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Finding the Tension

  1. Jun 25, 2015 #1
    Solved Thank You : )
    1. The problem statement, all variables and given/known data

    A rigid, uniform, horizontal bar of mass m1 and length L is supported by two identical massless strings. (Figure 1) Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass m2 is supported against gravity by the bar at a distance x from the left end of the bar, as shown in the figure.

    Throughout this problem positive torque is that which spins an object counterclockwise. Use g for the magnitude of the acceleration due to gravity.

    m1
    L
    dA < L/2
    m2
    dm2 = x
    g

    Find TA, the tension in string A.

    2. Relevant equations
    Στ = Iα = F × r
    Ibar = 1/12 m1L2

    3. The attempt at a solution
    the rod isn't moving so
    Στ = 0

    If I make the point of rotation the location of string B
    Στ = 0 = - τbar - τm2 + τA

    τbar = Fg × L/2 = m1g(L/2)
    τm2 = Fg × x = m2gx

    τA = T × dA = TdA(sin90)
    = τbar + τm2
    = m1g(L/2) + m2gx

    dA = this is a known variable so
    TAd = m1g(L/2) + m2gx
    TA = [m1g(L/2) + m2gx]/d

    Conclusion:
    Reason for error- the answer did not require the cross product to have sinθ
     
    Last edited: Jun 25, 2015
  2. jcsd
  3. Jun 26, 2015 #2
    All forces are 90 degrees cross to lengths. ##\sin(90)=1## for all. Your calculation is correct. The rope of T according to B point must be opposite to total masses rope ##\tau_{m_2}+\tau_{bar}##.
     
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