Determine the force

Question

A person lifts a 8-kg dumbbell. The combined mass and length of his hand and forearm are 4 kg and 0.4 m respectively. Determine the force exerted by his muscles 0.01 m from the pivot point? Assume that the center of gravity of the forearm is in the middle.

Answer 1

You can make a free-body diagram that looks something like this:

Where W_D is the weight of the dumbbell, W_A is the weight of the arm/hand, And F is the force the muscle applies. Now sum the moments around the pivot point:

∑M = 0 = 0.4•W_D + 0.2•W_A - 0.01F

W_D = mg = 8(9.8) = 78.4 N

W_A = mg = 4(9.8) = 39.2 N

So (0.4)(78.4) + 0.2(39.2) - 0.01F = 0

31.36 + 7.84 - 0.01F = 0

39.2 = 0.01F

F = 39.2/0.01 = 3920 N