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Forces due to Tension

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data


    Suppose that in the figure shown in the URL below, the masses of the blocks are M = 2.6 kg and m = 6.6 kg. What are (a) the magnitude of the acceleration and (b) the tension in the cord?


    http://www.google.com/imgres?num=10...5&start=0&ndsp=21&ved=0CHkQrQMwDQ&tx=48&ty=63

    2. Relevant equations

    F= ma



    3. The attempt at a solution

    I found the Fnet, y and Fnet, x for the hanging block (H) and for the sliding block (S).

    For S Fnet, y = May which becomes FN = FgS, and Fnet becomes T= Ma

    For H Fnet, y becomes T-FgH = may, which becomes T-mg= -ma


    I combined the two equations to cancel out T and solve for acceleration.

    T-mg+ma = T- Ma
    -mg+ma = - Ma
    -mg+ma+Ma = 0
    ma+Ma = mg
    a(m+M) = mg
    a = mg/(m+M)

    That gave me the correct answer for (a) but when I went to try to cancel out a and solve for tension

    T = Ma, T-mg = -ma
    a = T/M a= (T-mg)/(-m)

    a-T/M = a-(T-mg)/(-m)
    -T/M = -(T-mg)/(-m)
    (T-mg)/(-m) - (T/M) = 0
    M(T-mg)/(-m) - (-m)(T)/M = 0
    MT-Mmg/(m+M) -mT/(m+M) = 0
    T((M-mg)-m)/(m+M) = 0
    T= -M+mg +m/(m+M)

    That gives a negative number for tension. What did I do wrong?
     
    Last edited: Feb 19, 2012
  2. jcsd
  3. Feb 19, 2012 #2
    Check your algebra

    Solving "a-T/M = a-(T-mg)/(-m)" for T doesn't give -M+mg +m/(m+M)

    You can even see this from some dimensional analysis
    Tension has units of force but on the right hand side you have a mixture of force (mg) and mass terms (-M, -m/(m+M))
     
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