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College Physics (Urone)
Found in: Page 316
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

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Figure \({\rm{9}}{\rm{.1}}\)
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