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Expert-verified Found in: Page 316 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A sandwich board advertising sign is constructed as shown in Figure $${\rm{9}}{\rm{.35}}$$. 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 is exerted by each side on the hinge?

(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.

See the step by step solution

## Step 1: Equilibrium of Force

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

## Step 2: Diagram

The diagram is shown below Fig. 9.35

## Step 3: 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 $${\rm{21}}{\rm{.6}}\;{\rm{N}}$$.

## Step 4: 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. ### Want to see more solutions like these? 