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Q18PE

Expert-verifiedFound in: Page 317

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**In ****Figure 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 in ****Figure 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 force exerted by right hand is \(140\;{\rm{N}}\).
- (b) The force exerted by left hand is \(189\;{\rm{N}}\).
- (c) The condition of static equilibrium is followed here.

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

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

The diagram of the person is shown below:

- 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.

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