physics practice problems on drag force

Aerodynamic Drag

Practice problem 1.

Determine the drag coefficient of a 75 kg skydiver with a projected area of 0.33 m 2 and a terminal velocity of 60 m/s.
By how much would the skydiver need to reduce her project area so as to double her terminal velocity? How would she accomplish this?

Terminal velocity for a falling object occurs when the drag on the object equals its weight.

Solve for projected area, substitute values, and compute. (The density of air is in this book somewhere .)

This agrees with the range of values stated in the table on the discussion page of this topic.

We still start with the principle that drag equals weight.

But this time we'll solve for the terminal velocity instead of the drag coefficient.

Don't plug in any numbers, just look at the way terminal velocity is related to projected area. Projected area is in the denominator, under a radical sign. That means terminal velocity is inversely proportional to the square root of projected area. That means the skydiver would have to reduce her projected area to one-quarter of its original value.

The skydiver can do this by changing her orientation from horizontal to vertical — from spread eagle to head first.

practice problem 2

Table 1 time (s) distance (m) 1 00 4.5472 2 0 17.5392 3 0 38.0016 4 0 64.96 00 5 0 97.44 00 6 134.4672 7 175.0672 8 218.2656 9 263.088 0 10 308.56 00 11 353.7072 Voici des Tables faites sur cette hypothèse, par lesquelles on connoîtra combien une balle de plomb de six lignes de diamètre passera de pieds en chaque seconde en descendant; combien elle en passera dans tel nombre de secondes qu'on voudra choisir; quand elle cessera d'accélérer son mouvement; quelle sera sa vites se complette; & combien elle parcourra de pieds avant que de l'acquérir. These tables, made by applying this hypothesis, show how many feet a lead ball six lines [1.3536 cm] in diameter will fall in each second; how many feet it will fall in any number of seconds we choose; when and where it stops accelerating; what will be its final speed; & how many feet it will cover before acquiring it. Adapted from Edme Mariotte, 1673

Once this is graphed, you can kind of see what's going on. The last five data points line up pretty nearly perfectly. Make a line of best fit using those last five points. This part of the problem can be solved in a few different ways. I chose this method since it gives a reasonable answer with a minimal amount of effort.

Since speed is the rate of change of distance with time, the slope of the line added is the terminal velocity.

v mariotte = 44.76 m/s

This part of the problem can also be solved in a few different ways. Let's use as much algebra with symbols as we can. I chose this method since it involves the fewest number of calculations.

At terminal velocity, drag equals weight.

Replace those two symbols with appropriate contemporary equations.

1 2 ρ CAv 2 = mg

Add subscripts to identify the two very different materials.

1 2 ρ air CAv 2 = m lead g

Relate the mass of the lead sphere to its density

m lead = ρ lead V

Recall the equations for the projected area of a sphere (the area of a circle) and the volume of a sphere.

Now do a triple substitution.

1 2 ρ air C π r 2 v 2 = ρ lead 4 3 π r 3 g

Simplify a bit.

1 2 ρ air Cv 2 = ρ lead 4 3 rg

Solve for speed.

List the quantities with numbers and units.

Put them into the equation.

Compute the answer.

v modern = 57.67 m/s

The two answers we got are of the same order of magnitude — the minimum amount one needs to be considered "right" in physics.

44.76 m/s ≈ 57.67 m/s

I thought I might have more to say on this, but I don't. Given the state of development of fluid mechanics at the time (it was just being invented) I'd say Mariotte basically got it right . Your answer is allowed to differ only slightly from mine — let's say, to within an order of magnitude.

practice problem 3

Start with Newton's second law of motion. Identify the relevant forces. Weight ( W ) pulls the skydiver down. Drag ( R ) pushes her back up. We'll make down be the positive direction, since that's where the skydiver's headed.

Acceleration is the rate of change of velocity with time.

This is a first order differential equation — equation because there's an equals sign, differential because it contains two infinitesimals ( dt and dv ), and first order because the infinitesimals are not raised to anything higher than the first power. This kind of equation is best solved by separation of variables. Place all the time terms on one side (a rather lonely side) and all the velocity terms on the other side (a somewhat crowded side).

Integrate both sides. Time starts at zero and ends at some later time. Velocity starts at zero and ends at some later velocity. Note that the symbols t and v are doing double duty in the equation. They are acting as both variables and upper limits.

Complete the integral…

Evaluate over the limits…

Remember, we're trying to make velocity into a function of time. We need to get v out of the logarithm. Raise both sides of the equation into a power of e .

Finish things off with a little bit of algebra…

Let's check the limits of this equation to see if they make sense. This function returns the value v = 0 when t = 0 .

This agrees with our condition that the skydiver wasn't moving at first.

This function always increases, but it never quite reaches a final value. It approaches v = mg / b as we get closer and closer to t = ∞ .

This is our terminal velocity. We'd get the same thing if we set drag equal to weight and solved algebraically for speed.

One last test. What happens to our function if we let b = 0 ? What happens if we get rid of drag?

What is the limit of zero divided by zero? To answer that question, we'll use a little trick called L'Hôpital's rule — named for the French mathematician Guillaume de l'Hôpital (1661–1704). The ratio of two limits that approach zero is the same as the ratio of the limits of their first derivatives. (There's a bit more to this rule, but I'll leave it to your math teacher to fill in the details.)

For our problem, where the limiting variable is b , we'll let…

So now we need to take the limit of this instead…

When we get rid of drag, we get back the velocity-time equation for uniform acceleration — in other words, free fall…

The solution makes sense.

Repeat the previous approach using a drag that is proportional to speed squared.

Rearrange into a first order differential equation…

Separate the variables…

Integrate both sides.

Let me be honest here. I have no idea what to do next. I had to consult an expert . Apparently, the inetgral on the right works out to…

Make velocity the subject.

Have you ever seen anything that looks like this before? Did you notice the odd function? That is not a typo. I did not accidentally drop an "h" into "tan" (short for tangent). I really meant to write "tanh". Let me digress briefly.

The normal trig functions (sine, cosine, tangent, etc.) are ratios of the sides of a right triangle. A standard way to imagine this triangle is with one vertex at the center of a circle. In this configuration, the hypotenuse of our standard triangle is the radius of the circle in which the triangle is embedded. Take that radius and sweep it around the circumference. Watch the ratios evolve. This is the traditional way to graph them —as functions of an angle swept around a circle.

Well now, who says we have to do it this way? Who says we have to use a circle? Well now, I would if you asked me, since I tend to relate angles with fractions of a circle, but I'm just a physicist. Mathematicians on the other hand are some clever people. One of them actually had the temerity to respond to the question, "Why a circle?" with another, more interesting question, "Why not a hyperbola?" And the rest was history.

There are trig functions defined on a circle and trig functions defined on a hyperbola. The former are called sine (sin), cosine (cos), tangent (tan), and so on (cot, sec, and csc). The latter are called hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh), and hyperbolic so on (coth, sech, and csch). Here's what the mathematicians have determined these functions look like (where x is the "angle" measured in radians).

So what does that mean for us? Make the following replacement in the definition of tanh.

Look at the big beautiful pile of symbols we get.

Isn't that something? Astoundingly, this little mathematical diversion turns out to be useful. Hyperbolic trig functions are more than just an intellectual challenge. They wind up being the solution to some real world problems.

Let's not test the limits of this equation. There are so many terms to keep track of. It wouldn't be much fun. Instead, let's taste and compare. Graph the two solutions side by side.

Both solutions are exponential at their core. The speed squared model approaches its terminal velocity faster than the directly proportional model. This makes sense, since squaring the speed makes the drag increase more quickly. Whether the terminal velocities would differ is another matter. The graphs above assume that both models generate the same constant of proportionality. Since these are entirely hypothetical models, no real comparison can be made.

practice problem 4

My favorite data analysis software gave me the following coefficients when I graphed P vs. v max and did a power curve fit.

It looks like aerodynamic drag for cars is proportional to the square of speed. Bernoulli's equation tells us that drag is proportional to the square of speed and I see a power that's approximately 2 in the curve fit above. Isn't life grand when everything behaves exactly as you expect it to.

Halt! Proceed no further with this logic. It is wrong, wrong, wrong, wrong, wrong, wrong, wrong. The graph above shows the relationship between power and speed, not drag force and speed. The question now is how to make the jump from what we have to what we need.

Begin with the definition of work and play around with it a bit.

Replace the nonspecific force F with power law form of aerodynamic drag.

P = Fv = bv n v = bv n + 1

The curve fit that I did gave me one greater than the power in the general relation. This means that the aerodynamic drag on automobiles is proportional to speed not speed squared. Bernoulli's law does not apply for some reason. I don't know how to reconcile this discrepancy. I suspect that automotive engineers use…

for drag calculations, not the computationally simple, but physically unrealistic…

but then, I don't know any automotive engineers.

Learning Objectives

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

Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force F D F D is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as F D ∝ v 2 F D ∝ v 2 . When taking into account other factors, this relationship becomes

where C C is the drag coefficient, A A is the area of the object facing the fluid, and ρ ρ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as F D = bv 2 F D = bv 2 , where b b is a constant equivalent to 0 .5 CρA 0 .5 CρA . We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.

Drag force F D F D is found to be proportional to the square of the speed of the object. Mathematically

where C C is the drag coefficient, A A is the area of the object facing the fluid, and ρ ρ is the density of the fluid.

Athletes as well as car designers seek to reduce the drag force to lower their race times. (See Figure 5.7 ). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.

A two-person team in a bobsled race. The bobsled has an aerodynamic design and smooth runners so it can go as fast as possible.

The value of the drag coefficient, C C , is determined empirically, usually with the use of a wind tunnel. (See Figure 5.8 ).

A model plane is can be seen being tested in a wind tunnel.

The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table 5.2 lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).

Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (See Figure 5.9 ). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.

Three swimmers with are each wearing an L Z R Racer Suit, which is a swimsuit composed of elastane nylon and polyurethane. The seams of the suit are ultrasonically welded to reduce drag.

Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as given by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity ( v t v t ). Since F D F D is proportional to the speed, a heavier skydiver must go faster for F D F D to equal his weight. Let’s see how this works out more quantitatively.

At the terminal velocity,

Using the equation for drag force, we have

Solving for the velocity, we obtain

Assume the density of air is ρ = 1 . 21 kg /m 3 ρ = 1 . 21 kg /m 3 . A 75-kg skydiver descending head first will have an area approximately A = 0 . 18 m 2 A = 0 . 18 m 2 and a drag coefficient of approximately C = 0 . 70 C = 0 . 70 . We find that

This means a skydiver with a mass of 75 kg achieves a maximum terminal velocity of about 350 km/h while traveling in a headfirst position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.

Take-Home Experiment

This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity v v versus mass. Also plot v 2 v 2 versus mass. Which of these relationships is more linear? What can you conclude from these graphs?

Example 5.2

A terminal velocity.

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

At terminal velocity, F net = 0 F net = 0 . Thus the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find mg = 1 2 ρCAv 2 mg = 1 2 ρCAv 2 .

Thus the terminal velocity v t v t can be written as

All quantities are known except the person’s projected area. This is an adult (85 kg) falling spread eagle. We can estimate the frontal area as

Using our equation for v t v t , we find that

This result is consistent with the value for v t v t mentioned earlier. The 75-kg skydiver going feet first had a v = 98 m / s v = 98 m / s . He weighed less but had a smaller frontal area and so a smaller drag due to the air.

The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled “On Being the Right Size.”

To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law , which states that

where r r is the radius of the object, η η is the viscosity of the fluid, and v v is the object’s velocity.

Stokes’ Law

Good examples of this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about 1 μm 1 μm ) can be about 2 μm/s 2 μm/s . To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell. Sediment in a lake can move at a greater terminal velocity (about 5 μm/s 5 μm/s ), so it can take days to reach the bottom of the lake after being deposited on the surface.

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fishes, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a streamlined pattern (see Figure 5.10 ). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.

Geese flying across the sky in a V formation.

Galileo’s Experiment

Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each to reach the ground. Since stopwatches weren’t readily available, how do you think he measured their fall time? If the objects were the same size, but with different masses, what do you think he should have observed? Would this result be different if done on the Moon?

As an Amazon Associate we earn from qualifying purchases.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

© Jan 11, 2023 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

HIGH SCHOOL

GRADUATE SCHOOL

Search 50+ Tests

math tutoring

science tutoring

Foreign languages.

Mandarin Chinese

elementary tutoring

Computer Science

Search 350+ Subjects

AP Physics C: Mechanics : Forces

Study concepts, example questions & explanations for ap physics c: mechanics, all ap physics c: mechanics resources, example questions, example question #21 : forces.

$physics practice problems on drag force$

To find terminal velocity, set the magnitude of the drag force equal to the magnitude of the force of gravity since when these forces are equal and opposite, the object will stop accelerating:

$physics practice problems on drag force$

Example Question #22 : Forces

$physics practice problems on drag force$

Example Question #23 : Forces

A spherical asteroid has a hole drilled through the center as diagrammed below:

Hole through planet

Refer to the diagram above. An object that is much smaller than the asteroid is released from rest at the surface of the asteroid, at point a. How do the velocity and acceleration of the object compare at point b at the surface, and point a, located at the center of the asteroid?

$physics practice problems on drag force$

Because the gravitational force depends only on the mass beneath the object (which is the gravitational version of Gauss's Law for charge), the acceleration steadily decreases as the object falls, and drops to zero at the center. Nevertheless, the velocity keeps increasing as the object falls, it just does so more slowly.

With what minimum velocity must a rocket be launched from the surface of the moon in order to not fall back down due to the moon's gravity?

$physics practice problems on drag force$

Relevant equations:

$physics practice problems on drag force$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away.

For the minimum energy case as the rocket leaves the surface:

$physics practice problems on drag force$

Rearrange energy equation to isolate the velocity term.

$physics practice problems on drag force$

Substitute in the given values to solve for the velocity.

$physics practice problems on drag force$

Example Question #2 : Calculating Gravitational Forces

$physics practice problems on drag force$

Use the given values to solve for the force.

$physics practice problems on drag force$

Example Question #3 : Calculating Gravitational Forces

$physics practice problems on drag force$

Newton's law of universal gravitation states:

$physics practice problems on drag force$

We can write two equations for the gravity experienced before and after the doubling:

$physics practice problems on drag force$

The equation for gravity after the doubling can be simplified:

$physics practice problems on drag force$

Because the masses of the spheres remain the same, as does the universal gravitation constant, we can substitute the definition of Fg1 into that equation:

$physics practice problems on drag force$

The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.

Example Question #4 : Calculating Gravitational Forces

Two spheres of equal mass are isolated in space. If the mass of one sphere is doubled, by what factor does the gravitational force experienced by the two spheres change?

$physics practice problems on drag force$

Newton's law of universal gravitation states that:

$physics practice problems on drag force$

We can write two equations representing the force of gravity before and after the doubling of the mass:

$physics practice problems on drag force$

Substituting these defintions into the second equation:

$physics practice problems on drag force$

This equation simplifies to:

$physics practice problems on drag force$

Substituting the definition of Fg1, we see:

$physics practice problems on drag force$

Thus the gravitational forces doubles when the mass of one object doubles.

Example Question #5 : Calculating Gravitational Forces

$physics practice problems on drag force$

What does the spring scale read when the elevator is descending at constant speed?

$physics practice problems on drag force$

Since the elevator is descending at constant speed, no additional force is applied, therefore the force that the spring scale reads is only due to gravity, which is calculated by:

$physics practice problems on drag force$

Example Question #6 : Calculating Gravitational Forces

$physics practice problems on drag force$

With what minimum speed would a bullet have to be fired horizontally near the surface of this moon in order for it to never hit the ground?

(Note: You can treat the moon as a smooth sphere, and assume there’s no atmosphere.)

$physics practice problems on drag force$

To do this problem we have to realize that the force of gravity acting on the bullet is equal to the centripetal force. The equations for gravitational force and centripetal force are as follows:

$physics practice problems on drag force$

If we plug everything in, we get

$physics practice problems on drag force$

Example Question #24 : Forces

The force on an object due to gravity on the moon is one-sixth of that found on Earth. What is the acceleration due to gravity on the moon?

$physics practice problems on drag force$

We can use Newton's second law:

$physics practice problems on drag force$

Set up equations for the force on the moon and the force on Earth:

$physics practice problems on drag force$

Now we can use substitution:

$physics practice problems on drag force$

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

Learning Tools by Varsity Tutors

Find the Best Tutors

6 Applications of Newton’s Laws

6.4 drag force and terminal speed, learning objectives.

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

Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion.

Drag Forces

Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as cyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force [latex] {F}_{\text{D}} [/latex] is proportional to the square of the speed of the object. We can write this relationship mathematically as [latex] {F}_{\text{D}}\propto {v}^{2}. [/latex] When taking into account other factors, this relationship becomes

where C is the drag coefficient, A is the area of the object facing the fluid, and [latex] \rho [/latex] is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as [latex] {F}_{\text{D}}=b{v}^{2}, [/latex] where b is a constant equivalent to [latex] 0.5C\rho A. [/latex] We have set the exponent n for these equations as 2 because when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in Fluid Mechanics , for small particles moving at low speeds in a fluid, the exponent n is equal to 1.

Drag force [latex] {F}_{\text{D}} [/latex] is proportional to the square of the speed of the object. Mathematically,

where C is the drag coefficient , A is the area of the object facing the fluid, and [latex] \rho [/latex] is the density of the fluid.

Athletes as well as car designers seek to reduce the drag force to lower their race times ( (Figure) ). Aerodynamic shaping of an automobile can reduce the drag force and thus increase a car’s gas mileage.

A photograph of a bobsled on a track at the Olympics.

Figure 6.29 From racing cars to bobsled racers, aerodynamic shaping is crucial to achieving top speeds. Bobsleds are designed for speed and are shaped like a bullet with tapered fins. (credit: “U.S. Army”/Wikimedia Commons)

The value of the drag coefficient C is determined empirically, usually with the use of a wind tunnel ( (Figure) ).

A photograph of a model plane in a wind tunnel.

Figure 6.30 NASA researchers test a model plane in a wind tunnel. (credit: NASA/Ames)

The drag coefficient can depend upon velocity, but we assume that it is a constant here. (Figure) lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over [latex] 50% [/latex] of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).

Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned, as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics and won a gold medal in the 400-m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records ( (Figure) ). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.

A photograph of three swimmers wearing body suits.

Figure 6.31 Body suits, such as this LZR Racer Suit, have been credited with aiding in many world records after their release in 2008. Smoother “skin” and more compression forces on a swimmer’s body provide at least [latex] 10% [/latex] less drag. (credit: NASA/Kathy Barnstorff)

Terminal Velocity

Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the small buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as shown by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity [latex] ({v}_{\text{T}}). [/latex] Since [latex] {F}_{\text{D}} [/latex] is proportional to the speed squared, a heavier skydiver must go faster for [latex] {F}_{\text{D}} [/latex] to equal his weight. Let’s see how this works out more quantitatively.

At the terminal velocity,

Using the equation for drag force, we have

Solving for the velocity, we obtain

Assume the density of air is [latex] \rho =1.21\,{\text{kg/m}}^{3}. [/latex] A 75-kg skydiver descending head first has a cross-sectional area of approximately [latex] A=0.18\,{\text{m}}^{2} [/latex] and a drag coefficient of approximately [latex] C=0.70 [/latex]. We find that

This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.

Terminal Velocity of a Skydiver

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

At terminal velocity, [latex] {F}_{\text{net}}=0. [/latex] Thus, the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find [latex] mg=\frac{1}{2}\rho CA{v}^{2}. [/latex]

The terminal velocity [latex] {v}_{\text{T}} [/latex] can be written as

Significance

This result is consistent with the value for [latex] {v}_{\text{T}} [/latex] mentioned earlier. The 75-kg skydiver going feet first had a terminal velocity of [latex] {v}_{\text{T}}=98\,\text{m/s}\text{.} [/latex] He weighed less but had a smaller frontal area and so a smaller drag due to the air.

Check Your Understanding

Find the terminal velocity of a 50-kg skydiver falling in spread-eagle fashion.

The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You do not reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J. B. S. Haldane, titled “On Being the Right Size.”

“To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.”

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law.

Stokes’ Law

For a spherical object falling in a medium, the drag force is

where r is the radius of the object, [latex] \eta [/latex] is the viscosity of the fluid, and v is the object’s velocity.

Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about [latex] 1\,\text{μm}) [/latex] can be about [latex] 2\,\text{μm/s}\text{.} [/latex] To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell.

Sediment in a lake can move at a greater terminal velocity (about [latex] 5\,\text{μm/s}), [/latex] so it can take days for it to reach the bottom of the lake after being deposited on the surface.

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fish, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spearhead as the flock forms a streamlined pattern ( (Figure) ). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.

A photograph of geese flying in a V formation.

Figure 6.32 Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and also allows them a better way to communicate. (credit: “Julo”/Wikimedia Commons)

In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.

The Calculus of Velocity-Dependent Frictional Forces

When a body slides across a surface, the frictional force on it is approximately constant and given by [latex] {\mu }_{\text{k}}N. [/latex] Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by

where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.

Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in (Figure) . Newton’s second law in the vertical direction gives the differential equation

where we have written the acceleration as [latex] dv\text{/}dt. [/latex] As v increases, the frictional force – bv increases until it matches mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity [latex] {v}_{\text{T}}. [/latex] From the previous equation,

The free body diagram shows forces m times vector g pointing vertically down and b times vector v pointing vertically up. The velocity, vector v, is vertically down. The positive y direction is also vertically down.

Figure 6.33 Free-body diagram of an object falling through a resistive medium.

We can find the object’s velocity by integrating the differential equation for v . First, we rearrange terms in this equation to obtain

Assuming that [latex] v=0\,\text{at}\,t=0, [/latex] integration of this equation yields

where [latex] v\text{‘}\,\text{and}\,t\text{‘} [/latex] are dummy variables of integration. With the limits given, we find

Since [latex] \text{ln}A-\text{ln}B=\text{ln}(A\text{/}B), [/latex] and [latex] \text{ln}(A\text{/}B)=x\,\text{implies}\,{e}^{x}=A\text{/}B, [/latex] we obtain

Notice that as [latex] t\to \infty ,v\to mg\text{/}b={v}_{\text{T}}, [/latex] which is the terminal velocity.

The position at any time may be found by integrating the equation for v . With [latex] v=dy\text{/}dt, [/latex]

Assuming [latex] y=0\,\text{when}\,t=0, [/latex]

which integrates to

Effect of the Resistive Force on a Motorboat

A motorboat is moving across a lake at a speed [latex] {v}_{0} [/latex] when its motor suddenly freezes up and stops. The boat then slows down under the frictional force [latex] {f}_{R}=\text{−}bv. [/latex] (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?

which we can write as

Integrating this equation between the time zero when the velocity is [latex] {v}_{0} [/latex] and the time t when the velocity is [latex] v [/latex], we have

which, since [latex] \text{ln}A=x\,\text{implies}\,{e}^{x}=A, [/latex] we can write this as

Now from the definition of velocity,

With the initial position zero, we have

As time increases, [latex] {e}^{\text{−}bt\text{/}m}\to 0, [/latex] and the position of the boat approaches a limiting value

Although this tells us that the boat takes an infinite amount of time to reach [latex] {x}_{\text{max}}, [/latex] the boat effectively stops after a reasonable time. For example, at [latex] t=10m\text{/}b, [/latex] we have

whereas we also have

Now the boat’s limiting position is

In the both of the previous examples, we found “limiting” values. The terminal velocity is the same as the limiting velocity, which is the velocity of the falling object after a (relatively) long time has passed. Similarly, the limiting distance of the boat is the distance the boat will travel after a long amount of time has passed. Due to the properties of exponential decay, the time involved to reach either of these values is actually not too long (certainly not an infinite amount of time!) but they are quickly found by taking the limit to infinity.

Suppose the resistive force of the air on a skydiver can be approximated by [latex] f=\text{−}b{v}^{2} [/latex]. If the terminal velocity of a 100-kg skydiver is 60 m/s, what is the value of b?

Key Equations

Conceptual questions.

Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits.

The pros of wearing body suits include: (1) the body suit reduces the drag force on the swimmer and the athlete can move more easily; (2) the tightness of the suit reduces the surface area of the athlete, and even though this is a small amount, it can make a difference in performance time. The cons of wearing body suits are: (1) The tightness of the suits can induce cramping and breathing problems. (2) Heat will be retained and thus the athlete could overheat during a long period of use.

Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one?

As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference?

The oil is less dense than the water and so rises to the top when a light rain falls and collects on the road. This creates a dangerous situation in which friction is greatly lowered, and so a car can lose control. In a heavy rain, the oil is dispersed and does not affect the motion of cars as much.

Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of [latex] 0.140\,{\text{m}}^{2} [/latex].

115 m/s or 414 km/h

A 60.0-kg and a 90.0-kg skydiver jump from an airplane at an altitude of [latex] 6.00\,×\,{10}^{3}\text{m} [/latex], both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits.

A 560-g squirrel with a surface area of [latex] 930\,{\text{cm}}^{2} [/latex] falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance?

To maintain a constant speed, the force provided by a car’s engine must equal the drag force plus the force of friction of the road (the rolling resistance). (a) What are the drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is [latex] 0.70\,{\text{m}}^{2} [/latex]) (b) What is the drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is [latex] 2.44\,{\text{m}}^{2}) [/latex] Assume all values are accurate to three significant digits.

By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?

[latex] {(\frac{110}{65})}^{2}=2.86 [/latex] times

Calculate the velocity a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be [latex] 1.00\,×\,{10}^{3}\,{\text{kg/m}}^{3} [/latex], and the surface area to be [latex] \pi {r}^{2} [/latex].

Using Stokes’ law, verify that the units for viscosity are kilograms per meter per second.

Stokes’ law is [latex] {F}_{\text{s}}=6\pi r\eta v. [/latex] Solving for the viscosity, [latex] \eta =\frac{{F}_{\text{s}}}{6\pi rv}. [/latex] Considering only the units, this becomes [latex] [\eta ]=\frac{\text{kg}}{\text{m}·\text{s}}. [/latex]

Find the terminal velocity of a spherical bacterium (diameter [latex] 2.00\,\text{μm} [/latex]) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be [latex] 1.10\,×\,{10}^{3}\,{\text{kg/m}}^{3} [/latex].

Stokes’ law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density [latex] 7.8\,×\,{10}^{3}\,{\text{kg/m}}^{3} [/latex], diameter 3.0 mm) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.

[latex] 0.76\,\text{kg/m}·\text{s} [/latex]

Suppose that the resistive force of the air on a skydiver can be approximated by [latex] f=\text{−}b{v}^{2}. [/latex] If the terminal velocity of a 50.0-kg skydiver is 60.0 m/s, what is the value of b ?

A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. (a) Assuming the frictional force on the diamond obeys [latex] f=\text{−}bv, [/latex] what is b ? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?

a. 0.049 kg/s; b. 0.57 m

(a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of [latex] 0.400\,{\text{m/s}}^{2} [/latex] for 50.0 s? Assume a coefficient of friction of 1.0. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

A 75.0-kg woman stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with her weight. (The scale exerts an upward force on her equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

a. 1860 N, 2.53; b. The value (1860 N) is more force than you expect to experience on an elevator. The force of 1860 N is 418 pounds, compared to the force on a typical elevator of 904 N (which is about 203 pounds); this is calculated for a speed from 0 to 10 miles per hour, which is about 4.5 m/s, in 2.00 s). c. The acceleration [latex] a=1.53\,×\,g [/latex] is much higher than any standard elevator. The final speed is too large (30.0 m/s is VERY fast)! The time of 2.00 s is not unreasonable for an elevator.

(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s. (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

As shown below, if [latex] M=5.50\,\text{kg,} [/latex] what is the tension in string 1?

Mass M is suspended from strings 1 and 2. String 1 connects to a wall at a point below and to the left of the mass. String 1 makes an angle of 40 degrees below the horizontal. String 2 connects to a ceiling at a point above and to the right of the mass. String 2 makes an angle of 40 degrees to the right of vertical.

As shown below, if [latex] F=60.0\,\text{N} [/latex] and [latex] M=4.00\,\text{kg,} [/latex] what is the magnitude of the acceleration of the suspended object? All surfaces are frictionless.

Two blocks are shown. One block, labeled 2 M is on a horizontal table. A force F pulls on the 2 M block up and to the left at an angle of 30 degrees above the horizontal. On the opposite side, the block is connected to a string that pulls it to the right. The string passes over a pulley at edge of the table, then hangs straight down and connects to the second block, labeled M. Block 2 is not in contact with the ramp.

As shown below, if [latex] M=6.0\,\text{kg,} [/latex] what is the tension in the connecting string? The pulley and all surfaces are frictionless.

Two blocks, both mass M are connected by a string that passes over a pulley between the blocks. The upper block is on a surface that slopes down and to the right at an angle of 30 degrees to the horizontal. The pulley is attached to the corner at the bottom of the slope, where the surface then bends and goes vertically down. The lower mass hangs straight down. It is not in contact with the surface.

A small space probe is released from a spaceship. The space probe has mass 20.0 kg and contains 90.0 kg of fuel. It starts from rest in deep space, from the origin of a coordinate system based on the spaceship, and burns fuel at the rate of 3.00 kg/s. The engine provides a constant thrust of 120.0 N. (a) Write an expression for the mass of the space probe as a function of time, between 0 and 30 seconds, assuming that the engine ignites fuel beginning at [latex] t=0. [/latex] (b) What is the velocity after 15.0 s? (c) What is the position of the space probe after 15.0 s, with initial position at the origin? (d) Write an expression for the position as a function of time, for [latex] t>30.0\,\text{s}\text{.} [/latex]

A half-full recycling bin has mass 3.0 kg and is pushed up a [latex] 40.0\text{°} [/latex] incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?

A child has mass 6.0 kg and slides down a [latex] 35\text{°} [/latex] incline with constant speed under the action of a 34-N force acting up and parallel to the incline. What is the coefficient of kinetic friction between the child and the surface of the incline?

Additional Problems

The two barges shown here are coupled by a cable of negligible mass. The mass of the front barge is [latex] 2.00\,×\,{10}^{3}\,\text{kg} [/latex] and the mass of the rear barge is [latex] 3.00\,×\,{10}^{3}\,\text{kg}\text{.} [/latex] A tugboat pulls the front barge with a horizontal force of magnitude [latex] 20.0\,×\,{10}^{3}\,\text{N,} [/latex] and the frictional forces of the water on the front and rear barges are [latex] 8.00\,×\,{10}^{3}\,\text{N} [/latex] and [latex] 10.0\,×\,{10}^{3}\,\text{N,} [/latex] respectively. Find the horizontal acceleration of the barges and the tension in the connecting cable.

An illustration showing a tug boat pulling two barges. The barge directly attached to the tug boat has mass 2.00 times 10 to the third kilograms. The barge at the end, behind the first barge, has mass 3.00 times 10 to the third kilograms.

If the order of the barges of the preceding exercise is reversed so that the tugboat pulls the [latex] 3.00\,×\,{10}^{3}\text{-kg} [/latex] barge with a force of [latex] 20.0\,×\,{10}^{3}\,\text{N}, [/latex] what are the acceleration of the barges and the tension in the coupling cable?

An object with mass m moves along the x -axis. Its position at any time is given by [latex] x(t)=p{t}^{3}+q{t}^{2} [/latex] where p and q are constants. Find the net force on this object for any time t .

m (6 pt + 2 q )

A helicopter with mass [latex] 2.35\,×\,{10}^{4}\,\text{kg} [/latex] has a position given by [latex] \overset{\to }{r}(t)=(0.020\,{t}^{3})\hat{i}+(2.2t)\hat{j}-(0.060\,{t}^{2})\hat{k}. [/latex] Find the net force on the helicopter at [latex] t=3.0\,\text{s}\text{.} [/latex]

Located at the origin, an electric car of mass m is at rest and in equilibrium. A time dependent force of [latex] \overset{\to }{F}(t) [/latex] is applied at time [latex] t=0 [/latex], and its components are [latex] {F}_{x}(t)=p+nt [/latex] and [latex] {F}_{y}(t)=qt [/latex] where p , q , and n are constants. Find the position [latex] \overset{\to }{r}(t) [/latex] and velocity [latex] \overset{\to }{v}(t) [/latex] as functions of time t .

[latex] \overset{\to }{v}(t)=(\frac{pt}{m}+\frac{n{t}^{2}}{2m})\hat{i}+(\frac{q{t}^{2}}{2})\hat{j} [/latex] and [latex] \overset{\to }{r}(t)=(\frac{p{t}^{2}}{2m}+\frac{n{t}^{3}}{6m})\hat{i}+(\frac{q{t}^{3}}{60m})\hat{j} [/latex]

A particle of mass m is located at the origin. It is at rest and in equilibrium. A time-dependent force of [latex] \overset{\to }{F}(t) [/latex] is applied at time [latex] t=0 [/latex], and its components are [latex] {F}_{x}(t)=pt [/latex] and [latex] {F}_{y}(t)=n+qt [/latex] where p , q , and n are constants. Find the position [latex] \overset{\to }{r}(t) [/latex] and velocity [latex] \overset{\to }{v}(t) [/latex] as functions of time t .

A 2.0-kg object has a velocity of [latex] 4.0\hat{i}\,\text{m/s} [/latex] at [latex] t=0. [/latex] A constant resultant force of [latex] (2.0\hat{i}+4.0\hat{j})\,\text{N} [/latex] then acts on the object for 3.0 s. What is the magnitude of the object’s velocity at the end of the 3.0-s interval?

A 1.5-kg mass has an acceleration of [latex] (4.0\hat{i}-3.0\hat{j})\,{\text{m/s}}^{2}. [/latex] Only two forces act on the mass. If one of the forces is [latex] (2.0\hat{i}-1.4\hat{j})\,\text{N,} [/latex] what is the magnitude of the other force?

A box is dropped onto a conveyor belt moving at 3.4 m/s. If the coefficient of friction between the box and the belt is 0.27, how long will it take before the box moves without slipping?

Shown below is a 10.0-kg block being pushed by a horizontal force [latex] \overset{\to }{F} [/latex] of magnitude 200.0 N. The coefficient of kinetic friction between the two surfaces is 0.50. Find the acceleration of the block.

An illustration of a 10.0 kilogram block being pushed into a slope by a horizontal force F. The slope angles up and to the right at an angle of 30 degrees to the horizontal and the force F points to the right.

As shown below, the mass of block 1 is [latex] {m}_{1}=4.0\,\text{kg,} [/latex] while the mass of block 2 is [latex] {m}_{2}=8.0\,\text{kg}\text{.} [/latex] The coefficient of friction between [latex] {m}_{1} [/latex] and the inclined surface is [latex] {\mu }_{\text{k}}=0.40. [/latex] What is the acceleration of the system?

Block 1 is on a ramp inclined up and to the right at an angle of 37 degrees above the horizontal. It is connected to a string that passes over a pulley at the top of the ramp, then hangs straight down and connects to block 2. Block 2 is not in contact with the ramp.

A student is attempting to move a 30-kg mini-fridge into her dorm room. During a moment of inattention, the mini-fridge slides down a 35 degree incline at constant speed when she applies a force of 25 N acting up and parallel to the incline. What is the coefficient of kinetic friction between the fridge and the surface of the incline?

A crate of mass 100.0 kg rests on a rough surface inclined at an angle of [latex] 37.0\text{°} [/latex] with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is [latex] 80% [/latex] of that for the static case. (a) What is the coefficient of static friction? (b) What is the maximum force that can be applied upward along the plane on the rope and not move the block? (c) With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? (d) If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? (e) Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?

a. 0.60; b. 1200 N; c. [latex] 1.2\,{\text{m/s}}^{2} [/latex] and 1080 N; d. [latex] -1.2\,{\text{m/s}}^{2}; [/latex] e. 120 N

A car is moving at high speed along a highway when the driver makes an emergency braking. The wheels become locked (stop rolling), and the resulting skid marks are 32.0 meters long. If the coefficient of kinetic friction between tires and road is 0.550, and the acceleration was constant during braking, how fast was the car going when the wheels became locked?

A crate having mass 50.0 kg falls horizontally off the back of the flatbed truck, which is traveling at 100 km/h. Find the value of the coefficient of kinetic friction between the road and crate if the crate slides 50 m on the road in coming to rest. The initial speed of the crate is the same as the truck, 100 km/h.

The figure shows a truck moving to the right at 100 kilometers per hour and a 50 kilogram crate on the ground behind the truck.

A 15-kg sled is pulled across a horizontal, snow-covered surface by a force applied to a rope at 30 degrees with the horizontal. The coefficient of kinetic friction between the sled and the snow is 0.20. (a) If the force is 33 N, what is the horizontal acceleration of the sled? (b) What must the force be in order to pull the sled at constant velocity?

A 30.0-g ball at the end of a string is swung in a vertical circle with a radius of 25.0 cm. The rotational velocity is 200.0 cm/s. Find the tension in the string: (a) at the top of the circle, (b) at the bottom of the circle, and (c) at a distance of 12.5 cm from the center of the circle [latex] (r=12.5\,\text{cm}). [/latex]

a. 0.186 N; b. 774 N; c. 0.48 N

A particle of mass 0.50 kg starts moves through a circular path in the xy -plane with a position given by [latex] \overset{\to }{r}(t)=(4.0\,\text{cos}\,3t)\hat{i}+(4.0\,\text{sin}\,3t)\hat{j} [/latex] where r is in meters and t is in seconds. (a) Find the velocity and acceleration vectors as functions of time. (b) Show that the acceleration vector always points toward the center of the circle (and thus represents centripetal acceleration). (c) Find the centripetal force vector as a function of time.

A stunt cyclist rides on the interior of a cylinder 12 m in radius. The coefficient of static friction between the tires and the wall is 0.68. Find the value of the minimum speed for the cyclist to perform the stunt.

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0 cm. The body and spring are placed on a horizontal frictionless surface and rotated about the held end of the spring at 2.0 rev/s. How far is the spring stretched?

Railroad tracks follow a circular curve of radius 500.0 m and are banked at an angle of [latex] 5.00\text{°} [/latex]. For trains of what speed are these tracks designed?

A plumb bob hangs from the roof of a railroad car. The car rounds a circular track of radius 300.0 m at a speed of 90.0 km/h. At what angle relative to the vertical does the plumb bob hang?

An airplane flies at 120.0 m/s and banks at a [latex] 30\text{°} [/latex] angle. If its mass is [latex] 2.50\,×\,{10}^{3}\,\text{kg,} [/latex] (a) what is the magnitude of the lift force? (b) what is the radius of the turn?

a. 28,300 N; b. 2540 m

The position of a particle is given by [latex] \overset{\to }{r}(t)=A(\text{cos}\,\omega t\hat{i}+\text{sin}\,\omega t\hat{j}), [/latex] where [latex] \omega [/latex] is a constant. (a) Show that the particle moves in a circle of radius A . (b) Calculate [latex] d\overset{\to }{r}\text{/}dt [/latex] and then show that the speed of the particle is a constant [latex] {A}_{\omega }. [/latex] (c) Determine [latex] {d}^{2}\overset{\to }{r}\text{/}d{t}^{2} [/latex] and show that a is given by[latex] {a}_{\text{c}}=r{\omega }^{2}. [/latex] (d) Calculate the centripetal force on the particle. [ Hint : For (b) and (c), you will need to use [latex] (d\text{/}dt)(\text{cos}\,\omega t)=\text{−}\omega \,\text{sin}\,\omega t [/latex] and [latex] (d\text{/}dt)(\text{sin}\,\omega t)=\omega \,\text{cos}\,\omega t. [/latex]

Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of [latex] 2.0\,{\text{m/s}}^{2} [/latex] to the right, what is the magnitude F of the applied force?

Two blocks, 1.0 kilograms on the left and 3.0 kilograms on the right, are connected by a string and are on a horizontal surface. Force F acts on the 3.0 kilogram mass and points up and to the right at a angle of 60 degrees above the horizontal.

As shown below, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If [latex] F=10\,\text{N} [/latex] and [latex] M=1.0\,\text{kg} [/latex], what is the tension in the connecting string?

Two blocks, 2 M on the left and M on the right, are connected by a string and are on a horizontal surface. Force F acts on M and points to the right.

In the figure, the coefficient of kinetic friction between the surface and the blocks is [latex] {\mu }_{\text{k}}. [/latex] If [latex] M=1.0\,\text{kg,} [/latex] find an expression for the magnitude of the acceleration of either block (in terms of F , [latex] {\mu }_{\text{k}}, [/latex] and g ).

Two blocks, M on the left and 3 M on the right, are connected by a string and are on a horizontal surface. The following forces are indicated: f sub k 2 acting on M and pointing to the right, f sub k 1 acting on 3 M and pointing to the right, F acting on 3 M and pointing to the left, N sub 2 acting on M and pointing up, N sub 1 acting on 3 M and pointing up, M g acting on M and pointing down, , 3 M g acting on 3 M and pointing down.

Two blocks are stacked as shown below, and rest on a frictionless surface. There is friction between the two blocks (coefficient of friction [latex] \mu [/latex]). An external force is applied to the top block at an angle [latex] \theta [/latex] with the horizontal. What is the maximum force F that can be applied for the two blocks to move together?

Rectangular block M sub 2 is on a horizontal surface. Rectangular block M sub 1 is on top of block M sub 2. A force F pushes on block M sub 1. Force F is directed down and to the right, at a angle theta to the horizontal.

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?

A double-incline plane is shown below. The coefficient of friction on the left surface is 0.30, and on the right surface 0.16. Calculate the acceleration of the system.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.

Challenge Problems

In a later chapter, you will find that the weight of a particle varies with altitude such that [latex] w=\frac{mg{r}_{0}{}^{2}}{{r}^{2}} [/latex] where [latex] {r}_{0}{}^{} [/latex] is the radius of Earth and r is the distance from Earth’s center. If the particle is fired vertically with velocity [latex] {v}_{0}{}^{} [/latex] from Earth’s surface, determine its velocity as a function of position r . ( Hint: use [latex] {a}^{}dr={v}^{}dv, [/latex] the rearrangement mentioned in the text.)

[latex] v=\sqrt{{v}_{0}{}^{2}-2g{r}_{0}(1-\frac{{r}_{0}}{r})} [/latex]

A large centrifuge, like the one shown below, is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in the bottom accompanying figure. At what angle [latex] \theta [/latex] below the horizontal will the cage hang when the centripetal acceleration is 10 g ? ( Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free-body diagram of the forces to see what the angle [latex] \theta [/latex] should be.)

(a) A photograph of a high g training centrifuge. The astronaut sits in a cage at the end of a long arm that rotates in a horizontal plane. (b) An illustration of a top view of the centrifuge along with an illustration of the forces. The free body diagram shows the weight, w, pointing vertically down and the force F sub arm pointing up and to the left. The forces are then shown rearranged to form a right triangle. F sub arm is the hypotenuse of the triangle pointing up and left, w is the vertical side pointing down, and F sub c is the base pointing to the left. The F sub c arrow is then shown separately with the notation that vector F sub c equals F sub net.

A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. ( Hint: since the distance traveled is of interest rather than the time, x is the desired independent variable and not t . Use the Chain Rule to change the variable: [latex] \frac{dv}{dt}=\frac{dv}{dx}\,\frac{dx}{dt}=v\frac{dv}{dx}.) [/latex]

An airplane flying at 200.0 m/s makes a turn that takes 4.0 min. What bank angle is required? What is the percentage increase in the perceived weight of the passengers?

A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, [latex] {F}_{\text{D}} [/latex], is given by the formula [latex] {F}_{\text{D}}=\text{−}bv, [/latex] where b is a constant and v is the velocity. If [latex] b=0.750, [/latex] and the mass of the skydiver is 82.0 kg, first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.

a. 53.9 m/s; b. 328 m; c. 4.58 m/s; d. 257 s

In a television commercial, a small, spherical bead of mass 4.00 g is released from rest at [latex] t=0 [/latex] in a bottle of liquid shampoo. The terminal speed is observed to be 2.00 cm/s. Find (a) the value of the constant b in the equation [latex] v=\frac{mg}{b}(1-{e}^{\text{−}bt\text{/}m}), [/latex] and (b) the value of the resistive force when the bead reaches terminal speed.

A boater and motor boat are at rest on a lake. Together, they have mass 200.0 kg. If the thrust of the motor is a constant force of 40.0 N in the direction of motion, and if the resistive force of the water is numerically equivalent to 2 times the speed v of the boat, set up and solve the differential equation to find: (a) the velocity of the boat at time t ; (b) the limiting velocity (the velocity after a long time has passed).

a. [latex] v=20.0(1-{e}^{-0.01t}); [/latex] b. [latex] {v}_{\text{limiting}}=20\,\text{m/s} [/latex]

OpenStax University Physics. Authored by : OpenStax CNX. Located at : https://cnx.org/contents/[email protected]:[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

Diagram showing example of projected area used in drag force equation

Search Website

Real World Applications — for high school level and above

Education & Theory — for high school level and above

Kids Section

© Copyright 2009-2023 real-world-physics-problems.com

Library homepage

selected template will load here

This action is not available.

6.7: Drag Force and Terminal Speed

Learning Objectives

Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion.

Drag Forces

Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as cyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force \(F_D\) is proportional to the square of the speed of the object. We can write this relationship mathematically as \(F_D \propto v^2\). When taking into account other factors, this relationship becomes

\[F_{D} = \frac{1}{2} C \rho A v^{2}, \label{6.5}\]

where \(C\) is the drag coefficient, \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as \(F_D = bv^2\), where b is a constant equivalent to \(0.5C \rho A\). We have set the exponent n for these equations as 2 because when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in Fluid Mechanics , for small particles moving at low speeds in a fluid, the exponent n is equal to 1.

Definition: Drag Force

Drag force \(F_D\) is proportional to the square of the speed of the object. Mathematically,

\[F_{D} = \frac{1}{2} C \rho A v^{2},\]

where \(C\) is the drag coefficient, \(A\) is the area of the object facing the fluid, and \(\rho\) is the density of the fluid.

Athletes as well as car designers seek to reduce the drag force to lower their race times (Figure \(\PageIndex{1A}\)). Aerodynamic shaping of an automobile can reduce the drag force and thus increase a car’s gas mileage. The value of the drag coefficient \(C\) is determined empirically, usually with the use of a wind tunnel (Figure \(\PageIndex{1B}\)).

A photograph of a bobsled on a track at the Olympics. A photograph of a model plane in a wind tunnel.

The drag coefficient can depend upon velocity, but we assume that it is a constant here. Table \(\PageIndex{1}\) lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).

Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned, as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics and won a gold medal in the 400-m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (Figure \(\PageIndex{2}\)). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.

A photograph of three swimmers wearing body suits.

Terminal Velocity

Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the small buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as shown by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity (\(v_T\)). Since \(F_D\) is proportional to the speed squared, a heavier skydiver must go faster for F D to equal his weight. Let’s see how this works out more quantitatively.

At the terminal velocity,

\[F_{net} = mg - F_{D} = ma = 0 \ldotp\]

\[mg = F_{D} \ldotp\]

Using the equation for drag force, we have

\[mg = \frac{1}{2} C \rho A v_{T}^{2} \ldotp\]

Solving for the velocity, we obtain

\[v_{T} = \sqrt{\frac{2mg}{\rho CA}} \ldotp\]

Assume the density of air is \(\rho\) = 1.21 kg/m 3 . A 75-kg skydiver descending head first has a cross-sectional area of approximately A = 0.18 m 2 and a drag coefficient of approximately C = 0.70. We find that

\[v_{T} = \sqrt{\frac{2(75\; kg)(9.80\; m/s^{2})}{(1.21\; kg/m^{3})(0.70)(0.18\; m^{2})}} = 98\; m/s = 350\; km/h \ldotp\]

This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.

Example \(\PageIndex{1}\): Terminal Velocity of a Skydiver

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

At terminal velocity, \(F_{net} = 0\). Thus, the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find \(mg = \frac{1}{2} \rho C A v^{2}\).

The terminal velocity \(v_T\) can be written as

\[v_{T} = \sqrt{\frac{2mg}{\rho CA}} = \sqrt{\frac{2(85\; kg)(9.80\; m/s^{2})}{(1.21\; kg/m^{3})(1.0)(0.70\; m^{2})}} = 44\; m/s \ldotp\]

This result is consistent with the value for v T mentioned earlier. The 75-kg skydiver going feet first had a terminal velocity of v T = 98 m/s. He weighed less but had a smaller frontal area and so a smaller drag due to the air.

Exercise \(\PageIndex{1}\)

Find the terminal velocity of a 50-kg skydiver falling in spread-eagle fashion.

The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You do not reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J. B. S. Haldane, titled “On Being the Right Size.”

“To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.”

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law.

Stokes’ law

For a spherical object falling in a medium, the drag force is

\[F_{s} = 6 \pi r \eta v, \label{6.6}\]

where \(r\) is the radius of the object, \(\eta\) is the viscosity of the fluid, and \(v\) is the object’s velocity.

Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about \(1\, \mu m) can be about \(2\, \mu m/s. To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell.

Sediment in a lake can move at a greater terminal velocity (about 5 \(\mu\)m/s), so it can take days for it to reach the bottom of the lake after being deposited on the surface.

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fish, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spearhead as the flock forms a streamlined pattern (Figure \(\PageIndex{3}\)). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.

A photograph of geese flying in a V formation.

In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.

Video \(\PageIndex{1}\) : Fluid Mechanics - Drag force - Flow simulation

The Calculus of Velocity-Dependent Frictional Forces

When a body slides across a surface, the frictional force on it is approximately constant and given by \(\mu_{k}N\). Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by

\[f_{R} = -bv,\]

where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and \(v\) is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.

Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in Figure \(\PageIndex{4}\). Newton’s second law in the vertical direction gives the differential equation

\[mg - bv = m \frac{dv}{dt},\]

where we have written the acceleration as \(\frac{dv}{dt}\). As v increases, the frictional force \(–bv\) increases until it matches mg. At this point, there is no acceleration and the velocity remains constant at the terminal velocity v T . From the previous equation,

\[mg - bv_{T} = 0,\]

\[v_{T} = \frac{mg}{b} \ldotp\]

The free body diagram shows forces m times vector g pointing vertically down and b times vector v pointing vertically up. The velocity, vector v, is vertically down. The positive y direction is also vertically down.

We can find the object’s velocity by integrating the differential equation for \(v\). First, we rearrange terms in this equation to obtain

\[\frac{dv}{g- \left(\dfrac{b}{m}\right)v} = dt \ldotp \label{eq20}\]

Assuming that \(v = 0\) at \9t = 0\), integration of Equation \ref{eq20} yields

\[\int_{0}^{v} \frac{dv'}{g- \left(\dfrac{b}{m}\right)v'} = \int_{0}^{t} dt',\]

\[- \frac{m}{b} \ln \left(g - \dfrac{b}{m} v' \right) \Bigg|_{0}^{v} = t' \big|_{0}^{t} ,\]

where \(v'\) and \(t'\) are dummy variables of integration. With the limits given, we find

\[- \frac{m}{b} [ \ln \left(g - \dfrac{b}{m} v \right) - \ln g] = t \ldotp\]

Since \(\ln A − \ln B = \ln (\left(\frac{A}{B}\right)\), and \(\ln (\left(\frac{A}{B}\right) = x\) implies \(e^x = \dfrac{A}{B}\), we obtain

\[\frac{g - \left(\dfrac{bv}{m}\right)}{g} = e^{- \frac{bt}{m}},\]

\[v = \frac{mg}{b} \big( 1 - e^{- \frac{bt}{m}} \big) \ldotp\]

Notice that as t → \(\infty\), v → \(\frac{mg}{b}\) = v T , which is the terminal velocity.

The position at any time may be found by integrating the equation for v. With v = \(\frac{dy}{dt}\),

\[dy = \frac{mg}{b} \big( 1 - e^{- \frac{bt}{m}} \big)dt \ldotp\]

Assuming y = 0 when t = 0,

\[\int_{0}^{y} dy' = \frac{mg}{b} \int_{0}^{t} \big( 1 - e^{- \frac{bt}{m}} \big)dt',\]

which integrates to

\[y = \frac{mg}{b} t + \frac{m^{2}g}{b^{2}} \big( e^{- \frac{bt}{m}} - 1 \big) \ldotp\]

Example \(\PageIndex{2}\): Effect of the Resistive Force on a Motorboat

A motorboat is moving across a lake at a speed v 0 when its motor suddenly freezes up and stops. The boat then slows down under the frictional force \(f_R = −bv\).

In the both of the previous examples, we found “limiting” values. The terminal velocity is the same as the limiting velocity, which is the velocity of the falling object after a (relatively) long time has passed. Similarly, the limiting distance of the boat is the distance the boat will travel after a long amount of time has passed. Due to the properties of exponential decay, the time involved to reach either of these values is actually not too long (certainly not an infinite amount of time!) but they are quickly found by taking the limit to infinity.

Exercise \(\PageIndex{2}\)

Suppose the resistive force of the air on a skydiver can be approximated by \(f = −bv^2\). If the terminal velocity of a 100-kg skydiver is 60 m/s, what is the value of b?

Physics Formulas

Drag Force Formula

Top Banner

Drag Force (D) is defined as the force that resists the motion of a body with fluid. If the motion of the body exists in the fluid-like air it is known as aerodynamic drag. And, if the fluid is water it is known as hydrodynamic drag. The drag force always acts in the opposite direction to the flow of fluid.

Drag Coefficient Formula

Following is the formula used to calculate the drag coefficient:

Solved Examples

Question 1. A car travels with a speed of 80 km.h -1 with a drag coefficient of 0.25. If the cross-sectional area is 6 m 2 , calculate the drag force.

Solution: Given: Velocity, V= 80 km.h -1

Drag coefficient, C d = 0.25

Cross-sectional area, A= 6 m 2

Density of fluid, ρ =1.2 kg.m -3

Question 2. A plane moves with the velocity of 600 km.h -1 with a drag coefficient of 0.25. If the cross-sectional area of the plane is 110 m 2 , calculate the drag force. Solution: Given: Velocity, V=600 km.h -1

Drag coefficient, Cd= 0.25

Density of fluid, ρ=1.2 kg.m -3

Cross-sectional area, A=110 m 2

Stay tuned with BYJU’S to learn more about other Physics related concepts.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Share Share

Register with BYJU'S & Download Free PDFs

Find Study Materials for

Create Study Materials

Select your language

physics practice problems on drag force

Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken

Nie wieder prokastinieren mit unseren Lernerinnerungen.

Why is it harder to walk through water than through air? The answer is drag force. W ater contributes to a higher drag force than air, so it feels harder to walk through because a stronger force is acting against you. Drag force slows objects down as they move through a fluid. This article will discuss the definition, types, and examples of drag force, the equation for drag force, and then we will do an example of how to find drag force.

Drag Force in Physics

This article is about to dump a lot of information on you at once, so I'll try not to make it a drag.

In physics, drag force is the force that opposes the relative motion between an object and a fluid.

A fluid is anything that flows, such as a liquid or a gas. When the fluid is air, drag force is referred to as air resistance.

The object might move through the fluid, or the fluid might move around the object — either way, the drag force acts in the opposite direction of the relative movement . In this way, the drag force is similar to friction, but the motion is between a solid and a fluid instead of two solids. The image below shows a man running through air (a fluid). Since the man's motion is to the right, the drag force would act opposite to that motion, to the left, as shown by the arrows in the figure.

What if the man stands still and the air moves past him as wind, like in the image below? As you can see by the arrows in the figure, the drag force would act in the direction of the wind; this is because the relative motion between the man and the air is the same as in the image above, but instead of the man moving to the right, the air is just moving to the left.

What if the man was running in the direction of the wind? At this point, it depends on whether he runs faster or slower than the wind. It can help to think of the direction of drag as whichever direction you would feel pressure, or force, from the fluid. If he feels the breeze on his front, the drag force points towards his front; if he feels the breeze on his back, the drag force points towards his back.

If he runs at the same speed as the wind and thus feels no breeze, there would be no drag force since there would be no relative motion between him and the air around him. On the other hand, if he jumped off a balcony, he would feel wind upward so that the drag force would point up.

Drag force is an important consideration for engineering designs. Understanding drag and how to decrease it helps people design sturdier structures and bridges that hold up better to wind, more efficient cars and planes, and more efficient collection of wind energy and hydropower.

Types of Drag Force

There are different types of drag force, especially when considering the flight of airplanes:

Phew, that was a lot of information to trudge through, if word density could be considered, this article would have a massive drag force.

Examples of Drag Force

Here we go again — even more examples. It's like walking through honey.

Drag is present when there is relative motion between an object and a fluid. Some examples of drag force include the following:

Drag Force Equation

The common equation or formula for drag force is shown below:

$$D=\frac{1}{2}\\C\rho Av^2\mathrm{.}$$

This equation is only accurate under certain conditions: the motion is fast enough that the fluid behind the object is turbulent, the fluid is not denser than air, and the object is not tiny. As with other forces, drag force is measured in \(\mathrm{newtons}\) \(\mathrm{N}\).

Drag Force Formula

Above, you probably saw a bunch of variables that you have never seen before. To aid you in understanding what the drag force is, we'll go over each of these variables.

\(C\) is the coefficient of drag , which is a unitless number that has been determined experimentally. \(\rho\) represents the density of the fluid in \(\mathrm{kg/m^3}\); as the density increases, the drag force increases. In our opening example, water contributes to a higher drag force than air because it has a higher density.

\(A\) is the effective cross-sectional area of the object in \(\mathrm{m^2}\) — this is the area of the object that is perpendicular to the motion. So, for example, when sticking your hand outside a moving car's window, if you tilt your hand, so the side is facing the front of the car, you will feel less force than if you stick your hand out with the palm facing the front: this is because your hand has a smaller cross-section in the first orientation.

\(v\) is the relative velocity between the object and the fluid in \(\mathrm{m/s}\). If you put your hand in water and slam it down, you will feel more of a force fighting against you than if you slowly lower it. Drag force differs from friction because friction doesn't depend on the object's speed.

Stokes's Law

Have you enjoyed the terrible puns so far? I don't mean to stoak the fire burning in your heart even more from all these dad jokes, but I just have to.

When the conditions don't meet the requirements listed above, we can use Stokes's Law to find the friction force :

$$F_s = 6\pi \eta r v$$

where \(\eta \) is the viscosity of the fluid in SI units of Pascal-seconds \(\mathrm{Pa\,s}\), which is the same as kilograms per meter-second \(\mathrm{kg/m\,s}\), \(r\) is the radius of the object in meters, and \(v\) is the velocity in meters per second \(\mathrm{m/s}\). In this case, the drag force is proportional to the velocity rather than the velocity squared. With this equation, the drag force may be referred to as a viscous drag force , where the drag force is dependent on the fluid's viscosity.

Drag Force in Free-Fall — Terminal Velocity

If you drop a bouncy ball off the empire state building, initially the force acting against the ball is negligible since the velocity starts at zero. Instead, the main force acting on it is the force of gravity, which pulls it down and causes it to accelerate. As the velocity increases, the drag force acting against the ball's fall will increase. This opposite force causes the ball's acceleration to slow until, eventually, the drag force equals the force of gravity and the ball no longer accelerates. At this point, the ball will reach a constant speed at its terminal velocity .

Example Problem Using Drag Force

Now, it's time for an example.

A box falls through the air at a speed of \(12\,\mathrm{m/s}\). Its dimensions are \(0.5\,\mathrm{m} \cdot 0.5\,\mathrm{m} \cdot 2\,\mathrm{m}\), oriented vertically as shown in the image below. The density of air is \(1.225\,\mathrm{kg/m^3}\), and the drag coefficient for the box is \(2.1\). What is the drag force?

Most of the variables are fairly self-explanatory, and we can plug them straight into our drag force equation. The trickiest one to know what numbers to use is the area. We need the effective cross-sectional area, which is the area of the box that is facing the motion — in this case, we can think of it as the area facing the wind as it falls. This area is \(0.5\,\mathrm{m} \cdot 0.5\,\mathrm{m}\), or \(0.25\,\mathrm{m^2}\). Now we can write our equation

$$D=\frac{1}{2}\\ C\rho A v^2\mathrm{,}$$

and plug everything in

$$D=\frac{1}{2}\\(2.1)(1.225\,\mathrm{kg/m^3})(0.25\,\mathrm{m^2})(12\,\mathrm{m/s^2})^2\mathrm{,}$$

which gives us the answer

$$D=46.3\,\mathrm{N.}$$

If we double-check the units in our equation, we can see that they all cancel out to give us \(\mathrm{kg\,m/s^2}\), which is the same as \(\mathrm{newtons}\).

Hopefully, finishing this article wasn't too much of a drag —more like walking through a luscious breeze rather than cold honey. If your answer was the honey, don't panic, you're almost done: here are the most important points you need to remember.

Drag Force - Key takeaways

Frequently Asked Questions about Drag Force

--> what is drag force.

Drag force is the force that opposes the relative motion between an object and a fluid.

--> How to calculate drag force?

For most simple scenarios, drag force can be found by multiplying one half, the coefficient of drag, the density of the fluid the object moves through, the cross-sectional area of the object, and the object's velocity together.

--> What are the types of drag force?

The types of drag force include parasitic drag, form drag, skin friction drag, interference drag, induced drag, and wave drag. These types of drag are applicable to flying an airplane.

--> What is the importance of drag force?

Drag force is an important consideration in engineering design. It impacts the performance of products such as cars, planes, ships, structures, wind turbines, and water turbines.

--> What is viscous drag force?

Viscous drag force occurs when the fluid is too viscous for the normal drag force equation to apply. In this case, the viscosity of the liquid is an important factor in determining the drag force.

Final Drag Force Quiz

What is the drag force?

The force that opposes the motion between an object and a fluid.

In what direction does the drag force act?

Opposite to the relative motion.

What condition does not apply to being able to use the drag force equation accurately?

The coefficient of drag is less than \(2\).

What is Stokes's Law?

The equation used to find the drag force when the situation doesn't meet the qualifications required to use the drag force equation.

Which of the following is not an example of drag force?

A box resists being pulled across the floor with a rope.

What does \(A\) represent in the drag force equation?

The effective cross-sectional area of the object.

If the density of the fluid increases, the drag force decreases.

If the relative velocity between the object and fluid increases, the drag force increases.

If the effective cross-sectional area of the object decreases, the drag force increases.

$$ D=\frac{1}{2}C\rho Av^2$$

Objects with a higher drag coefficient will create more drag than those with a lower coefficient.

A model rocket is placed in a wind tunnel with the point towards the direction of the wind. It has a diameter of \(5\,\mathrm{cm}\) and a length of \(10\,\mathrm{cm}\). What is the effective cross-sectional area?

\(19.6\,\mathrm{cm^2}\).

Bob wants to reduce drag to get better miles per gallon on his car. What might he do?

Buy a skinnier car.

If the relative velocity between the object and fluid is multiplied by \(2\), what happens to the drag force?

It is multiplied by 2.

of the users don't pass the Drag Force quiz! Will you pass the quiz?

More explanations about Dynamics

Discover the right content for your subjects, business studies, combined science, english literature, environmental science, human geography, macroeconomics, microeconomics, no need to cheat if you have everything you need to succeed packed into one app.

Be perfectly prepared on time with an individual plan.

Test your knowledge with gamified quizzes.

Create and find flashcards in record time.

Create beautiful notes faster than ever before.

Have all your study materials in one place.

Upload unlimited documents and save them online.

Study Analytics

Identify your study strength and weaknesses.

Weekly Goals

Set individual study goals and earn points reaching them.

Smart Reminders

Stop procrastinating with our study reminders.

Earn points, unlock badges and level up while studying.

Magic Marker

Create flashcards in notes completely automatically.

Smart Formatting

Create the most beautiful study materials using our templates.

Join millions of people in learning anywhere, anytime - every day

Sign up to highlight and take notes. It’s 100% free.

This is still free to read, it's not a paywall.

You need to register to keep reading, get free access to all of our study material, tailor-made.

Over 10 million students from across the world are already learning smarter.

StudySmarter bietet alles, was du für deinen Lernerfolg brauchst - in einer App!

IMAGES

Drag Force: Definition, Formula, and Examples
Spice of Lyfe: Formula For Drag Force Physics
Drag Force, APC
What Is Net Force : Monster Force Zero (2020) Review
Physics Formulas
Drag Force and Terminal Speed

VIDEO

What's this crazy Walkabout thing? #shorts
How to download best Free Physics Practice Problems for JEE 2023
Velocity dependent forces
Drag force
Drag and lift force in fluid mechanics (हिन्दी)
AP Physics 5.2

COMMENTS

APC Practice Problems 06
FOS4 – Practice Problems –Drag Force – APC. 1) A 0.500-kg object is suspended from the ceiling of an accelerating boxcar as shown.
Aerodynamic Drag
Practice. practice problem 1. Two related questions… Determine the drag coefficient of a 75 kg skydiver
Drag Force and Terminal Speed
Balanced Forces Equilibrium Example · Drag Force Differential Equation · Newton's Law of Motion - First, Second & Third - Physics · Centripetal
5.2 Drag Forces
Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid).
Drag Force
assesses the students' ability to apply calculus in the solution of a physics problem.
AP Physics C: Mechanics : Forces
An object of mass \displaystyle \small m is dropped from a tower. The object's drag force is given by \displaystyle \small F_{d
6.4 Drag Force and Terminal Speed
We have set the exponent n for these equations as 2 because when an object is moving at high velocity through air, the magnitude of the drag force is
Drag Force
Drag Force. A drag force is the resistance force caused by the motion of a body through a fluid, such as water or air. A drag force acts opposite to the
6.7: Drag Force and Terminal Speed
The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a
Drag Force Formula with Solved Numericals
Question 1. A car travels with a speed of 80 km.h-1 with a drag coefficient of 0.25. If the cross-sectional area is 6
Drag Force: Definition, Examples & Formula
Example Problem Using Drag Force.