Blocks moving on frictionless incline at constant speed, accleration

[joemama69]

Homework Statement

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 acceleration a = g/2

Homework Equations

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)

 

[Doc Al]

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)

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

Block C

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

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

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

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

 

[joemama69]

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

but since a is 0, doesn't 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

 

[Doc Al]

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

Right.

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

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?

 

[joemama69]

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)

 

[Doc Al]

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

All good.

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

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