Posted by Jessica on Thursday, July 12, 2012 at 1:37pm.
The sum of torques about the shoulder is zero =>
mg•0.35 +Mg•0.7 -F•sinα•0.15 =0
F= g(m•0.35 +M•0.7)/ sinα•0.15=
=9.8(2.5•0.35+20•0.7)/0.24•0.15=
=4049 N.
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