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

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,

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