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Tension of dangling box

  1. Feb 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A 50 kg woman dangles a 50 kg mass from the end of a rope. If she stands on a frictionless surface and hangs the mass over a cliff with a pulley as shown, the tension in the rope will be...


    2. Relevant equations

    F = m*a

    3. The attempt at a solution
    (no calculator allowed, so estimating)
    There are 3 forces acting in this diagram...gravity pulls down on the dangling mass, tension pulls up on the dangling mass, and the woman is pulled horizontally by tension.
    So I want to find the tension pulling up on the mass as well as the tension pulling on the woman, and then add them together.

    For the box:
    F(g) = 50 kg * 10 m/s^2 = 500 N
    This would equal the tension on the woman. But how do I find the tension on the box?

    I'm not even sure this first step was correct. Corrections to my general interpretation are very welcome!
     
  2. jcsd
  3. Feb 11, 2014 #2

    PhanthomJay

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    No that is not correct. You need to isolate the box and woman separately in free body diagrams, identify the forces acting on the box and use newton 2nd law, and do same for woman. The tensions don't add, they are equal when the pulley is massless and frictionless.
     
  4. Feb 11, 2014 #3
    So for the box, there are 2 forces acting:
    Force of gravity = 50kg * (10) = 500N
    Tension = 50 kg * a
    How do I know what acceleration for tension is?

    For the woman, I think there is only one force acting, which would be the tension of the rope.
    T = 50kg * a

    But setting the tensions equal at this point does not give me a...I'm sorry, I'm stuck again.
     
  5. Feb 11, 2014 #4
    Or...since the surface is frictionless, the only force on the woman is in the horizontal direction.
    If that force = 500 due to F(g), does she accelerate at a rate of 10 m/s^2?
    Then tension would be 500N.

    But what about the box?
     
  6. Feb 11, 2014 #5

    PhanthomJay

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    No the tension is not 500 N ,let's step back a minute and look at the net force acting on the woman . You correctly identified just one force acting on her horizontally, the unknown tension force, T. So that is the net force acting on her, and thus, applying Newton 2, T = ma . You have that correct. That is one equation, with 2 unknowns, T and a. So you need another equation, found by looking at the forces acting on the block. What 2 forces act on the block? What is the net force acting on the block, in letter form? Now use Newton 2 to get that 2nd equation and solve 2 equations with 2 unknowns for T and a.
     
  7. Feb 11, 2014 #6
    F(net) on the block
    F = T - mg = m*a - mg = 50*a - 500
    So now I don't know F(net) of the block, or its a. I'm tempted to say that it's -10, but that's acceleration due to gravity. I don't know T either, so I'm not sure where to go again. Am I completely overlooking something?
    I know a is included in T, but since I don't know either, do I technically have 3 unknowns? T, a, and F(net)?
     
  8. Feb 11, 2014 #7

    PhanthomJay

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    Which way will the block move, down or up? Down of course. That means the weight is greater than T. So the NET force acting on the block is mg - T. Down. So use newton 2 , mg -T = ma. But you have an equation for 'a' easily derived from your first equation from the FBD of the woman. Use it in this 2nd equation to solve for T..
     
  9. Feb 11, 2014 #8
    This may be a really silly, but do you assume the block will move down because there is no way it can move up or stay still? If the surface the woman was standing on included friction, would we then worry about if the block could would stay still or not?

    Thank you very much! That was very helpful!
     
  10. Feb 11, 2014 #9
    Tension in a Rope

    Think of the woman and block as a closed system. What external forces act on this system? Gravity pulling the woman down, the ground beneath her pushing back up on her, and gravity pulling the block down. The net external force on the system equals the force of gravity pulling the block down. Fnet = Wblock = 50 kg * 9.8 m/s2 = 490 N. The acceleration of the system equals the quotient of the net external force on the system and its mass. asystem = 490 N / 100 kg = 4.9 m/s2. Assuming the rope is inelastic, the accelerations of the woman and the block are equal in magnitude and perpendicular in direction. That is, the woman accelerates 4.9 m/s2 right, the block 4.9 m/s2 down. The force of tension in the rope is equal to the force that pulls upward on the block or the force that pulls the woman to the right. The force that pulls upward on the block plus the force of gravity on the block equals the net force on the block. Fpull, on block, by rope + Fgravity, on block, by Earth = Fnet on block. Then Fpull, on block, by rope = Fnet on block - Fgravity, on block, by Earth. So, Fpull, on block, by rope = 50 kg * -4.9 m/s2 - 50 kg * -9.8 m/s2, so Fpull, on block, by rope = 245 N upward, and Fpull, on woman, by rope = 245 N right. The tension in the rope is 245 N.
     
  11. Feb 12, 2014 #10

    PhanthomJay

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    good question. The block must accelerate down at the same rate as the woman because the woman is accelerating from the net pulling force T acting on her, and they are connected with the ideal rope. If there were friction however, she might not accelerate and instead she and block might remain still if the friction force was large enough to balance the tension force acting on her. More likely that she would still accelerate with friction but at a slower rate than the frictionless case.
     
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