# Pulley Block Mass Problem

1. Oct 6, 2007

### sylenteck0

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 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
Fs=mgsin$$theta$$

Sin$$theta$$= Fs/mg

3. The attempt at a solution

I tried using substition in order to find the second mass, but that route bore no fruit.

9.8 x 10 x sin 30= 49
Fs=49
T- mgsin$$theta$$=ma
T-49=ma
T=ma+49

-T+m(subscript 2)g= M(subscript2)a (we substitute t for ma+49)
-(m(subscript 1)a+49)+m(subscript 2)g= M(subscript 2)a
-msub1-49+msub2g=msub2a

all that boils down to is

(9.8m(sub2)-49)/(10+msub2)= a

How exactly am I supposed to find the second mass without knowing the acceleration?

2. Oct 6, 2007

### bob1182006

when you found the Fs that was static friction? because the block is under a constant velocity so it is actually kinetic friction.

and also you did 9.8x10xsin30 but you didn't include the frictional constant in there.

3. Oct 6, 2007

### sylenteck0

Alrighty. But even if I do include the kinetic friction, how would I find a?

4. Oct 6, 2007

### bob1182006

well the problem states that Block A (m2) is sliding down at a constant velocity. constant velocity means what type of acceleration?

and if block A has acceleration (a) and we're assuming that the string doesn't stretch what does that say about block B's acceleration?

5. Oct 6, 2007

### sylenteck0

Constant velocity means no acceleration
And ah, ok :)

6. Oct 6, 2007

### bob1182006

yes also your calculation for friction was wrong it should be:

$$\mu_kN$$

and you know N as one of the components of mg and you know the downward is mgsintheta

7. Oct 6, 2007

### sylenteck0

Right right. Thanks so much :)