# Homework Help: Two masses two pulleys and an inclined plane

1. Nov 6, 2016

### Valerion

1. The problem statement, all variables and given/known data
A block of mass m1 is at rest on an inclined plane that makes an angle (theta) with the horizontal. The coefficient of static friction between the block and the inclined planed is µs. A massless, inextensible string is attached to one end of the block, passes over a fixed pulley, pulley 1, around a second freely suspended pulley, pulley 2, and is finally attached to a fixed support. The pulleys are massless, but a second block of mass m2 is hung from the suspended pulley. Gravity acts downwards.

Now assume that the block on the incline plane is sliding up the plane. The coefficient of kinetic friction is µk. Find the magnitude of the acceleration of the block on the inclined plane a_x1 Express your answer in terms of some or all of the variables m1, m2, theta, µk and g

2. Relevant equations

F = ma

3. The attempt at a solution

I choose coordinate systems. For block 1 I chose the positive X axis downward along the surface of the inclined plane and the positive Y axis up, perpendicular to the surface of the inclined plane with the x origin at the top of the incline. For block 2 I chose the positive Y-direction downwards with the origin at the ceiling.

I made free-body diagrams of both block 1, 2 and the suspended pulley which gave me the following equations:

I can treat the suspended pulley and block 2 as one system because the length between them is constant so they move with the same acceleration a_2

-T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = m_1 a_x1
N = m_1 gcos(theta)
-2T + m_2 g = m_2 a_2

The length of the cord connecting block 1 with the ceiling is constant. So after taking two derivatives of the length. I get the relation between a_x1 and a_2 which is a_1 = -2a_2

T = ((m_2*g- m_2*a_2)/2 )

T = µ_k*m_1*g*cos(theta) + m_1*g*sin(theta) - m_1*a_1

((m_2*g-m_2*a_2 /2 ) = µ_k*m_1*g*cos(theta) + m_1*g*sin(theta) - m_1*a_1

2*(µ_k*m_1*g*cos(theta) + m_1*gsin(theta) - m_1*a_1 = (m_2*g-m_2*a_2)

2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta)) - m_2*g = -m_2*a_2 + 2*m_1*a_1

substitute a_1 / -2 for a_2 and factor a_1 out

2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta)) -m_2*g = (m_2/2 + 2*m_1) * a_1

a_1 = (2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta))-m_2*g ) / (m_2/2 + 2*m_1)

For some reason this answer is not correct and I have been looking for that reason for several hours but I can't seem to find it. I think its a math error rather than a physics error. Can someone help me out?

Last edited: Nov 6, 2016
2. Nov 6, 2016

### PhanthomJay

Looks like you have signage errors. You have one block moving down the plane and the other hanging block moving downward. If the hanging block moves down the other moves up the plane, as stated. Acceleration of each block is in the direction of the net force.

3. Nov 6, 2016

### Valerion

So the acceleration of block 1 has a negative sign and the acceleration of block 2 has a positive sign. Then,

-T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = m_1 a_x1 should be -T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = - m_1*a_x1 ?

4. Nov 6, 2016

### PhanthomJay

Yes , and a2 is one half of a1.

5. Nov 6, 2016

### Valerion

Ok, I think I got it. a_1 = (2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta))-m_2*g ) / (-m_2/2 - 2*m_1)

6. Nov 6, 2016

### PhanthomJay

Yes that looks correct. I find it easier to assume the positive direction in the direction of the acceleration, positive up incline for m1 and positive down for m2. Helps a bit with signage errors

7. Nov 6, 2016