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Homework Help: Tension Problem - Kinematics

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 921 N and 1330 N, respectively. The acceleration of the cable is 0.620 m/s2, upward. What is the tension in the cable (a) below the worker and (b) above the worker?

    2. Relevant equations


    3. The attempt at a solution

    The tension below the worker is (921+1330)*.62=1395.62 N ; I came across this answer by mistake but found it to be correct. However, how does one find the tension in the rope above the worker? Why wouldn't it be identical to the tension below him, since both weights are being pulled downward by gravity?
  2. jcsd
  3. Sep 9, 2007 #2


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    Can you describe or post the picture?
  4. Sep 9, 2007 #3
    | <-- massless cable
    | ^ upward acceleration = .620 m/s^2
  5. Sep 9, 2007 #4


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    Try drawing free body diagrams. The cable under the worker is only supporting the crate, whilst the cable above the worker supports both the worker and the crate. The acceration would the vector sum of the upward one provided by the cable and the downward one of the earth's gravity.
  6. Sep 9, 2007 #5
    What equation should I be using to find tension? Is it the same as the usual force equation?
  7. Sep 9, 2007 #6


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    I'm not getting 1395.62 N... I get a close number but not the same number.

    Use the freebody diagram of the crate first. What are the forces acting on the crate? sum of forces = ma (where m is the mass of the crate)
  8. Sep 9, 2007 #7
    Forces acting on the crate are the pull of gravity (giving it a weight of 1330 N) and the acceleration upward due to the rope (135.7 (mass of crate) * .62 m/s^2) = 84.14 N.

    1330-84.14 = 1245.86, so this is the net force acting on the crate?
  9. Sep 9, 2007 #8


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    net force acting on the crate is ma where m is the mass of the crate and a is the acceleration. careful about signs... the acceleration is upward. use upward positive, downward negative:

    Ftension - 1330 = ma

    so solve here for Ftension.
  10. Sep 9, 2007 #9
    Then 135.7*.62 + 1330 = tension. Why do you add 1330 to the tension?
  11. Sep 9, 2007 #10


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    I don't understand. We didn't add to the tension... the tension is 135.7*.62 + 1330 = 1414.1N.

    A way to think about it... The tension has to compensate for the weight of the crate, and also allow for the 0.62m/s^2 acceleration... If there was no weight... then the tension would just be ma.
  12. Sep 9, 2007 #11
    So tension above the worker then is his mass (93.98kg) * .62 + the forces of his weight and the crate's weight?

    93.98 * .62 + 921 + 1330 =2309.27 N
  13. Sep 9, 2007 #12


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    careful... don't assume anything. go with only what you know for sure.

    What exactly are the forces acting on the worker? There's the tension in the rope above... the tension in the rope below acting downward (which you calculated) and the worker's own weight.

    So do F = ma, for the worker using the sum of those 3 forces.
  14. Sep 9, 2007 #13
    So for the tension above the worker, F = (93.98 + 135.7)*.62 = 142.40 N

    For the tension below, it's -1414.1 N

    And for the weight of the worker, it's -921 N.

    -921 - 1414.1 + 142.40 = -2192.7 N

    Is that more on the right track?
  15. Sep 9, 2007 #14


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    Remember ma is the net force... it is the sum of all the individal forces acting on the worker... and the mass is only 93.98.

    ma should be on the right side of your equation. The 3 forces being added should be on the left.

    [tex]\Sigma{F} = m_{worker}a[/tex]

    [tex]\Sigma{F} = 93.98(0.62)[/tex]

    so put in the 3 forces on the left hand side being careful about signs... you'll have tension above as a variable which you can solve for.
  16. Sep 9, 2007 #15
    But why doesn't the tension above the worker also include the weight of the crate? If both are below the point where tension is being measured, then both weights are pulling downward on the rope. Which means that above and below should be equal, but they're not. I don't quite understand that concept.
  17. Sep 9, 2007 #16


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    Ok... hope I didn't confuse you... there are two ways to solve the problem... I think you were thinking of the second way... which is actually the easier way. I didn't realize... sorry about that.

    You're right... you can take the freebody diagram of the entire system of the worker and the crate together... draw a circle around the system... and the only forces acting on that system are the tension above... the worker's weight and the crate's weight. But then the mass you need to use is the sum of the two masses.

    Ok. The first way which I was saying taking the freebody diagram of the worker:

    Tabove - 921 - 1414 = 93.68*0.62

    that gives Tabove = 2393N

    Now the other way taking the freebody diagram of the worker-crate system (the 1414N rope is inside the freebody diagram... it isn't an external force on this system, so it doesn't get counted... we only count the forces acting on the system from the outside...externally):

    Tabove - 921 - 1330 = (93.68+135.7)*0.62

    also gives Tabove = 2393N

    Both ways are equivalent... Try both ways yourself...
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