# Blocks moving on frictionless incline at constant speed, accleration

1. Oct 18, 2009

### joemama69

1. The problem statement, all variables and given/known data

There is no friction

Part A) Find the mass of block C so that block B moves up the incline with a constant speed

Part B) fint he mass of block C so that Block B moves up the incline with a constant accleration a = g/2

2. Relevant equations

3. The attempt at a solution

Block A

Fy = T(2) -m(a)g = m(a)a T(2) = m(a)a + m(a)g

Block B

Fx = T(1) - m(b)gsinQ - T(2) = m(b)a T(1) = m(b)a + m(b)gsinQ + T(2)

Block C

Fy = T(1) - m(c)g = m(c)a

Do i just plug them all tother and solve for m(c)

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2. Oct 19, 2009

### Staff: Mentor

Re: Tension

OK. (Looks like you switched labels for the strings.)

Careful with signs. Make sure "a" is going in the same direction in all equations.

You could. Just plug the given values into each equation and see what you end up with.

3. Oct 25, 2009

### joemama69

Re: Tension

since it is moving with constant speed, that makes a = 0.

but since a is 0, doesnt that put it in equalibrium and make its not moving

Part B a = g/2

A = T(1) - m(A)g = 1/2 m(A)g
B = T(2) - T(1) - m(B)gcosQ = 1/2 m(B)g
C = m(c)g - T(2) = 1/2 m(c)g

T(1) = 1/2 m(A)g + m(A)g
T(2) = m(c)g - 1/2 m(c)g

m(c)g - 1/2 m(c)g - 1/2 m(A)g - m(A)g - m(B)gcosQ = 1/2 m(B)g

m(c) = m(B) + 3m(A) + 2m(B)cosQ

Last edited: Oct 25, 2009
4. Oct 25, 2009

### Staff: Mentor

Re: Tension

Right.
Since a = 0 it is in equilibrium, but that doesn't mean it's not moving. (It's moving with constant velocity.) What's the net force in this case?

5. Oct 25, 2009

### joemama69

Re: Tension

so then Part A

A = T(1) - m(A)g = 0
B = T(2) - T(1) - m(B)gcosQ = 0
C = m(C)g - T(2) = 0

T(1) = m(A)g
T(2) = m(C)g

m(A)g - m(C)g - m(B)gcosQ = 0

m(C) = m(B)cosQ - m(A)

6. Oct 26, 2009

### Staff: Mentor

Re: Tension

All good.

You have the first two terms reversed. (Compare to your equation for B above.)