Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

A person carries a plank of wood \({\rm{2}}\;{\rm{m}}\) long with one hand pushing down on it at one end with a force \({{\rm{F}}_{\rm{1}}}\) and the other hand holding it up at \({\rm{50}}\;{\rm{cm}}\) from the end of the plank with force \({{\rm{F}}_{\rm{2}}}\). If the plank has a mass of \({\rm{20}}\;{\rm{kg}}\) and its center of gravity is at the middle of the plank, what are the magnitudes of the forces \({{\rm{F}}_{\rm{1}}}\) and \({{\rm{F}}_{\rm{2}}}\)?

The upward force is \({{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}\).

The downward force is \({{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Force

An external factor that can change the rest or motion of a body is the force.

Step 2: Diagram

The mass is \({\rm{m = 20}}\;{\rm{kg}}\).

The length of the plank is \({{\rm{r}}_{\rm{1}}}{\rm{ = 2}}{\rm{.00}}\;{\rm{m}}\).

The distance between the forces is \({\rm{r = 0}}{\rm{.5}}\;{\rm{m}}\).

The free-body Diagram

Step 3: Explanation

Zero linear acceleration and zero rotational acceleration are the first and second conditions, respectively.

Torque expression,

\({\rm{\tau = }}{{\rm{r}}_ \bot }{\rm{ \times F (1)}}\)

We begin by calculating the torque exerted by the plank's weight by putting our values for \({\rm{w = mg}}\) and \({{\rm{r}}_{\rm{w}}}{\rm{ = l/2}}\) into equation \({\rm{(1)}}\), yielding:

\(\begin{array}{c}{{\rm{\tau }}_{\rm{w}}}{\rm{ = - }}\left( {\frac{{{\rm{2}}{\rm{.0\;m}}}}{{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{20\;kg \times 9}}{\rm{.8\;m/}}{{\rm{s}}^{\rm{2}}}} \right)\\{\rm{ = - 392\;N \times m}}\end{array}\)

\({{\rm{F}}_{\rm{1}}}\) produces a torque of,

\(\begin{array}{c}{{\rm{\tau }}_{\rm{1}}}{\rm{ = }}{{\rm{r}}_{\rm{1}}}{\rm{ \times }}{{\rm{F}}_{\rm{1}}}\\{\rm{ = 0 \times }}{{\rm{F}}_{\rm{1}}}\\{\rm{ = 0}}\end{array}\)

And \({{\rm{F}}_{\rm{2}}}\) produces a torque of,

\(\begin{array}{c}{{\rm{\tau }}_{\rm{2}}}{\rm{ = (0}}{\rm{.50\;m) \times }}{{\rm{F}}_{\rm{2}}}\\{\rm{ = 0}}{\rm{.50 }}{{\rm{F}}_{\rm{2}}}\end{array}\)

The second condition is satisfied since the system is in equilibrium. So, we get,

\(\begin{array}{c}\sum {\rm{\tau }} {\rm{ = }}{{\rm{\tau }}_{\rm{1}}}{\rm{ + }}{{\rm{\tau }}_{\rm{2}}}{\rm{ + }}{{\rm{\tau }}_{\rm{w}}}\\{\rm{ = 0}}\\{\rm{ - 392 + 0 + 0}}{\rm{.50}}{{\rm{F}}_{\rm{2}}}{\rm{ = 0}}\\{{\rm{F}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{392}}}}{{{\rm{0}}{\rm{.50}}}}\\{\rm{ = 784\;N}}\\{{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}\end{array}\)

The first condition has been met. So, we get,

\(\begin{array}{c}\sum {\rm{F}} {\rm{ = - }}{{\rm{F}}_{\rm{1}}}{\rm{ + }}{{\rm{F}}_{\rm{2}}}{\rm{ - mg}}\\{\rm{ = 0}}\\{{\rm{F}}_{\rm{1}}}{\rm{ = }}{{\rm{F}}_{\rm{2}}}{\rm{ - mg}}\\{\rm{ = 784 - (20 \times 9}}{\rm{.8)}}\\{\rm{ = 588\;N}}\\{{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}\end{array}\)

Hence, the forces are \({{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}\) and \({{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}\).

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