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Torque degrees homework

  1. Apr 27, 2008 #1
    Please can someone help me with these two problems:

    A man is holding an 8.00 kg vacuum cleaner at arm's length, a distance of 0.550 m from his shoulder. What is the torque on the shoulder joint if the arm is held at 30.0 degrees below the horizontal?


    A 15.0 kg child sitting on a playground teeter-totter, 1.50 m from the pivot. What force, applied 0.300 m on the other side of the pivot, is needed to make the child life off the ground?
     
  2. jcsd
  3. Apr 27, 2008 #2

    Astronuc

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    Staff: Mentor

    Students are expected to show effort and work out the problems.

    What is the formula for a torque?

    What is the force (weight) of a 15 kg mass in the earth's gravity field? What is the moment produced by that force at 1.5 m? The other moment must exceed that moment in order to lift the child.
     
  4. Apr 27, 2008 #3
    oh sorry about that.

    1) Well I know that T=Fsin(angle)r
    and when I set it up T=(8.00kg x 9.8)(sin 30.0)(0.550m)=21.6 Nm

    That is the answer I originally had but according to the sheet I am working from the answer is 37.4 Nm...did I do something wrong in my calculations or in seting up the problem?


    2) I=MR^2=(15.0kg)(1.50m)^2=33.75
    I am confused as to how to find the moment of the other force...
     
  5. Apr 27, 2008 #4

    alphysicist

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    Homework Helper

    Hi sruthis,

    For 1) I believe the problem is the angle/trig function you chose. How did you choose sin(30 degrees)?
     
  6. Apr 27, 2008 #5
    well the torque equation uses sin and when you draw out a picture you will notice that sine works best for the Y component of the Force.
     
  7. Apr 27, 2008 #6

    alphysicist

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    There are several different ways of thinking about how to calculate the torques, so I'm not sure which one you're thinking about.

    The definition of a torque that I think they first teach is:

    [tex]
    \tau =\pm F r \sin\theta
    [/tex]

    where theta is the angle between the force and the moment arm, and the sign is given by the direction of the torque.

    So here the force is vertically downwards, and the moment arm is 30 degrees below the horizontal. What's the angle between those two directions?

    (Some other ways of figuring torques is by finding the component of F that is perpendicular to r, or alternatively finding the component of r that is perpendicular to F. They will all give the same answer, of course.)
     
  8. Apr 27, 2008 #7
    Oh i forgot to factor in the perpendicular. Thank you!
     
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