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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

Calculate the magnitude and direction of the force on each foot of the horse in Figure \({\rm{9}}{\rm{.31}}\) (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is \({\rm{500}}\;{\rm{kg}}\). (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal.

  1. The force on the foot is \({\rm{2553}}{\rm{.93}}\;{\rm{N}}\) inclined \({\rm{73}}{\rm{.73^\circ }}\) to the ground.
  2. The coefficient of friction is \({\rm{0}}{\rm{.292}}\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept

Torque depends on the force on an object and the torque about a point is always constant.

Step 2: Calculation of the horizontal force

  1. The mass of the horse and the rider is \({{\rm{m}}_{\rm{1}}}{\rm{ = 500}}\;{\rm{kg}}\).

The perpendicular distance from the center of mass of horse and rider to toe of the horse is \({{\rm{r}}_{\rm{1}}}{\rm{ = 0}}{\rm{.35}}\;{\rm{m}}\).

The perpendicular distance from toe of the horse to the force exerted by wall is \({{\rm{r}}_{\rm{2}}}{\rm{ = 1}}{\rm{.2}}\;{\rm{m}}\) .

The torque due to the weight of the rider and the horse is,

\(\begin{array}{c}{\rm{\tau = }}{{\rm{m}}_{\rm{1}}}{\rm{g}}{{\rm{r}}_{\rm{1}}}\\{\rm{ = 500 \times 9}}{\rm{.81 \times 0}}{\rm{.35}}\\{\rm{ = 1716}}{\rm{.75}}\;{\rm{N \times m}}\end{array}\)

The torque at the toe of horse due to the horizontal force is,

\({\rm{\tau ' = }}{{\rm{F}}_{{\rm{wall}}}}{\rm{ \times 1}}{\rm{.2}}\)

Now, in equilibrium, the torques will be the same. So, the horizontal force is,

\(\begin{array}{c}{{\rm{F}}_{{\rm{wall}}}}{\rm{ = }}\frac{{{\rm{1716}}{\rm{.75}}}}{{{\rm{1}}{\rm{.2}}}}\\{\rm{ = 1430}}{\rm{.42}}\;{\rm{N}}\end{array}\)

Thus, the horizontal force is \({\rm{1430}}{\rm{.42}}\;{\rm{N}}\).

Step 3: Calculation of the components of the force

The center of mass of the horse is midway between the feet. The weight of horse and the rider is balanced by the force on feet.

Let, the force on each foot is F and the inclination with the ground is \({\rm{\theta }}\).

The horizontal component of the force on feet balances the force on the wall. The vertical component balances the weight of the rider and the horse.

So,

\(\begin{array}{c}{\rm{2Fcos\theta = 1430}}{\rm{.42}}\\{\rm{Fcos\theta = 715}}{\rm{.1}}\end{array}\)

And,

\[\begin{array}{c}{{\rm{m}}_{\rm{1}}}{\rm{g = 2Fsin\theta }}\\{\rm{Fsin\theta = }}\frac{{{\rm{500 \times 9}}{\rm{.81}}}}{{\rm{2}}}\\{\rm{Fsin\theta = 2450}}\end{array}\)

\[{\rm{tan\theta = }}\frac{{{\rm{2450}}}}{\begin{array}{l}{\rm{715}}{\rm{.1}}\\{\rm{\theta }} \approx {\rm{73}}{\rm{.73^\circ }}\end{array}}\)

The force is,

\(\begin{array}{c}{\rm{F = }}\frac{{{\rm{715}}{\rm{.1}}}}{{{\rm{cos73}}{\rm{.73^\circ }}}}\\{\rm{ = 2553}}{\rm{.93}}\;{\rm{N}}\end{array}\)

Hence, the force is \({\rm{2553}}{\rm{.93}}\;{\rm{N}}\).

Step 4: Calculation of the coefficient of friction

  1. The reaction force on 1 foot is,

\(\begin{array}{c}{\rm{F = }}\frac{{{\rm{500 \times 9}}{\rm{.81}}}}{{\rm{2}}}\\{\rm{ = 2452}}{\rm{.5}}\;{\rm{N}}\end{array}\)

The coefficient of friction is,

\(\begin{array}{c}{\rm{\mu = }}\frac{{{\rm{715}}{\rm{.21}}}}{{{\rm{2452}}{\rm{.5}}}}\\{\rm{ = 0}}{\rm{.292}}\end{array}\)

Hence, the coefficient of friction is \({\rm{0}}{\rm{.292}}\).

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