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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 Moment_{total} = Moment 1 + Moment 2
M_{1} = (21.41N)*(0.054m) coming out of the page = 1.156 J
M_{2} = (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 khat direction in this case.
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