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Torque and Equilibrium

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the mass necessary for equilibrium to occur in the following image. Assume that the mass and the friction of the pulleys are negligible.

    giiiiirl-1.png

    2. Relevant equations

    None directly provided for this problem.

    3. The attempt at a solution

    I figured for this problem I would set the sum of the net torques on the pulley equal to zero, because then the pulley is not spinning and the system is in equilibrium.

    Here is my equation:

    gravity = 9.8 m/s

    Since torque is the cross product of the force times the perpendicular distance from the axis of rotation, I used counterclockwise torque (provided by the truck) = F*rsin(x), but since the force is already perpendicular to the axis of rotation in the image I just used F*r, where the F = 1500*9.8*sin(45). The clockwise torque on the pulley is m*g*3r

    ƩT = 1500kg*9.8*sin(45)*r - m*g*3r = 0

    add m*g*3r to both sides and then divide by r to obtain:

    ƩT = 1500kg*9.8*sin(45) = m*g*3

    solving for the mass I get 353 kg, but the answer is 178 kg.

    I noticed that multipying the right side by 6 instead of 3 yields the right answer, I just don't know why or what I'm doing wrong.
     
  2. jcsd
  3. Dec 11, 2012 #2

    gneill

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    Staff: Mentor

    How many cables support the truck's weight?
     
  4. Dec 11, 2012 #3
    I don't know, the question is literally this :

    "Find the mass necessary for the truck to be balanced on the slope in the figure. Assume that the mass and frictions of the pulleys are negligible."

    That is all it says, no information regarding how many cables or the like is given.
     
  5. Dec 11, 2012 #4

    gneill

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    Staff: Mentor

    Doesn't the picture make it clear? How many cables with tension T are holding the truck?

    (Think: mechanical advantage)
     
  6. Dec 11, 2012 #5
    Ah, I thought you were asking me as if you needed more information! So since two cables are supporting the truck, I divide my answer by two and get approximately 178 kg.
     
  7. Dec 11, 2012 #6

    gneill

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    Staff: Mentor

    That looks about right. Well done.
     
  8. Dec 11, 2012 #7
    Thank you for the eye opener :-)
     
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