Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In the Figure, a lead brick rests horizontally on cylinders A and B. The areas of the top faces of the cylinders are related by A_A=2A_B; the Young’s moduli of the cylinders are related by E_A=2E_B. The cylinders had identical lengths before the brick was placed on them. What fraction of the brick’s mass is supported (a) by cylinder A and (b) by cylinder B? The horizontal distances between the center of mass of the brick and the centerlines of the cylinders are d_A for cylinder A and d_B for cylinder B. (c) What is the ratio d_A/d_B ?
Figure:

The fraction of bricks mass supported by cylinder is 0.80
The fraction of bricks mass supported by cylinder is 0.20
The ratio $d_{A} / d_{B}$ is 0.25

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the given quantities

The areas of faces of the cylinders are related by $A_{A} = 2 A_{B}$ and Young moduli of the cylinders by $E_{A} = 2 E_{B}$

Step 2: Determine the concept of torque and Young’s modulus

Using formula for Young’s modulus and torque, we can find the magnitude of the forces on the log from wire A and wire B and the ratio $d_{A} / d_{B}$ respectively.

Formula:

$E = \frac{F / A}{I / L}$ ….. (i)

Here, E is Young’s modulus, F is force, A is area, I is change in length, and L is original length.

Consider the formula for the torque:

data-custom-editor="chemistry" $ζ = Fd$ ….. (ii)

Here, data-custom-editor="chemistry" $ζ$ is torque, F is force, d is perpendicular distance.

Step 3: (a) Determine the fraction of brick mass supported by cylinder A

From equation (i) for cylinder A and solve as:

$I_{A} = \frac{F_{A} L_{A}}{A_{A} E_{A}}$ …… (iii)

Similarly, for cylinder B solve as:

data-custom-editor="chemistry" $I_{B} = \frac{F_{B} L_{B}}{A_{B} E_{B}}$ …… (iv)

The change in the length of both cylinders is the same. Therefore, equate equations (iii) and (iv) and simplify them further as,

$\begin{array}{rcl} \frac{F_{A}}{F_{B}} & = & \frac{A_{A} E_{A}}{A_{B} E_{B}} \\ = & \frac{(2 A_{B}) (2 E_{B})}{A_{B} E_{B}} \\ = & 4 \end{array}$

Consider the equation as:

data-custom-editor="chemistry" $F_{A} + F_{B} = W$

Solve further as:

$\begin{array}{rcl} \frac{F_{A}}{W} & = & \frac{4}{5} \\ = & 0.80 \end{array}$

The fraction of bricks mass supported by cylinder is 0.80

Step 4: (b) Determine the fraction of brick mass supported by cylinder B

Consider the ratio:

$\begin{array}{rcl} \frac{F_{B}}{W} & = & \frac{1}{5} \\ = & 0.20 \end{array}$

The fraction of bricks mass supported by cylinder is 0.20

Step 5: (c) Calculation of horizontal distance between center of mass

Applying torque equation about the center of mass is written as follows:

$F_{A} d_{A} = F_{B} d_{B}$

Substitute the values and solve as:

$\frac{d_{A}}{d_{B}} = \frac{F_{B}}{F_{A}} \frac{d_{A}}{d_{B}} = \frac{1}{4} \frac{d_{A}}{d_{B}} = 0.25$ N

Therefore, the ratio $\frac{d_{A}}{d_{B}}$ is 0.25.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Determine the given quantities

Step 2: Determine the concept of torque and Young’s modulus

Step 3: (a) Determine the fraction of brick mass supported by cylinder A

Step 4: (b) Determine the fraction of brick mass supported by cylinder B

Step 5: (c) Calculation of horizontal distance between center of mass

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Determine the given quantities

Step 2: Determine the concept of torque and Young’s modulus

Step 3: (a) Determine the fraction of brick mass supported by cylinder A

Step 4: (b) Determine the fraction of brick mass supported by cylinder B

Step 5: (c) Calculation of horizontal distance between center of mass

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