_2\) drops a distance of \(\Delta y \), \(m_1\) must rise only half that far.

What does this say about the comparison of the speeds and accelerations of the two blocks? Well, they are required to move simultaneously, so every unit of length dropped by \(m_2\) is matched by a rise of \(m_1\) by half as much, which means that \(m_1\) always moves at half the speed and accelerates half as much as \(m_2\). If this system is not balanced (as it was above), then applying Newton's second law to both blocks includes two accelerations, but these are constrained to be related to each other by a factor of two, providing us with an additional constraint equation :

\[ 2\left| a_1 \right| = \left| a_2 \right|\]

What's with the absolute values, you ask? Well, these variables can have positive or negative values, and we must be careful when it comes to signs. in particular, we have to look at how our constraint relates to our choice of coordinate systems for the two blocks. In Figure 2.4.2 , we chose "up" as the positive direction for both blocks. So we need to ask ourselves, "If one block experiences a positive displacement, what is the sign of the displacement of the other block?" In this case it's clear that the displacement of the two blocks have opposite signs. Therefore the constraint equation for the block accelerations is:

\[ 2a_1 = -a_2\]

Note that it is perfectly fine to set up different coordinate systems for the two blocks – each FBD is entitled to its own individual coordinate system. How the coordinate systems relate to each other affects the equation of constraint. So for example, if we had instead chosen downward to be the +\(y\)-direction for block #2 (but left upward as positive for the other block), then there would be no need for the minus sign in the constraint equation – positive displacements of one block correspond to positive displacements of the other block. We see that there is therefore no "correct" choice of coordinate system, but we must take care when the time comes to combine the equations from the two FBDs that the constraint equations relate the variables correctly. We will see an even more striking example of coordinate system choice in the next example.

Example \(\PageIndex{1}\)

The heights of the two blocks in the diagram below differ by 36cm. When they are released from rest, the higher block falls while the lower block rises. One of the blocks has a mass that is three times the mass of the other block, the pulleys are massless and frictionless, and the string doesn't stretch.

• Which is the heavier block? Explain.
• Find the distance that the lower block rises when the two blocks are aligned.
• Find the time it takes for the two blocks to be aligned.

a. The tension exerted by the string threaded through the pulleys is the same everywhere, so there is three times as much tension force acting up on the lower block as there is up on the higher one. If the lower block was three times heavier than the higher block, then the system would be balanced, and neither mass could be accelerating. [You should try doing the math of part (c) with the masses the other way, and demonstrate for yourself that it the acceleration would have to be zero.] Given that the system is accelerating, it must be the higher block with the greater mass.

b. As the top block pulls the string down on one side of the large pulley, the same amount of string that is gained on the left side of the pulley is lost from the right side. The string on the right side of the large pulley is divided between the three segments holding up the other block. Therefore the lower block moves up one third as far as the higher block moves down. If the lower block rises a distance \(y\), then the higher block drops a distance \(3y\), and since they reach the same height, the sum of those changes is 36cm, which means the lower block rises a distance of \(y=9\)cm.

c. To find the the time it takes them to align, we need to use their accelerations. We know their relative accelerations already: The lower block accelerates at one third the rate of the higher block. We'll therefore call the higher block's acceleration "\(3a\)," making the lower block's acceleration equal to \(a\). But we need Newton's laws in order to go any further. FBD of the two systems involved look like this:

The higher mass is three times the lower mass, so we will call \(m_2\) simply "\(m\)," which makes \(m_1\) equal to \(3m\). Plugging everything into Newton's second law for both FBDs gives these equations:

\[ \left. \begin{array}{l} a_1 = \dfrac{F_{net \; 1}}{m_1} \;\;\; \Rightarrow \;\;\; 3a = \dfrac{-T + 3mg}{3m} \\ a_2 = \dfrac{F_{net \;2}}{m_2} \;\;\; \Rightarrow \;\;\; a = \dfrac{3T - mg}{m} \\ \end{array} \right\} \;\;\; \Rightarrow \;\;\; a = \frac{2}{7} g \nonumber \]

With the acceleration of the lower block, the distance it travels, and the fact that it starts from rest, we can compute the time it takes to make the trip:

\[ y = v_o t + \frac{1}{2}at^2 \;\;\; \Rightarrow \;\;\; t = \sqrt{ \dfrac{7y}{g}} = \boxed{0.25s} \nonumber \]

Inclined Planes

Pulleys (and in particular the block & tackle) are an example of something often referred to as a simple machine . This is because we can use such a device to lift a heavy weight with a force less than the weight itself. Another example of a simple machine is the inclined plane . These devices also allow one to accomplish the task of raising a heavy object to a higher position using less force than would be necessary in a direct lift. The main feature of problems involving inclined planes is dealing with the coordinate system used for the object on the inclined plane. Let’s look at an example. As usual, we start with the very simplest of examples, and complicate it as we go – a block of mass \(M\) on a frictionless plane inclined at an angle \(\theta\):

Figure 2.4.3 – Block on an Inclined Plane

We begin our analysis, as always, with a free-body diagram. No FBD is complete without a choice of coordinate system, so we must choose one here. If we choose our coordinate system to be horizontal and vertical as we usually do, then when the block slides down the plane its acceleration will have both \(x\) and \(y\) components. This is fine of course, but it can be a bit cumbersome to work with. There is nothing sacred about the horizontal and vertical directions, so why not choose a coordinate system that is parallel and perpendicular to the plane?

Figure 2.4.4 – Coordinate System for an Inclined Plane

The middle diagram in Figure 2.4.4 shows the tail-to-head sum of the normal and gravity forces resulting in a net force parallel to and down the plane, as it should be, since we know that is the direction the block will accelerate. The right diagram shows the gravity force broken into its \(x\) and \(y\) components in the chosen coordinate system (you should verify for yourself the geometry that leads to concluding that the angle in this diagram is the same angle that the incline makes with the horizontal). Note that since the block does not accelerate perpendicular to the plane, we can conclude that \(N=Mg\cos{\theta}\). Also it's clear that the net force is the \(x\)-component of the gravity force, resulting in an acceleration down the plane of simply \(a=g\sin{\theta}\).

Suppose we wanted to raise this block to some new height. To get it moving up the plane, we would need to apply a force that exceeds the net force shown above, which is less than the force we would have to apply to lift it straight up. Like the case of the block & tackle, this "simple machine" helps us to get a job done while using less force than is required if it is done more directly. In the case of the block & tackle, to raise the mass some distance, the other end of the rope had to be pulled a greater distance. In this case we see a similar thing – the distance we must push the block is the hypotenuse of the triangle in order to raise it the height of the vertical leg of the triangle. This is a common theme in simple machines – less force is required, but it must be applied over a longer distance. In the next chapter we will see why this is the case.

We can complicate this simple example considerably. The most natural adjustment is to incorporate friction. Because of the nature of the two types of friction, adding static friction (which enters as an inequality) can be particularly troublesome. To see why, let's consider the following problem:

The system depicted in Figure 2.4.5 shows two blocks that remain at rest which are attached by a massless string over a massless, frictionless pulley. The plane is inclined at an angle \(theta\) up from the horizontal, and its surface is rough (i.e. not frictionless). The mass of the hanging block is given, as is the angle of incline and the coefficient of static friction. From these quantities, determine the minimum possible value of the mass of the block on the plane.

Figure 2.4.5 – A Mechanical System

Starting (as always) with a FBD (including a coordinate system) for each block, we have:

Figure 2.4.6 – Free-Body Diagrams of Blocks

Let's take a moment to comment on the direction of the static friction force. Recall that a static friction force merely reacts to the "attempted" motion of the object along the surface. In this case, if the block was "trying" to slide down the plane, then the static friction force must be up the plane. Here it is drawn pointing down the plane, which means the other forces present must be such that they would accelerate it up the plane... How do we know this is the case? The answer lies in the statement of the question: We are looking for the minimum mass for the block on the plane. Imagine putting in a block whose mass balances the system. if a small mass is added to or subtracted from the block, the system may still remain at rest, as the static friction keeps the balance. If we add too much mass to that block, the static friction will reach its limit and the block will begin sliding down, while if we take away too much mass, it will slide up the plane. The static friction will oppose the intended motion, so for the minimum mass, the static friction force must point down the plane.

While we are able to determine the direction of the static friction for this problem, in many problems this is not possible. If you know that the static friction force is (or could be) present, just draw it in with either direction. When the problem is complete, if you solve for the value of this force, it will come out positive if you chose the correct direction, and negative if you did not. The FBD is just a tool, and in the end you end up with the answer, so don't waste energy worrying about getting the direction correct on the diagram.

Breaking the vectors into components in our chosen coordinate systems and applying Newton's second law (for zero acceleration) gives:

\[ \begin{array}{l} block\;on\;plane:\;\; \begin{array}{l} x - direction\;forces: \;\; 0 = a_x = \dfrac{f - T + Mg\sin{\theta}}{M} \\ y - direction\;forces:\;\; 0 = {a_y} = \dfrac{N - Mg\cos{\theta}}{M} \end{array} \\ hanging\;block:\;\;\; y - direction\;forces:\;\; 0 = {a_y} = \dfrac{T - mg}{m} \end{array} \]

Next, apply the constraint that relates the maximum static friction force (which occurs when the minimum mass is on the plane) and the normal force:

\[ f \le \mu_S N \;\;\; \Rightarrow \;\;\; (maximum \; f \; for \; minimum \; M) \;\;\; \Rightarrow \;\;\; f = \mu_S N\]

The rest is algebra with four simultaneous equations, the result of which is:

\[M_{min} = \frac{m}{\mu _S \cos \theta + \sin \theta }\]

We should now check to see if this answer makes sense. If the angle \(theta\) is \(90^o\), then both masses are hanging, and there is not friction force (because there is no normal force). For the system not to accelerate, the two masses must be equal. Plugging in \(\theta=90^o\) indeed results in \(M=m\). The \(\theta\ = 0^o\) case (a horizontal surface, where the normal force equals the weight of the block on the surface) will require that the friction force equals the weight of the hanging block. That is, we must have \(mg=f=\mu_S N=\mu_S Mg\;\; \Rightarrow \;\; m=\mu_S M\), which is what we get when we plug in zero for \(\theta\).

This problem could just as easily have asked for the maximum possible value for \(M\). It is left as an exercise for the reader to try this.

There was a lot involved with this problem, but the key is to take it one step at a time and follow the following prescribed steps:

• draw a diagram
• isolate the relevant objects and draw free-body diagrams for them
• choose coordinate systems for the diagrams that are convenient
• break forces into components in the coordinate system chosen
• sum the forces in the \(x\) and \(y\) directions and apply Newton’s second law for both directions
• apply the constraints
• solve the algebra

Example \(\PageIndex{2}\)

A rope is fastened to a 50.0kg block in two places and passes through a system of two pulleys, as shown in the diagram below. The block rests on a rough (coefficient of static friction is 0.400) horizontal surface. The bigger pulley is then pulled upward with gradually increasing force. Both pulleys are massless and frictionless, and the rope is also massless. The smaller pulley is fastened to the floor and the both pulleys are positioned such that the rope is perpendicular to the floor on one end and parallel to it on the other. When the pull force reaches a certain magnitude, the block just barely begins to slide to the right. Compute the magnitude of this pull force.

This problem doesn't feature an inclined plane (though it could!), but it is a good example of the importance of following the prescription listed above. Start with the free-body diagrams and coordinate systems. The FBD of the smaller pulley will yield us nothing useful, so there are just two FBDs to draw. Note that the tension on the side of the block comes from the same rope as the tension on the top of the block, so they are equal:

The block is not accelerating at all (nor is the pulley), so the sum of the forces in each of the \(x\) and \(y\) directions comes out to zero.

\[ \begin{array}{l} block:\;\; \begin{array}{l} x - direction\;forces: \;\; 0 = T - f \\ y - direction\;forces:\;\; 0 = T+N-mg \end{array} \\ pulley:\;\; y - direction\;forces:\;\; 0 = pull - 2T \end{array} \nonumber \]

If we have to pull "just hard enough" to get the block moving, then this occurs when the horizontal pull equals the maximum static friction force, which gives us a constraint equation:

\[ f = \mu_S N\nonumber\]

Note that the block will have to start sliding before it starts rising, because rising requires than the normal force equals zero, and it will slide when the static friction force is small-but-non-zero. Now solve the equations simultaneously to get:

\[ pull = \boxed{280N} \nonumber \]

Incorporating Motion

We’ve done two examples involving systems for which the acceleration is zero. But of course it’s possible that a problem could actually involve accelerating objects. Sometimes we are asked to find this acceleration, and other times the acceleration is a piece of information that is given. The acceleration can be given directly, or possibly it can be calculated in another way, perhaps from kinematics or if the motion is circular. Knowing something about the motion of the object falls under the “constraints” category, because the motion is specified (constrained), bringing in equations that don't result from Newton's second law. We'll look at an example that employs the steps to solving mechanics problems that also includes the added constraint of circular motion. Before we do, give the following example a try:

Example \(\PageIndex{3}\)

A block is attached to one end of a massless spring, the other end of which is attached to a vertical fixed peg in a frictionless horizontal surface. The block is spun around a circle, and the spring stretches as a result of this motion. In fact, the faster the motion, the more the spring stretches. To stretch any spring, both ends need to be pulled simultaneously. Clearly the peg is pulling on one end of the spring as the block goes in the circle, but what force is pulled the block outward to stretch the spring?

The block is not pulled outward! It is only pulled inward (by the spring). It is not the block that needs to be pulled outward to stretch the spring, but rather the spring that needs to be pulled that way. The spring pulls the block inward (keeping it accelerating centripetally), and the third-law-pair force of the block on the spring is what pulls the spring outward.

This points out possibly better than any other example the importance of isolating objects with force diagrams. The block here is not a conduit for some mysterious force pulling out on the spring – it is the object pulling out on the spring. You thoroughly need to trust the third law here to get the force between the spring and the block, and you need to thoroughly trust the second law to realize that the block does not require another force on it outward to balance the spring force, because it is accelerating.

Now for the promised example that incorporates motion following the step-by-step prescription given earlier. What makes this problem interesting is the information that is hidden within the wording...

A rock on a string flies around in a circle in a vertical plane (in the presence of the earth's gravity) such that it just barely gets by the top (the string remains stretched to its full length, but there is no tension) as it continues in its circular path. Find the speed of the rock in terms of the length of the string.

We'll choose down as the positive \(y\)-direction.

Unnecessary here.

• sum the forces in the \(x\) and \(y\) directions and apply Newton's second law in both directions

\[a = \dfrac{\sum {F_y} }{m} = \dfrac{T + mg}{m}\]

The first constraint is that the rock barely makes it around. What does this mean? To answer this, think about what would happen if the rock was moving any slower... It would fall out of the circle, which means the string would not remain straight. This would mean that the tension is zero. So the condition of "just making it around" is equivalent to requiring that the tension vanishes. The second constraint is that the rock is traveling in a circle, which requires that the acceleration is centripetal, with the radius of the circle equalling the length of the string (note there is no horizontal force at the top, so there is no tangential component to the acceleration). Stating both mathematically:

\[T=0,\;\;\;\;\; a=\dfrac{v^2}{l} \]

• do the algebra

Simple enough:

\[v=\sqrt{gl} \]

While this is a pretty simple example in terms of the steps taken, it points out one very important aspect of these problems. If you don’t spend some time thinking about what is physically happening, you are likely to overlook the “hidden” information in the wording of the problem. This isn’t intended as a trick to trip you up – this is exactly what you run into in the real world when you need to solve a real problem. You need to be able to convert descriptive aspects of a system into mathematically-analyzable quantities.

Example \(\PageIndex{4}\)

A tetherball swings around a pole, making a full circle once every 1.5s. The total length of the rope is 2.4m. Calculate the angle θ that the rope makes with the pole. The rope has negligible mass.

Start with a force diagram of the ball, including a coordinate system:

Next sum the forces along the \(x\) and \(y\) axes and apply Newton's second law:

\[ \begin{array}{l} a_x = \dfrac{T \sin \theta}{m} \\ a_y = \dfrac{T \cos \theta - mg}{m} \end{array} \nonumber \]

The acceleration in the \(x\)-direction is centripetal, with the radius of its circular motion equal to the horizontal leg of the right triangle whose hypotenuse is the length of the string. The speed is constant, so it equals the circumference of the circle divided by the time for the trip. Putting all this together gives:

\[ \left. \begin{array}{l} a_x = \dfrac{v^2}{R} \\ R = l \sin \theta \\ v = \dfrac{2 \pi R}{t} \end{array} \right\} \;\;\; \Rightarrow \;\;\; a_x = \dfrac{4 \pi^2 l \sin \theta}{t^2} \nonumber \]

The acceleration in the \(y\)-direction is zero, so plugging the accelerations into the equations from Newton's second law and eliminating \( \frac{T}{m} \) from the two equations gives:

\[ \left. \begin{array}{l} a_x = \dfrac{4 \pi^2 l \sin \theta}{t^2} = \dfrac{T \sin \theta}{m} \;\;\; \Rightarrow \;\;\; \dfrac{T}{m} = \dfrac{4 \pi^2 l }{t^2} \\ a_y = 0 = \dfrac{T \cos \theta - mg}{m} \;\;\; \Rightarrow \;\;\; \dfrac{T}{m} = \dfrac{g}{\cos \theta} \end{array} \right\} \;\;\; \Rightarrow \;\;\; \theta = \cos ^{-1} \left(\dfrac{g t^2}{4 \pi^2 l} \right) = \boxed{76.5^o} \nonumber \]

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6.1 Solving Problems with Newton’s Laws

Learning objectives.

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

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. 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, so the following techniques should reinforce skills you have already begun to develop.

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.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have 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 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to 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. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (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 in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (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 the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this 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. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. 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; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

• We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
• Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

• We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
• Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

• ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
• Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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1.7: Problem Solving Process

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Learning how to use a structured problem solving process will help you to be more organized and support your future courses. Also, it will train your brain how to approach problems. Just like basketball players practice jump shots over and over to train their body how to act in high pressure scenarios, if you are comfortable and familiar with a structured problem solving process, when you’re in a high pressure situation like a test, you can just jump into the problem like muscle memory.

6 Step Problem Solving Method:

• Write out the answer with all necessary information that is given to you. It feels like it takes forever, but it’s important to have the problem and solution next to each other.
• Draw the problem, this is usually a free-body diagram (don’t forget a coordinate frame). Eventually, as you get further into the course, you might need a few drawings. One would be a quick sketch of the problem in the real world, then modelling it into a simplified engineering drawing, and finally the free-body diagram.
• Write out a list of the known/given values with the variable and unit, i.e m = 14 kg   (variable = number unit)
• Write out a list of the unknown values that you will have to solve for in order to solve the problem
• You can also add any assumptions you made here that change the problem.
• Also state any constants, i.e. g = 32.2 ft/m 2   or g = 9.81 m/s 2
• This step helps you to have all of the information in one place when you solve the problem. It’s also important because each number should include units, so you can see if the units match or if you need to convert some numbers so they are all in English or SI. This also gives you the variables side by side to ensure they are unique (so you don’t accidentally have 2 ‘d’ variables and can rename one with a subscript).
• Write a simple sentence or phrase explaining what method/approach you will be using to solve the problem.
• For example: ‘use method of joints’, or equilibrium equations for a rigid body, MMOI for a certain shape, etc.
• This is going to be more important when you get to the later chapters and especially next semester in Dynamics where you can solve the same problem many ways. Might as well practice now!
• This is the actual solving step. This is where you show all the work you have done to solve the problem.
• When you get an answer, restate the variable you are solving for, include the unit, and put a box around the answer.
• Write a simple sentence explaining why (or why not) your answer makes sense. Use logic and common sense for this step.
• When possible, use a second quick numerical analysis to verify your answer. This is the “gut check” to do a quick calculation to ensure your answer is reasonable.
• This is the most confusing step as students often don’t know what to put here and up just writing ‘The number looks reasonable’. This step is vitally important to help you learn how to think about your answer. What does that number mean? What is it close to? For example, if you find that x = 4000 m, that’s a very large distance! In the review, I would say, ‘the object is 4 km long which is reasonable for a long bridge’. See how this is compared to something similar? Or you could do a second calculation to verify the number is correct, such as adding up multiple parts of the problem to confirm the total length is accurate i.e. ‘x + y + z = total, yes it works!’

Additional notes for this course:

• It’s important to include the number and label the steps so it’s clear what you’re doing, as shown in the example below.
• It’s okay if you make mistakes, just put a line through it and keep going.
• Remember your header should include your name, the page number, total number of pages, the course number, and the assignment number. If a problem spans a number of pages, you should include it in the header too.

Key Takeaways

Basically: Use a 6-step structured problem solving process: 1. Problem, 2. Draw, 3. Known & Unknown, 4. Approach, 5. Analysis (Solve), 6. Review

Application: In your future job there is likely a structure for analysis reports that will be used. Each company has a different approach, but most have a standard that should be followed. This is good practice.

Looking ahead: This will be part of every homework assignment.

Written by Gayla & Libby

• © 2018

Solving Practical Engineering Mechanics Problems

• Sayavur I. Bakhtiyarov 0

New Mexico Institute of Mining and Technology, USA

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Part of the book series: Synthesis Lectures on Mechanical Engineering (SLME)

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Table of contents (6 chapters)

Front matter.

Sayavur I. Bakhtiyarov

Back Matter

Engineering Mechanics is one of the fundamental branches of science which is important in the education of professional engineers of any major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics, acoustics, vibrations, etc. are based on Engineering Mechanics course. In order to absorb the materials of Engineering Mechanics, it is not enough to consume just theoretical laws and theorems—student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many problems independently. This book is a part of a four-book series designed to supplement the Engineering Mechanics courses in the principles required to solve practical engineering problems in the following branches of mechanics: Statics, Kinematics, Dynamics, and Advanced Kinetics. Each book contains 6-8 topics on its specific branch and each topic features 30 problems to be assigned as homework, tests, and/or midterm/final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided.

This second book in the series contains six topics of Kinematics, the branch of mechanics that is concerned with the analysis of motion of both particle and rigid bodies without reference to the cause of the motion. This book targets undergraduate students at the sophomore/junior level majoring in science and engineering.

Book Title : Solving Practical Engineering Mechanics Problems

Book Subtitle : Kinematics

Authors : Sayavur I. Bakhtiyarov

Series Title : Synthesis Lectures on Mechanical Engineering

DOI : https://doi.org/10.1007/978-3-031-79609-8

Publisher : Springer Cham

eBook Packages : Synthesis Collection of Technology (R0) , eBColl Synthesis Collection 8

Copyright Information : Springer Nature Switzerland AG 2018

Softcover ISBN : 978-3-031-79608-1 Published: 10 April 2018

eBook ISBN : 978-3-031-79609-8 Published: 31 May 2022

Series ISSN : 2573-3168

Series E-ISSN : 2573-3176

Edition Number : 1

Number of Pages : XI, 143

Topics : Engineering, general , Electrical Engineering , Engineering Design , Nanotechnology and Microengineering

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4.8: Principle of Moments: 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.

The JoVE video player is compatible with HTML5 and Adobe Flash. Older browsers that do not support HTML5 and the H.264 video codec will still use a Flash-based video player. We recommend downloading the newest version of Flash here, but we support all versions 10 and above.

Consider a pole placed in a 3-dimensional system with a cable attached. When the tension of 15 kN is applied to the cable, determine the moment about the z-axis passing through the base.

The moment can be determined using two methods.

In the first method, calculate the projection of the force along the unit vector and multiply it by the force's magnitude to obtain the force vector.

The moment about the origin is the cross-product of the position vector and the force. The moment along the z-axis can be obtained by the dot product of the moment about the origin and the unit vector along the z-axis.

Alternatively, resolve the force vector into its components. The components along the y-axis and z-axis exert no moment as they pass through and are parallel to the z-axis, respectively.

The tension in the x-direction can be obtained by multiplying the tension with the direction cosine with respect to the x-axis.

Recall the moment of force equation, and by substituting the terms, the moment about the z-axis can be determined.

The principle of moments is a fundamental concept in physics and engineering. It refers to the balancing of forces and moments around a point or axis, also known as the pivot. This principle is used in many real-life scenarios, including construction, sports, and daily activities like opening doors and pushing objects.

One such scenario involves a pole placed in a three-dimensional system with a cable attached. When a tension is applied to the cable, the moment about the z-axis passing through the base needs to be determined. There are two methods to approach this problem.

The first method involves calculating the projection of the force along the unit vector and multiplying it by the force's magnitude to obtain the force vector. The moment about the origin can be calculated by taking the cross-product of the position vector and the force vector. Once the moment vector is determined, the moment along the z-axis can be evaluated by taking the dot product of the moment about the origin and the unit vector along the z-axis.

Alternatively, resolve the force vector into its components. The components along the y-axis and z-axis exert no moment as they pass through and are parallel to the z-axis, respectively. The tension in the x-direction can be calculated by multiplying the tension with the direction cosine with respect to the x-axis. Then, use the moment of force equation and substitute the appropriate terms to determine the moment about the z-axis.

It is important to note that the principle of moments is crucial in understanding the behavior of structures and machines. Engineers and designers use it extensively to ensure their creations are stable, safe, and effective. By balancing the forces and moments around a pivot point, they can calculate the stresses and strains that a structure or machine will endure and identify potential failure points.

Overall, the principle of moments is a powerful tool that helps to solve problems related to forces, torques, and motion. Whether it is regarding building bridges, analyzing sports techniques, or simply opening a door, one can rely on this principle to ensure that things are done correctly and efficiently.

• Hibbeler, R.C. (2016). Engineering Mechanics ‒ Statics and Dynamics. Hoboken, New Jersey: Pearson Prentice Hall. pp 133-138
• Beer, F.P.; Johnston, E.R.; Mazurek, D.F; Cromwell, P.J. and Self, B.P. (2019). Vector Mechanics for Engineers ‒ Statics and Dynamics . New York: McGraw-Hill. pp 96 -99
• Meriam, J.L.; Kraige, L.G. and Bolton, J.N. (2020). Engineering Mechanics ‒ Statics . Hoboken, New Jersey: John Wiley. pp 78 and 79

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Mechanics: Fluids

Problem Solving in Mechanical Engineering With Real World Examples

• Mechanical Engineering

Mechanical engineering is all about solving problems by using science and math. Engineers have to come up with better designs and improve how things are made. They make sure everything works well and lasts long. It’s important because they need to know a lot about their field and think both creatively and logically to find solutions to real problems.

For instance, they might work on making heating and cooling systems use less energy, find ways to cut down on waste when making products, or create new materials for planes and spaceships. These examples show how crucial mechanical engineers are in making technology and industries better.

In simpler terms, mechanical engineers are like problem-solving wizards. They use their deep knowledge and smart thinking to tackle challenges, like making a car engine that uses less fuel or a machine that makes fewer errors. They’re always learning and inventing to make sure the things we use every day are the best they can be.

This is key because their work helps us save money, be safer, and even protect the environment. It’s how they play a big part in pushing technology forward and keeping industries running smoothly.

Understanding Fundamental Principles

In mechanical engineering, it’s crucial to really get thermodynamics, materials science, and how to analyze structures. Knowing these core ideas helps you figure out how forces and materials work together, how energy moves and changes, and how to make sure structures are strong enough to handle different kinds of pressure.

When dealing with complicated systems, you break them down to understand how they work under different situations. For example, when choosing materials, you look closely at their strength, how much they can bend, and how well they conduct heat to make sure they will work well and last a long time. Thermodynamics helps make energy systems work better and use less power. Every solution is carefully made using these ideas to make sure the engineering designs do what they’re supposed to and are safe.

As an example, when building a bridge, engineers will use materials science to pick the right steel that can support the weight of cars over time without bending too much. They’ll apply thermodynamics to design any moving parts, like a drawbridge, to work efficiently with minimal energy waste. By focusing on these principles, engineers make sure the bridge is not only functional, allowing people to cross safely, but also stands the test of time.

Analyzing Complex Systems

To really get how complex machinery works, engineers take it apart to look at each piece. This helps them see how all the parts fit together and make the machine do its job. They start by figuring out where the system begins and ends, then they take a closer look at the smaller parts, like the sensors and motors, and the computer brain that controls everything.

Engineers use special tools and tests, like checking what could go wrong and how likely it is (that’s called FMEA), running computer models, and seeing how changes affect the system. By doing all this, they make sure the machine is safe, reliable, and works well because they’ve checked everything carefully, not left it to luck.

It’s important that they do this because it helps prevent accidents and breakdowns. For example, think about a car: if engineers didn’t test all the parts, like brakes and airbags, we wouldn’t trust them to keep us safe on the road. So, they use these tools to make sure everything is in top shape. This kind of detailed work means that when you use something like a car or a dishwasher, it’s been checked to work properly and safely.

Innovating in Product Design

Creating new and better product designs starts with really understanding how current products work. Mechanical engineers look at these products in detail to figure out how they can make them work better, use less energy, and give people a better experience when using them.

The first step is to carefully study what the product is supposed to do, how people use it, and where it can be improved. Engineers have to take apart complicated parts and processes to spot opportunities for new ideas. They use practical engineering knowledge to make designs that are not just better, but also cost less and are better for the environment. They make sure every part of the new product has a reason to be there and helps make the product both new and useful.

Being committed to making these kinds of advances is a big part of what mechanical engineering is all about in product design.

For example, when engineers worked on a new blender, they saw that the old design was hard to clean. They redesigned the blades to be detachable, which made cleaning easier and the blender more efficient. This change also saved materials, making the blender more eco-friendly.

This kind of thoughtful redesign shows how engineers can make our everyday products better.

Optimizing Manufacturing Processes

In manufacturing, engineers focus on improving the process to achieve faster production, reduced waste, and cost savings. They analyze production methods, examining data and observing operations to identify bottlenecks and inefficiencies. Strategies such as lean manufacturing or Six Sigma are employed to optimize operations and enhance overall efficiency.

Improvements can be made by rearranging machine placement to minimize material movement, implementing proactive maintenance practices to prevent breakdowns, and introducing new technologies like robots. Additionally, engineers work on optimizing the timing of supply deliveries to minimize storage costs. These deliberate actions lead to smarter and more cost-effective manufacturing processes, giving companies a competitive edge.

Ensuring Quality and Reliability

After improving how things are made, it’s crucial to make sure the products are of high quality and can be relied upon. To do this, it’s essential to have a well-thought-out plan for checking the quality and making sure it’s consistent.

Engineers need to create detailed tests that really show what conditions and pressures the products will face in the real world. For example, they might use Failure Mode and Effects Analysis (FMEA) to find and fix possible weaknesses before they cause problems. They also keep an eye on the production process using Statistical Process Control (SPC) to ensure everything stays the same.

Moreover, they use reliability engineering to figure out how to make products last longer. This is all about cutting down on mistakes and making sure the product is as good as it can be, which makes customers happy and maintains the manufacturer’s good name.

To give a specific example, a car manufacturer might use crash tests to simulate real-life accidents. This helps them understand how the car would perform and what they need to improve to ensure passenger safety. By doing this, they not only meet safety standards but also build trust with their customers who know the vehicles are tested thoroughly.

To wrap things up, solving problems in mechanical engineering isn’t simple—it’s a detailed task that involves really understanding the basics, figuring out complicated machinery, being creative when making products, making sure manufacturing is as good as it can be, and always aiming for the highest quality and dependability. It’s vital that all these pieces work together to tackle the tough problems we see in the real world. Mechanical engineers must think things through step by step and apply what they know to keep coming up with new and better ways to move technology forward and make industries run more smoothly.

For example, when engineers work on a new car engine, they need to know exactly how each part works. They must come up with smart designs that make the engine more powerful without using more fuel. They also have to refine the way the engine is built so that the factory can make it without wasting time or materials. Plus, they have to test the engine over and over to make sure it will last a long time and won’t break down. This kind of detailed work is what pushes us ahead, making cars more efficient and reliable for everyone.

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AI Physics mechanics solver

Entering the expression, choose a prompt command, about ai physics mechanics solver.

MathCrave AI physics mechanics solver is an incredible tool that utilizes MathCrave artificial intelligence to assist students in solving complex mechanics problems. With its advanced algorithms and deep understanding of physics concepts, this solver provides step-by-step solutions, enabling learners to grasp the fundamental principles and enhance their problem-solving skills. The AI physics mechanics solver is a valuable resource that offers convenience, accuracy, and efficiency, fostering a deep understanding of mechanics and enabling users to excel in their studies or professional endeavors.

AI Physics Mechanics Solver Solves Problems On:

1. Introduction to mechanics

2. Scalars and vectors

3. Kinematics

4. Motion in one dimension

5. Motion in two dimensions

6. Projectile motion

7. Relative motion

8. Newton's laws of motion

9. Force and motion

10. Friction

11. Circular motion and centripetal force;

12. Work, energy, and power

13. Conservation of energy

14. Potential energy and conservative forces

15. Kinetic energy and the work-energy theorem

16. Linear momentum and collisions

17. Conservation of momentum

18. Center of mass and its motion

19. Rotational mechanics

20. Torque and rotational equilibrium

21. Angular velocity and angular acceleration

22. Moment of inertia

23. Newton's law of universal gravitation

24. Gravitational field strength

25. Kepler's laws of planetary motion

26. Circular orbits and satellites

27. Oscillatory motion

28. Simple harmonic motion

29. Damped harmonic motion

30. Forced oscillations and resonance

31. Waves and periodic motion

32. Wave characteristics and properties

33. Wave speed, frequency, and wavelength

34. Reflection, refraction, and diffraction

35. Interference and standing waves

36. Doppler effect

37. Fluid mechanics

38. Pascal's principle and hydrostatic pressure

39. Archimedes' principle and buoyancy

40. Bernoulli's principle and fluid dynamics

41. Heat and thermodynamics

42. Temperature and thermal equilibrium

43. Heat transfer mechanisms

44. Laws of thermodynamics

45. Entropy and the second law of thermodynamics

Practice Questions on Mechanics

1. What is the difference between scalar and vector quantities in kinematics?

2. How is displacement different from distance in kinematics?

3. What is the equation for calculating average velocity in kinematics?

4. How does acceleration affect an object's motion in kinematics?

5. What is the difference between instantaneous velocity and average velocity in kinematics?

6. How do you calculate the time it takes for an object to reach a certain velocity in kinematics?

7. How does the concept of projectile motion relate to kinematics?

8. Explain the relationship between displacement, velocity, and time in kinematics

9. What is the equation for calculating acceleration in kinematics?

10. How does the concept of relative motion apply to kinematics?

Newton's Laws of Motion

1. What is Newton's first law of motion and how does it relate to inertia?

2. How do Newton's laws of motion explain the concept of force?

3. Explain the relationship between mass and acceleration according to Newton's second law of motion

4. How does Newton's third law of motion describe the interaction between two objects?

5. How does the force of gravity relate to Newton's laws of motion?

6. What is the difference between static friction and kinetic friction in relation to Newton's laws?

7. How does Newton's second law of motion explain the concept of momentum?

8. Explain how Newton's laws of motion apply to objects in circular motion

9. How does Newton's third law of motion explain the recoil of firearms?

10. How does Newton's laws of motion apply to the motion of a car on a banked curve?

Work and Energy:

1. What is the relationship between work and energy?

2. How do you calculate the work done by a force on an object?

3. Explain the difference between kinetic energy and potential energy

4. How does the principle of conservation of energy apply to work and energy?

5. How is mechanical energy conserved in a system?

6. What is the difference between positive work and negative work in relation to energy?

7. How does the concept of power relate to work and energy?

8. Explain how work is done against gravity and how it affects an object's potential energy

9. How does the law of conservation of energy apply to simple machines?

10. How does the concept of work-energy theorem relate to the conservation of energy?

Conservation Laws:

1. What is the law of conservation of momentum and how does it apply to collisions?

2. How does the concept of impulse relate to the law of conservation of momentum?

3. Explain the conservation of angular momentum in rotational motion

4. How does the law of conservation of energy relate to the conservation of momentum?

5. How does the law of conservation of charge apply to electric circuits?

6. What is the law of conservation of mass and how does it apply to chemical reactions?

7. How does the law of conservation of linear momentum apply to rocket propulsion?

8. Explain the concept of elastic collisions and how they relate to the conservation of momentum

9. How does the law of conservation of angular momentum explain the behavior of spinning tops?

10. How does the conservation of momentum apply to the motion of a pendulum?

Rotational Motion:

1. What is the difference between linear motion and rotational motion?

2. How do you calculate the rotational velocity of an object?

3. Explain the relationship between torque and rotational motion

4. How does the concept of angular acceleration relate to rotational motion?

5. What is the difference between rotational inertia and moment of inertia?

6. How does angular momentum relate to rotational motion?

7. Explain the concept of centripetal force in relation to rotational motion

8. How does the principle of conservation of angular momentum apply to rotational motion?

9. What is the difference between rotational kinetic energy and linear kinetic energy?

10. How does the concept of rotational equilibrium apply to objects at rest or in motion?

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