Question

The coefficients of friction between a block of mass 0.5 kg and a surface are μs = 0.75 and μk = 0.45. Assume the only other force acting on the block is that due to gravity. What is the magnitude of the frictional force (in newtons) on the block if the block is at rest and the surface is horizontal? Use g = 9.79 m/s2.

Answer 1

The maximum frictional force will be .75(mg), but since no other forces are acting on the block, the actual frictional force has 0 magnitude. Seems like kinda a trick question.

Answer 2

really? it wouldn't be fs=UsN

Answer 3

so just fs=(.75)(mg)=(.75)(.5)(9.79)?

Answer 4

The normal force is mg. The coefficient is .75 (static).

Answer 5

yep. that looks right

Answer 6

so its not 0?

Answer 7

Well, static friction only resists other forces. So if there are no other forces laterally there is no force to resist and the static frictional force is technically 0. It will continue to increase to a maximum force of (.75)(.5)(.9.79) if any force tries to push the block. Once a lateral force exceeds that amount the block will move and the frictional force will be kinetic instead of static.

Answer 8

ok. so because its not on an incline and there's no other forces pushing it, its zero?

Answer 9

How does that fit into the formula though?

Answer 10

I would say yes, but that's sorta the trick question thing. If you think it's likely to be a trick question then 0 is correct. If you think that the instructor meant what is the maximum frictional force which will resist any lateral forces then the answer is (.75)(.5)(9.79)

Answer 11

well the next question asks what it would be if it was still at rest but there's an incline?

Answer 12

Ah.. then yes. It's a trick question. It's 0 in the first case.

Answer 13

oh ok. :) Thank you!