91Ó°ÊÓ

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.