# Homework Help: Friction force in Newton's law

1. Oct 12, 2011

### theunloved

In Figure 6-63, a block weighing 22 N is held at rest against a vertical wall by a horizontal force of magnitude 60 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second force is applied to the block and directed parallel to the wall with these magnitudes and directions: (a) 34 N, up, (b) 12 N, up, (c) 48 N, up, (d) 62 N, up, (e) 10 N, down, and (f) 18 N, down. In each experiment, what is the frictional force on the block, including sign? In which does the block move (g) up the wall and (h) down the wall ? (i) In which is the frictional force directed down the wall ?

How I attempted to solve:
Taking upwards as positive, x+ from left to right

If the block is at rest, then:

fs <= us N = 0.55 * 60 = 33N
fk = uk N = 0.55 * 60 = 22.8N

Part a.
P = 34 N, up.
P is greater than fs at max, so the block must move, and friction is going down.

P - mg - fk = ma

Ok, from here, I really kinda confused. If you don't know the sign of frictional force, how are you gonna write down Newton's second law ? and also, we don't know acceleration too, how are we gonna calculate fk ???

Part b, e, f, the block can't move, because P is smaller than fs at max....

2. Oct 12, 2011

### grzz

Where is the figure?

3. Oct 12, 2011

### Staff: Mentor

Don't forget about gravity when deciding if the block must move and determining the direction of friction. You need to compare P - mg with friction, not just P.

The proper reasoning as above will tell you the direction and thus the sign of the friction force.
Why would you need the acceleration?