# Homework Help: Help Needed for Two-Body Inclined Plane and Pulley Problem! please!

1. Oct 12, 2009

### katzmeow9

1. The problem statement, all variables and given/known data
Two blocks are connected over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. Angle theta of the incline is 30 degrees. Block A slides down the incline at constant speed. What is the mass of Block B?

2. Relevant equations
a = Fnet/m
fk = [coefficient of kinetic friction] * N
Fg = mg

3. The attempt at a solution
I found N [normal force] to equal 84.87 N and fk to equal 16.97 N, but those could be wrong. Also, when tilting the plane, Fgx equals 49 N and a = 0. I don't know if those are correct, because I started to get confused at the end. My answer is 3.27 kg (after getting a weight of 32.03 N). I doubt that's the right answer because I started to mix up which forces should be equal to each other.

Unfortunately I can't find the image of the diagram online, but Block A is on an incline in this direction / with the angle 30 degrees above the horizontal on the positive x axis. the pulley is attached to the top right corner (at the highest point) and Block B is hanging vertical. It is SIMILAR to this picture, except block a is on the incline sliding downward and block b is hanging vertically, I assume being pulled upward.

I need help tonight!

2. Oct 12, 2009

### rl.bhat

Hi catzmeow9, welcome to PF.
Downward motion of A can prevented by the frictional force and the weight of B.
Find down ward force on A.
Find normal reaction and the frictional force. Then proceed.

3. Oct 12, 2009

### katzmeow9

hmm....I THINK that's what I did....would it be possible for anyone to check my answer?
I didn't explain my work very well but I can barely follow it myself... I found the force of gravity on A, and used that to find the force of gravity on A that points down the slope. Normal force should be equal (but opposite direction) to the force of gravity that's in the y direction of the shifted coordinate plane...so using that I found the kinetic friction, then used Fgx - (tension on rope + fk) = 0 to find the tension on the rope. At that point I was confused about whether that tension would equal the same as the tension int he rope for Block B, and then would therefore equal the gravitational force on B, since acceleration is 0. I assumed it would, and used the same value as the tension force for the weight of B (32.03 N). I then used that to find the mass of 3.27 kg.
Is this correct?

4. Oct 12, 2009