a pole-vaulter is holding the vaulting pole parallel to the ground. the pole is

Question

a pole-vaulter is holding the vaulting pole parallel to the ground. the pole is 5 m long. the vaulter grips the pole with his right hand at the top end of the pole and with his left hand 1 m from the top end of the pole. the mass of the pole is 2.5 kg,the forces of the vaulter must excert on the pole to maintain it in this position are quite large. how large are they? ( assume the vulter excerts only vertical forces on the horizontal pole aand that the center of gravity of the pole is located at the center of its length)

Explanation / Answer

Fr+Fl - Fg = 0

Fr+Fl = Fg = 2.5*9.8 = 24.5 N

from top end

net torque = 0

Fr*0.1 + Fl*1 - Fg*2.5 = 0


Fr*0.1 + Fl*1 = Fg*2.5

0.1*(24.5-Fl) + Fl = 24.5*2.5

2.45 - 0.1Fl + Fl = 61.25

0.9Fl = 61.25-2.45

Fl = 65.33 N.........upwards

Fr = 24.5-Fl

Fr = -40.83 N....downwards

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