Static and Kinetic Friction:?

Question

A 10 kg crate is at rest on a horizontal table (mus =0.8, muk =0.3). Then various amounts of horizontal pushing force are applied to it.

a.) The largest pushing force that can be applied before the crate starts moving is of size:___N

b) if the horizonal pushing force was 10N, the static friction force is ___N. what is the net force on the crate? __N and what is the acceleration of the crate___m/s/s

c) if the horizonal pushing force was 115N, the static friction force is ___N. what is the net force on the crate? __N and what is the acceleration of the crate___m/s/s

Answer 1

a. 78.4 N

b. 0.102, 0 N, 0

c. won't be static the crate will move.  85.6 N, 8.56 m/s^2

Answer 2

Hi D-

I'm not a fan of just handing out answers with no explanation.  How will you learn the stuff?

When the box is at rest, it will "accept" the largest "push" without moving.  F(push)<F(static friction) will keep the box stationary.

F max = F (normal) * mus = Weight * mus= mg*mus = 10*10*.8 = 80 N.  Above this and the box will move (and you MUST change over to muk).

a) 80N

If the horizontal force of pushing is less than or equal to 80N, then the box will push back with an equal but opposite force.

b) 10N

At 115N, you are over 80N, so the KINETIC friction opposes the motion with 10*10* .3 N = 30 N

Net force = 115N - 30N = 85N

From F=ma, we get 85 = 10 * a and a=8.5 m/s^2 for acceleration.

These problems are all the same whether you are on an inclined plane or not.

1) determine if the force pushing (including gravity doen plane) > Static friction.  If not, no motion, no acceleration, forces are in balance,

2)( If F push > f friction (static)), go calculate friction kinetic and now..

F net = F push-F friction kinetic = ma

I know that you will "get it".  good luck in school!

-Fred