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Q15PE

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Found in: Page 317

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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# What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.35 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge?

1. The coefficient of friction is $$0.551$$.
2. The force exerted by each side on the hinge is $${\rm{21}}{\rm{.6}}\;{\rm{N}}$$
See the step by step solution

## Step 1: Given Data

Given:

The sign’s mass is $$m = 8\;{\rm{kg}}$$.

The diagram is shown below:

## Step 2: Calculation of the force

The net force is zero in equilibrium. So,

$${N_L} + {N_r} - {w_s} = 0$$

Normal forces should be equal by symmetry. So,

$${N_L} = {N_r} = N$$

We write,

\begin{align}2N - {w_s} &= 0\\N = \frac{{{w_s}}}{2}\\N &= \frac{{8 \times 9.8}}{2}\\N &= 39.2\;{\rm{N}}\end{align}

The total tension on each leg is$$39.2\;{\rm{N}}$$.

So, this tension is equal to the weight of the sign board.

$$39.2 \times 2 = 78.4\;{\rm{N}}$$

The net force,

$${F_t}\sin \varphi - N - w = 0$$

And

$${F_t}\cos \varphi = T$$

So,

\begin{align}\sin \varphi &= 0\\\varphi &= 0^\circ \end{align}

 $${F_t}\cos \varphi = T = 21.6\;{\rm{N}}$$

If the chain breaks, the tension will be zero and the force on the hinge will be balanced by the friction.

\begin{align}\mu N &= {F_t}\cos \varphi \\\mu &= \frac{{{F_t}\cos \varphi }}{N}\\\mu &= \frac{{21.6 \times 1}}{{39.2}}\\\mu &= 0.551\end{align}

## Step 3: Calculation of the force exerted by each hinge

b.

The force calculated in the part (a) is,

$${F_t}\cos \varphi = T = 21.6\;{\rm{N}}$$

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