# 05.02.03 - Solving a Problem with Static Friction (Physics)

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Okay, so what can we do with static friction?

Well, let's consider my couch problem from before. I did say that for a short time I was able to push it at a constant velocity. Before that I had to get the &@#^ thing moving. I mentioned that the coefficient of static friction is usually larger than the coefficient of kinetic friction. This is mostly because the surfaces have really good contact when they are not sliding against one other, so every ridge on one surface can hit up against every ridge on the other surface, but if they are sliding, basically just the tops of the ridges bump into each other; the surfaces don't get really good contact. (Most lubricants work by actually preventing any contact between the two surfaces.)

Let's figure out how hard this was to get the couch moving.

In order to push my couch to the other side of the room, I first had to move it from where it was. If the couch still has a mass of 55 kg, and the coefficient of static friction, μs, is 0.74, what is the minimum horizontal force I needed to apply to get it moving?

Same story:

Step #1 Draw a picture.

Step #2 Draw and label the forces

We know that the frictional force is resisting our attempts to move the couch. Since I am pushing to the left, it must be acting to the right.

Step #3 Choose a coordinate system

Let's keep the one we had before. No really good reason.

Step #4 Apply Newton's Second Law

Here we have a similar set of forces. In the "x" direction we have the force of friction, , and the force I applied to the couch, F. In the "y" direction we have the normal force, N, and the force of gravity, . Since we picked the positive direction to be the direction I am pushing, the applied force is positive and the frictional force is negative. Likewise, we chose up to be positive, so the normal force is positive, and gravity is negative. The couch is obviously not jumping off the ground, so the acceleration in the "y" direction is zero. What about the "x" direction?

Here is where the wording is important. I asked what is the minimum force I need to apply to get the couch moving. Basically, I want a non-zero acceleration, but just barely a non-zero acceleration. The way we treat this mathematically is we go ahead and say that the acceleration is zero, but we know we need just a bit more force than that to actually get it going.

Okay, ready to apply Newton's 2nd Law?

Step #5 Substitute known forces. So, we know that

But we actually know something else. Remember, the wording in the problem is important. Since I asked for the minimum force I need to overcome friction, this means that I am also asking for the maximum value the frictional force can have. So I also have

Be very careful with that equation. It is only true when the code words "minimum" or "maximum" are in the problem.

So, make those substitutions.

Step #6 Solve the algebra

This is exactly the same set of equations I had before, but now I am solving for F instead of μ. Do you think you can follow this?

Solve "y" equation for N

Substitute into "x" equation, and solve for F.

Step #7 Plug in numbers

Wow! Do you think I need some help with that?

Step #8 Answer the question

In order to push my couch to the other side of the room, I first had to move it from where it was. If the couch still has a mass of 55 kg, and the coefficient of static friction, μs, is 0.74, what is the minimum horizontal force I needed to apply to get it moving?

Okay, here we are going to remember significant figures. I should report my answer to 2 sig figs, but 399 rounds up to 400, which looks like something with only 1 sig fig. Therefore, what I really should do is rewrite this answer in scientific notation. My answer really ought to be:

Or if I want to be perfectly clear I should say, I need to exert a horizontal force that is greater than 4.0x102 N in order to get the couch moving.

(And while we are discussing sig figs... I have been treating the value as if it were exact, and using it in both calculations with 3 sig figs and in calculations with 2 sig figs. The truth is, 9.8 m/s2 is not quite exact, but it is close enough that it introduces very little error, so I can go ahead and use it as if it were.)