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Blocks moving on frictionless incline at constant speed, accleration

  1. Oct 18, 2009 #1
    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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  3. Oct 19, 2009 #2

    Doc Al

    User Avatar

    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.
     
  4. Oct 25, 2009 #3
    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
  5. Oct 25, 2009 #4

    Doc Al

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    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?
     
  6. Oct 25, 2009 #5
    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)
     
  7. Oct 26, 2009 #6

    Doc Al

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    Staff: Mentor

    Re: Tension

    All good.

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