Question

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?

Short answer

(a) The tension is\({\rm{21}}{\rm{.6}}\;{\rm{N}}\).

(b) The force exerted by the hinge is \({\rm{21}}{\rm{.6}}\;{\rm{N}}\) and only horizontally.

Step-by-step solution

Equilibrium of Force

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

Diagram

The diagram is shown below

Fig. 9.35

Calculation of the force

1. The net force is zero in equilibrium. So,

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

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 \({\rm{21}}{\rm{.6}}\;{\rm{N}}\).

Calculation of the force

1. The net force,

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

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.