Lousy static friction.

I don’t know why I can get the kinetic friction problems, but not these ones. Especially the first one. I know I’m going to kick myself when someone explains it.

  1. A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.31. Suddenly, the plane accelerates, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

  2. The drawing shows a circus clown who weighs 760 N. The coefficient of static friction between the clown’s feet and the ground is 0.42. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

  1. 0.31
  2. > (760*0.42)

Tried that already for the second one, it said I was wrong.
Edit: just checked the first one, that’s wrong too.

What do the answers say?

I don’t have the answers, but I can submit mine up to 5 times, since the problems are online.

Well, then that just sucks.

The coffee’s coefficent of friction is 0.31, so Force_normal multiplied by the coefficient of friction should give you Force_friction. The maximum amount of Force applied to the coffee before it starts moving should then be Force_normal *0.31, which is mass_coffee * 9.8 * 0.31. Now, they’re asking for acceleration, and f = ma, so conversely a = f/m, so a = mass_coffee * 9.8 * 0.31 / mass_coffee, so the answer should be 9.8 * 0.31 because I’m stupid.

The clown weighs 760N, so the Force_normal is 760N. The Force_applied should equal Force_normal. Ah, but for every newton of force he applies to pulling the rope, he is decreasing the normal force. Therefore if he applies force x then:
0.42(760 - x) = x
319.2 - 0.42x = x
319.2 = 1.42x
x = 224.79N
so x = 224.79N because I’m stupid.

Okay, answer to the second one is 225N. I got that earlier, but thought that sig figs mattered and entered 230.

The first one, however, seems to not be 3.1 m/s^2.

Remember that the coefficient of friction is equal to the Force of Friction / Force of Normal.

So, for #1,

0.31 = <strike>m<sub>cup</sub></strike> a / <strike>m<sub>cup</sub></strike> g

0.31 = a / g
g * 0.31 = a
9.8 * 0.31 = a
a = 3.038m/s<sup>2</sup>

That works. Thanks. All done now.