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Hammer/nail Moment

  • Thread starter BMcC
  • Start date
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The handle of a hammer is pushed with a force F = 103 N, at an angle θ = 12°, to pull out the nail at B. What is the moment of this force about point A, if L1 = 0.054 m and L2 = 0.32 m?

My attempt

So what I've been doing is finding the Force that is perpendicular to the two lengths, 0.054m and 0.32m.

The force perpendicular to 0.054m is Fsinθ = 21.41 N
The force perpendicular to 0.32m is Fcosθ = 100.75 N

Now I've been using the formula Momenttotal = Moment 1 + Moment 2

M1 = (21.41N)*(0.054m) coming out of the page = 1.156 J
M2 = (100.75N)*(0.32) going into the page = 32.24 J

Then I've been going 32.24 - 1.156 = 31.084 J (going into the page).



This answer hasn't been working for me. Any ideas? I know for certain that the moment is indeed pointing into the page, in the negative k-hat direction in this case.
 

Attachments

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Both components seem to rotate the hammer clockwise around point A. So they should point the same way. just try to imagine how the hammer will rotate if only one component is acting.
 

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