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Moment about a point of a hammer

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data

    Probs._48_9.jpg
    In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 590 lb⋅in about point A.(Figure 1)

    Determine the required magnitude of force F.
    2. Relevant equations


    3. The attempt at a solution

    M=590lb*in

    -590= -Fcos(30)(18)-Fsin(30)(5)

    I'm not sure how to deal with F. I know it's simple if given the force to determine the moment but I'm a bit stumped trying to go the other way.
     
  2. jcsd
  3. Oct 6, 2016 #2

    haruspex

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    How would you solve 3=x+2x?
     
  4. Oct 6, 2016 #3
    So, F=72.9431?
     
  5. Oct 6, 2016 #4

    haruspex

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    Not what I get. Please post your working.
     
  6. Oct 6, 2016 #5

    andrewkirk

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    Let ##X## be the point where the force is applied to the handle.

    The angle you need to use is not 30 degrees but the angle ##\alpha## between the force vector and the perpendicular to the line segment ##\overline{XA}##. You can calculate that angle using geometry, from the information given.

    The Torque (moment) will be the force multiplied by ##\cos\alpha## and the length of ##\overline{XA}##.

    By the way, that's one weird hammer! Based on the drawing and the dimensions given, the claw is about eight inches long! That's a gemmy, not a hammer. The drawing is not to scale. The 18 in length is not 18/5 times the 5 inch length. That's why the hammer doesn't look weird.
     
  7. Oct 6, 2016 #6

    Sorry, -590=-F18cos(30)-5Fsin30)=32.6175
     
  8. Oct 6, 2016 #7

    haruspex

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    Yes.
     
  9. Oct 6, 2016 #8

    haruspex

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    the method Robb used is valid, and quite efficient.
     
  10. Oct 6, 2016 #9
    Dang, nothing like making a mountain out of a mole hill!
     
  11. Oct 6, 2016 #10

    andrewkirk

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    So it is. I hadn't noticed that.
     
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