Question

Block A, with a mass of 10 kg, rests on a 36.9° incline. It is attached to a string that is parallel to the incline and passes over a massless, frictionless pulley at the top. Block B, also with a mass of 10 kg, is attached to the dangling end of the string. The two blocks remain at rest. The force of friction on the block A is:
a. 16 N
b. 8 N
c. 4 N
d. 12 N

I got 39.16N as the answer. But it seems that it is wrong.
This is my working equation wherein the system is in equilibrium given that the blocks are at rest.
f = T - m1*g*sin36.9
T = m2*g

Any solution for this?

Answer 1

Since all numbers are known for m1,m2,g and θ, we can calculate that
force pulling block up-plane
= T = m2g=98.1 N
Force pulling block down-plane
= mg sin(&theta)
=58.90 N
The difference is due to friction, of the amount = 39.20 N

I believe you answer is correct. Book answers could differ because when they revise questions, they might have forgotten to revise the answers. I suggest you double check the correspondence of question numbers, or check with your teacher.

Answer 2

Thanks. :D