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Multiple Pulleys and Masses with Friction

  1. Mar 19, 2012 #1
    Multiple Pulleys and Masses with Friction (Picture Included)

    1. The problem statement, all variables and given/known data
    [​IMG]

    I have solved the question but I am still iffy as to whether my solution is correct or not.

    2. Relevant equations
    Let m1 = 10kg.
    Let m2 = 20kg.

    (1) m1a1 = T1 - μm1g
    (2) m2aa = m2g - T2
    (3) 2T1 = T2
    (4) 2a2 = a1


    3. The attempt at a solution
    T2 = m2g
    T2 = (20kg)(9.8m/s2)
    T2 = 196N

    T1 = [itex]\frac{T_{2} + μm_{1}g}{2}[/itex]
    T1 = [itex]\frac{196N + (0.32)(10kg)(9.8m/s^{2})}{2}[/itex]
    T1 = 113.68N

    Sub (3) and (4) into (2)
    m2a2 = m2g - T2
    m2([itex]\frac{a_{1}}{2}[/itex]) = m2g - 2T1

    Called this new equation (5).
    (5) m2([itex]\frac{a_{1}}{2}[/itex]) = m2g - 2T1


    (1) m1a1 = T1 - μm1g
    (5) m2([itex]\frac{a_{1}}{2}[/itex]) = m2g - 2T1

    (1) + (5)
    m1a1 + m2([itex]\frac{a_{1}}{2}[/itex]) = T1 - μm1g + m2g - 2T1
    a1(m1 + m2([itex]\frac{1}{2}[/itex]) = -μm1g + m2g - T1
    a1(10kg + 20kg([itex]\frac{1}{2}[/itex]) = -(0.32)(10kg)(9.8m/s2)+(20kg)(9.8m/s2)-113.68N
    a1 = 2.5448 m/s2

    from (4)
    2a2 = a1
    a2 = [itex]\frac{a_{1}}{2}[/itex]
    a2 = [itex]\frac{2.548m/s^{2}}{2}[/itex]
    a2 = 1.274 m/s2

    Therefore, T1 = 113.68N, T2 = 196N, a1 = 2.548 m/s2, and a2 = 1.274 m/s2.

    My question is if my solution is correct. Please check and reply back with suggestions or corrections. Thank you so much.
     
    Last edited: Mar 19, 2012
  2. jcsd
  3. Mar 19, 2012 #2
    Please help me with this question. have i done this question correctly?
     
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