PhysicsScience

The Physics Philes, lesson 9: Minor Successes!

In which practice problems are done, help is acquired, and more practice problems are done.

If you will remember, last week I tried to explain friction forces. Except I didn’t include any math, because I didn’t really know what I was doing. This week, you’re in luck! I worked really hard and asked for help (thanks, Carl!), and I think I understand! But that’s not all! I actually solved a problem that had no numbers in it. Guys, I might be a genius.

But enough bragging! Time to do a few simple of friction practice problems! Once again, I evidently don’t know how HTML works, so subscripts will be screwy. Please be patient.

Let’s say that you’re trying to move a box that weighs 500 N across a flat, level floor. To initially get the crate moving you have to pull the box with a 230 N horizontal force. When the box starts to move, you can keep it moving at a constant velocity with a force of 200 N. (Newtons first law, anyone?) What are the coefficients of static and kinetic friction?

Let’s draw this out. Here is the situation we’re dealing with:

Person pulling a box.

Here are our free-body diagrams of the forces involved: one for kinetic friction force and one for static friction force.

Static friction diagram

Kinetic friction force diagram.

In the diagrams, n = normal force, T = tension force, (f sub s) max = static friction force at its maximum, f sub k = kinetic friction force, and w = weight. The target variables for this problems are the coefficients of static and kinetic friction (µ sub s and µ sub k). Since the box is in equilibrium (either at rest or moving at a constant velocity) we need to use Newton’s first law of motion.

We know a couple of the forces for each diagram. In the diagram for static friction, we know that the weight of the box is 500 N and the tension force is 230 N. We also know the weight and tension forces for the diagram for kinetic friction; they are 500 N and 230 N, respectively. We also know the equations for finding static and kinetic friction:

Equation for static friction: f sub s ≤ µ sub s(n)

Equation for kinetic friction: f sub k = µ sub k(n)

That equation for static friction looks pretty weird, but it’s really very simple. Static friction is only equal to the coefficient of friction times the normal force when the tension force hits that critical point when the object breaks free. Prior to that point, static friction is less than the coefficient of friction time the normal force. Or we could say it like this:

(f sub s)max = µ sub s(n)

Only when the static friction force is at its maximum can it be equal to the coefficient of friction times the normal force.

We will need four equations to solve this problem: the component equations for Newton’s first law and the equations for the static and friction forces:

∑F sub x = 0

∑F sub y = 0

(f sub s)max = µ sub s(n)

f sub k = µ sub k(n)

Let’s work on filling in the free-body diagram for static friction first. We need to find the tension force T and the normal force n. This diagram illustrates the forces at work just before the box starts to move. Let’s solve for T:

∑F sub x = T + (-(f sub s)max) = 0

T + (-(f sub s) max) + (f sub s)max = 0 + (f sub s)max

T = (f sub s)max

Because T = 230 N, that means (f sub s)max also equals 230 N

Now let’s solve for n:

∑F sub y = n + (-w) = 0

n + (-w) + (w) = 0 + w

n = w

Because w = 500 N, and because the normal force equals the wight, the normal force also equals 500 N.

Now that we know the tension force T and the normal force n, we can figure out the coefficient of static friction! We just need to plug in the values:

(f sub s)max = µ sub s(n)

230 N = µ sub s(500 N)

230 N/500 N = µ sub s

µ sub s = 0.46

That was super easy, right? We just have to do all of that again to find the answers for our kinetic friction force diagram. Since it’s exactly the same as what I did above, just with different numbers, I’ll abridge the math work. Here I solve for T:

∑F sub x = T + (-f sub k) = 0

200 N + (-f sub k) = 0

f sub k = T = 200 N

Here I solve for n:

∑F sub y = n + (-w) = 0

n + (-500 N) = 0

n = w = 500

Now that we know the tension force T and the normal force n, we can solve for the coefficient of kinetic friction:

f sub k = µ sub k(n)

200 N = µ sub k(500 N)

200 N/500 N = µ sub k

µ sub k = 0.40

Voila! We’ve solved for the coefficients of static and kinetic friction! You’ll notice that the kinetic number is lower than the static number. This makes sense because it’s easier to keep something moving than to get it moving in the first place.

But what if the static friction isn’t at its maximum? I’m glad you asked. Let’s say that we have the same problem, except the tension force T is only 50 N. What would be the static friction force? Here is the free-body diagram of the problem:

Notice that the static friction force is denoted as f sub s, not f sub s max. This is because the static friction force is not at its maximum.

We’re looking for the static friction force. Since the box is not moving (it’s in equilibrium) we’ll once again use Newton’s first law. We just need to plug in the numbers:

∑F sub x = T + (-f sub s) = 0

50 N = f sub s

We know from the previous problem that the maximum static friction force for this box is 230 N, so this tension force of 50 N won’t move the box. The static friction can resist any horizontal force up to 230 N.

Finally, let’s do a rolling friction problem. The equation for the coefficient of rolling friction is the horizontal force exerted on the body F divided by the normal force n. Let’s find out how this works. This is the last problem, so hang in there!

We have a 12,000 N car. If the coefficient of rolling friction is 0.015, what horizontal force is needed to move the car at a constant speed on a level road? (We’re not taking air resistance into consideration here.)

Because the car will be moving at a constant velocity, we need to use Newton’s first law. Here is the free-body diagram for this problem:

Notice the four forces acting on the body: the rolling friction force, the normal force, the unknown force F, and the weight.

Using the y-component equation of Newton’s first law, we know that normal force n is equal in magnitude to the weight. So if n = 12,000 N, all we need to do is plug the values into the rolling friction equation:

µ sub r = f sub r / n

0.015 = f sub r / 12,000 N

f sub r = 0.015(12,000 N)

f sub r = 180 N

The x-component equation of Newton’s first law tells us that the magnitude of the force needed to make the care move at a constant speed is equal to f sub r. Therefore, we would need to exert a force of 180 n on the car to keep the car moving at a constant velocity.

Whew! This post was very long today and very, very full of math. Hopefully it is correct and clear. As always, leave questions, corrections, or praise for my mad math abilities in the comments.

Featured image credit: The Eggplant

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

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