InFigure 9.21, the cg of the pole held by the pole vaulter is \(2.00\;{\rm{m}}\)from the left hand, and the hands are \(0.700\;{\rm{m}}\) apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole inFigure 9.19, show that the second condition for equilibrium(net\(\tau = 0\))is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above.

The distance from the cg to the left handis \(2.00\;{\rm{m}}\).

The hands are \(0.700\;{\rm{m}}\) apart.

The diagram of the person is shown below:

1. From the condition of equilibrium under the forces we get,

\({F_L} = {F_R} + w\;......(1)\)

Considering the equilibrium under torque we get,

\(\begin{align}{F_R} \times 0.7 &= w \times 2\\{F_R} &= \frac{{5 \times 9.8 \times 2}}{{0.7}}\\{F_R} &= 140\;{\rm{N}}\end{align}\)

b. From equation (1) we can write,

\(\begin{align}{F_L} &= {F_R} + w\\ &= 140 + 5 \times 9.8\\ &= 189\;{\rm{N}}\end{align}\)

When the weight is supported by each hand, we can write,

\({F_L} + {F_R} = w\)

The sum of the force of the left hand and the right hand are the same. So,

\(\sum F = 0\)

Let, the distance from the distance of the right hand from the CG is \({r_r}\) and the distance of the left hand from CG is \({r_l}\).

For the equilibrium of torque,

\({F_R}{r_r} - {F_L}{r_l} = 0\)

Given that the force acts due to the left hand and the right hand are the same and they are at the same distances. So,

\(\sum {\tau = 0} \)

Hence, the condition of static equilibrium is followed here.