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Q15PE

Expert-verifiedFound in: Page 317

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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? **

- The coefficient of friction is \(0.551\).
- The force exerted by each side on the hinge is \({\rm{21}}{\rm{.6}}\;{\rm{N}}\)

Given:

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

The diagram is shown below:

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.

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

b.

The force calculated in the part (a) is,

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

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