a vaulter is holding a horizontal 3.00-kg pole, 4.50 m long. his front arm lifts straight up on the pole, o.750 m from the end, and his back arm pushes straight down on the end of the pole.

Question

how much force does his FRONT arm exert on the pole?

(Unit = N)

Answer 1

Sketch the situation:

W = mg = (3.00)(9.81) = 29.43 N

The pole pivots around the front arm (F_FA in the sketch). So, since there is no motion, we can say the sum of the moments (aka torques) equals zero. So:

W(2.25-0.75) = .75F_BA

1.5W = .75F_BA

F_BA = 1.5(29.43)/0.75) = 58.86 N

Then summing the forces in the y-direction:

F_FA = W + F_BA

F_FA = 29.43 + 58.86 = 88.3 N