Blocks moving on frictionless incline at constant speed, accleration

AI Thread Summary
The discussion focuses on solving two parts of a physics problem involving blocks on a frictionless incline. In Part A, the goal is to determine the mass of block C required for block B to move up the incline at constant speed, leading to the conclusion that the system is in equilibrium with net forces equal to zero. Part B requires finding the mass of block C for block B to accelerate at g/2, with the equations indicating that tensions and gravitational forces must balance. Participants emphasize the importance of consistent sign conventions and clarify that constant speed implies zero acceleration, not a lack of movement. The final equations derived for both parts reflect the relationships between the masses and the incline's angle.
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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 accleration 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)
 

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joemama69 said:
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.
 


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
 
Last edited:


joemama69 said:
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?
 


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)
 


joemama69 said:
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.)
 
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