Blocks moving on frictionless incline at constant speed, accleration

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Homework Help Overview

The discussion revolves around a physics problem involving blocks moving on a frictionless incline. The original poster seeks to determine the mass of block C required for block B to move up the incline at constant speed and with a specific acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between tension and gravitational forces acting on the blocks. They discuss the implications of constant speed and acceleration on the system's equilibrium and the equations governing the motion of the blocks.

Discussion Status

Participants are actively engaging with the problem, checking each other's equations and reasoning. Some have provided guidance on ensuring consistency in the direction of acceleration and forces, while others are questioning the assumptions about equilibrium and motion.

Contextual Notes

There is an ongoing discussion about the implications of constant speed versus acceleration on the system's state, with some participants noting the need for careful attention to signs in the equations. The problem constraints include the absence of friction and the specific conditions for motion.

joemama69
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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 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)
 

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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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