A sandwich board advertising sign is constructed as shown inFigure\({\rm{9}}{\rm{.35}}\). The sign’s mass is \({\rm{8}}{\rm{.00}}\;{\rm{kg}}\). (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force isexerted by each side on the hinge?

At equilibrium, the total force is zero on a system.

The diagram is shown below

Fig. 9.35

\({{\rm{N}}_{\rm{L}}}{\rm{ + }}{{\rm{N}}_{\rm{r}}}{\rm{ - }}{{\rm{w}}_{\rm{s}}}{\rm{ = 0}}\)

Normal forces should be equal by symmetry. So,

\({{\rm{N}}_{\rm{L}}}{\rm{ = }}{{\rm{N}}_{\rm{r}}}{\rm{ = N}}\)

We write,

\(\begin{array}{c}{\rm{2N - }}{{\rm{w}}_{\rm{s}}}{\rm{ = 0}}\\{\rm{N = }}\frac{{{{\rm{w}}_{\rm{s}}}}}{{\rm{2}}}\\{\rm{N = }}\frac{{{\rm{8 \times 9}}{\rm{.8}}}}{{\rm{2}}}\\{\rm{N = 39}}{\rm{.2}}\;{\rm{N}}\end{array}\)

The tension is calculated as,

\(\begin{array}{c}{\rm{ - T \times 0}}{\rm{.500 - w}}\frac{{{\rm{1}}{\rm{.10}}}}{{\rm{4}}}{\rm{ + N}}\frac{{{\rm{1}}{\rm{.10}}}}{{\rm{2}}}{\rm{ = 0}}\\{\rm{T = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.500}}}}\left( {\frac{{{\rm{39}}{\rm{.2 \times 0}}{\rm{.55}}}}{{\rm{2}}}} \right)\\{\rm{T = 21}}{\rm{.6}}\;{\rm{N}}\end{array}\)

Hence, the tension is \({\rm{21}}{\rm{.6}}\;{\rm{N}}\).

\({{\rm{F}}_{\rm{t}}}{\rm{sin\varphi - N - w = 0}}\)

And

\({{\rm{F}}_{\rm{t}}}{\rm{cos\varphi = T}}\)

So,

\(\begin{array}{c}{\rm{sin\varphi = 0}}\\{\rm{\varphi = 0^\circ }}\end{array}\)

\(\)\(\begin{array}{c}{{\rm{F}}_{\rm{t}}}{\rm{cos\varphi = T}}\\{\rm{ = 21}}{\rm{.6}}\;{\rm{N}}\end{array}\)

Hence, the force is \({\rm{21}}{\rm{.6}}\;{\rm{N}}\)and only horizontally.