Chapter 12, Problem 44SP

A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3.0 cm and a cross-sectional area of 15 ${\text{cm}}^{\text{2}}$. The shear modulus for this rubber is 2.0 MPa. (a) If a sideways force of 300 N is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

To determine

The distance moved by the rubbers along sideways if a sideways force of $300\text{ N}$ is applied to the motor.

Answer to Problem 44SP

Solution:

$0.075\text{ cm}$

Explanation of Solution

Given data:

There are four cylindrical rubber blocks and a $60\text{-kg}$ motor sits on them.

The sideway force on the motor is $300\text{ N}$.

The height of the cylinder is $3.0\text{ cm}$.

The cross-sectional area of the cylinder is $15{\text{ cm}}^2$.

The shear modulus of the rubber is $2.0\text{ mPa}$.

Formula used:

Write the expression for Young’s modulus $Y$:

$Y=\frac{F{L}_0}{A\Delta L}$

Here, $F$ is the force, $L_0$ is the original length, $A$ is the crosssection of the area, and $\Delta L$ is the change in length.

Explanation:

Understand that the rubber blocks are at the same distance from the sideways force. Therefore, the force will equally be distributed to each rubber blocks.

Therefore, the forcesacting on each block is as follows:

$\begin{array}{c}F=\frac{300\text{ N}}{4}\\ =75\text{ N}\end{array}$

Recall the expression for Young’s modulus $Y$:

$Y=\frac{F{L}_0}{A\Delta L}$

Substitute $15{\text{ cm}}^2$ for $A$, $75\text{ N}$ for $F$, $3.0\text{ cm}$ for $L_0$, and $2.0\text{ mPa}$ for $Y$

$\begin{array}{c}2.0\text{ mPa}=\frac{\left(75\text{ N}\right)\left(3.0\text{ cm}\right)}{15{\text{ cm}}^2\left(\Delta L\right)}\\ \Delta L=\frac{\left(75\text{ N}\right)\left(3.0\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{\left(15{\text{ cm}}^2\right)\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\left(2.0\text{ mPa}\right)\left(\frac{{10}^6\text{ Pa}}{1\text{ mPa}}\right)}\\ \Delta L=7.5\times {10}^{-4}\text{ m}\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ \Delta L=0.075\text{ cm}\end{array}$

Conclusion:

The distance moved due to the sideways force is $0.075\text{ cm}$.

To determine

The frequency with which the $60\text{ kg}$ motor vibrates.

Answer to Problem 44SP

Solution:

$13\text{ Hz}$

Explanation of Solution

Given data:

There are four cylindrical rubber blocks and a $60\text{-kg}$ motor sits on them.

The sideway force on the motor is $300\text{ N}$.

The height of the cylinder is $3.0\text{ cm}$.

The area of the cylinder is $15{\text{ cm}}^2$.

The shear modulus of the rubber is $2.0\text{ mPa}$.

The mass of the motor is $60\text{ kg}$.

Formula used:

Write the expression for spring constant:

$k=\frac{F}{\Delta L}$

Here, $F$ is the force and $\Delta L$ is the change in length.

Write the expression for frequency:

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Here, $m$ is the mass and $k$ is the spring constant.

Explanation:

The forces that causes the back and forth motion is equivalent to the total forces acting as the sideways force of $300\text{N}$.

Recall the expression for equivalent spring constant:

$k=\frac{F_{\text{s}}}{\Delta L}$

Here, $F_{\text{s}}$ is the total of all forces and $\Delta L$ is the change in the length.

Substitute $300\text{ N}$ for $F_{\text{s}}$ and $0.075\text{ cm}$ for $\Delta L$

$\begin{array}{c}k=\frac{\text{300 N}}{0.075\text{ cm}}\\ =\frac{300\text{ N}}{0.075\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\\ =4.0\times {10}^5\text{ N}\cdot \text{m}\end{array}$

Recall the expression for frequency:

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Substitute $60\text{ kg}$ for $m$ and $4.0\times {10}^5\text{ N}\cdot \text{m}$ for $k$

$\begin{array}{c}f=\frac{1}{2\pi }\sqrt{\frac{4.0\times {10}^5\text{ N}\cdot \text{m}}{60\text{ kg}}}\\ =13\text{ Hz}\end{array}$

Conclusion:

The frequency of the motor with which itvibrates back and forth is $13\text{ Hz}$.