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Found in: Page 344

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 AA=2AB; the Young’s moduli of the cylinders are related by EA=2EB. 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 dA for cylinder A and dB for cylinder B. (c) What is the ratio dA/dB ?Figure:

1. The fraction of bricks mass supported by cylinder is 0.80
2. The fraction of bricks mass supported by cylinder is 0.20
3. The ratio ${\mathrm{d}}_{\mathrm{A}}/{\mathrm{d}}_{\mathrm{B}}$ is 0.25
See the step by step solution

## Step 1: Determine the given quantities

The areas of faces of the cylinders are related by ${\mathrm{A}}_{\mathrm{A}}=2{\mathrm{A}}_{\mathrm{B}}$ and Young moduli of the cylinders by ${\mathrm{E}}_{\mathrm{A}}=2{\mathrm{E}}_{\mathrm{B}}$

## 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 ${{\mathbf{d}}}_{{\mathbf{A}}}{\mathbf{/}}{{\mathbf{d}}}_{{\mathbf{B}}}$ respectively.

Formula:

$\mathrm{E}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{I}/\mathrm{L}}$ ….. (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" $\mathrm{\zeta }=\mathrm{Fd}$ ….. (ii)

Here, data-custom-editor="chemistry" $\mathrm{\zeta }$ 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:

${\mathrm{I}}_{\mathrm{A}}=\frac{{\mathrm{F}}_{\mathrm{A}}{\mathrm{L}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{A}}{\mathrm{E}}_{\mathrm{A}}}$ …… (iii)

Similarly, for cylinder B solve as:

data-custom-editor="chemistry" ${\mathrm{I}}_{\mathrm{B}}=\frac{{\mathrm{F}}_{\mathrm{B}}{\mathrm{L}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}$ …… (iv)

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

$\begin{array}{rcl}\frac{{\mathrm{F}}_{\mathrm{A}}}{{\mathrm{F}}_{\mathrm{B}}}& =& \frac{{\mathrm{A}}_{\mathrm{A}}{\mathrm{E}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}\\ & =& \frac{\left(2{\mathrm{A}}_{\mathrm{B}}\right)\left(2{\mathrm{E}}_{\mathrm{B}}\right)}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}\\ & =& 4\end{array}$

Consider the equation as:

data-custom-editor="chemistry" ${\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}=\mathrm{W}$

Solve further as:

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

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:

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

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:

${\mathrm{F}}_{\mathrm{A}}{\mathrm{d}}_{\mathrm{A}}={\mathrm{F}}_{\mathrm{B}}{\mathrm{d}}_{\mathrm{B}}$

Substitute the values and solve as:

$\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=\frac{{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{F}}_{\mathrm{A}}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=0.25$N

Therefore, the ratio $\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}$ is 0.25.