Chapter 14, Problem 36PQ

A square plate with sides of length 4.0 m can rotate about an axle passing through its center of mass and perpendicular to the plate as shown in Figure P14.36. There are four forces acting on the plate at different points. The rotational inertia of the plate is 24 kg·m². Is the plate in equilibrium?

FIGURE P14.36

To determine

Whether the plate is in equilibrium, if rotational inertia of the plate is $24\text{kg}\cdot {\text{m}}^2$ .

Answer to Problem 36PQ

Since net torque acting on the plate is zero, the pate is in equilibrium.

Explanation of Solution

Take right direction as $+x$ direction, upward direction as $+y$ direction with center of mass of the plate as origin.

From figure P14.36, write the expression for the vector form of force of magnitude $60.0\text{N}$ .

  ${\overrightarrow{\text{F}}}_1=-\left(60.0\text{N}\right)\widehat{\text{j}}$                                                                                                        (I)

Here, ${\overrightarrow{\text{F}}}_1$ is the force of magnitude $60.0\text{N}$ and $\widehat{\text{j}}$ is the unit vector along $y$ axis.

Write the expression for the vector form of force of magnitude $30.0\text{N}$ .

  ${\overrightarrow{\text{F}}}_2=\left(30.0\text{N}\right)\widehat{\text{i}}$                                                                                                         (II)

Here, ${\overrightarrow{\text{F}}}_2$ is the force of magnitude $30.0\text{N}$ and $\widehat{\text{i}}$ is the unit vector along $x$ axis .

Write the expression for the vector form of force of magnitude $20.0\text{N}$ .

  ${\overrightarrow{\text{F}}}_3=\left(20.0\text{N}\right)\widehat{\text{j}}$                                                                                                       (III)

Here, ${\overrightarrow{\text{F}}}_3$ is the force of magnitude $20.0\text{N}$ .

The force of magnitude $77.8\text{N}$ makes an angle $45°$ with $x$ axis .

Write the expression for the vector form of force of magnitude $77.8\text{N}$ .

  $\begin{array}{c}{\overrightarrow{\text{F}}}_4=\left(77.8\text{cos}\text{45}°\text{N}\right)\widehat{\text{i}}\text{+}\left(77.8\text{sin}\text{45}°\text{N}\right)\widehat{\text{j}}\\ \text{=}\left(55.01\text{N}\right)\widehat{\text{i}}\text{+}\left(55.01\text{N}\right)\widehat{\text{j}}\end{array}$                                                             (IV)

Here, ${\overrightarrow{\text{F}}}_4$ is the force of magnitude $77.8\text{N}$ .

Write the position vector of point where ${\overrightarrow{\text{F}}}_1$ is acting.

  ${\overrightarrow{\text{r}}}_1=\left(2.0\text{m}\right)\widehat{\text{i}}+\left(2.0\text{m}\right)\widehat{\text{j}}$                                                                                          (V)

Here, ${\overrightarrow{\text{r}}}_1$ is the position vector of point where ${\overrightarrow{\text{F}}}_1$ is acting.

Write the position vector of point where ${\overrightarrow{\text{F}}}_2$ is acting.

  ${\overrightarrow{\text{r}}}_2=\left(-2.0\text{m}\right)\widehat{\text{i}}+\left(2.0\text{m}\right)\widehat{\text{j}}$                                                                                       (VI)

Here, ${\overrightarrow{\text{r}}}_2$ is the position vector of point where ${\overrightarrow{\text{F}}}_2$ is acting.

Write the position vector of point where ${\overrightarrow{\text{F}}}_3$ is acting.

  ${\overrightarrow{\text{r}}}_3=\left(-2.0\text{m}\right)\widehat{\text{i}}+\left(-2.0\text{m}\right)\widehat{\text{j}}$                                                                                      (VII)

Here, ${\overrightarrow{\text{r}}}_3$ is the position vector of point where ${\overrightarrow{\text{F}}}_3$ is acting.

Write the position vector of point where ${\overrightarrow{\text{F}}}_4$ is acting.

  ${\overrightarrow{\text{r}}}_4=\left(2.0\text{m}\right)\widehat{\text{i}}+\left(-2.0\text{m}\right)\widehat{\text{j}}$                                                                                   (VIII)

Here, ${\overrightarrow{\text{r}}}_4$ is the position vector of point where ${\overrightarrow{\text{F}}}_4$ is acting.

Write general expression for $\overrightarrow{\text{r}}$ .

  $\overrightarrow{\text{r}}=r_x\widehat{\text{i}}+r_y\widehat{\text{j}}\text{+}{\text{r}}_z\widehat{\text{k}}$                                                                                                   (IX)

Here, $r_x,r_y$, $r_z$ are the $x,y\text{and z}$ component of $\overrightarrow{\text{r}}$ and $\widehat{\text{i}}$ , $\widehat{\text{j}}$ , $\widehat{\text{k}}$ are unit vector along $x$, $y$, $z$ directions.

Write the expression for torque.

  $\overrightarrow{\tau }=\overrightarrow{\text{r}}\times \overrightarrow{\text{F}}$                                                                                                                  (X)

Here, $\overrightarrow{\tau }$ is the torque on an object, $\overrightarrow{\text{r}}$ is the position vector of force from axis of rotation and $\overrightarrow{\text{F}}$ is the force acting on the object.

Write the general form of $\overrightarrow{\text{F}}$ .

  $\overrightarrow{\text{F}}=F_x\widehat{\text{i}}+F_y\widehat{\text{j}}+F_z\widehat{\text{k}}$                                                                                               (XI)

Here, $F_x$ is the $x$ component of force, $F_y$ is the $y$ component of force and $F_z$ is the $z$ component of force.

Substitute (IX) and (XI) in equation (X) to get $\overrightarrow{\tau }$ .

  $\begin{array}{c}\overrightarrow{\tau }=\left(r_x\widehat{\text{i}}+r_y\widehat{\text{j}}\text{+}{\text{r}}_z\widehat{\text{k}}\right)\times \left(F_x\widehat{\text{i}}+F_y\widehat{\text{j}}+F_z\widehat{\text{k}}\right)\\ =\left(r_y{F}_z-r_z{F}_y\right)\widehat{\text{i}}+\left(r_z{F}_x-r_x{F}_z\right)\widehat{\text{j}}+\left(r_x{F}_y-r_y{F}_x\right)\widehat{\text{k}}\end{array}$                                               (XII)

Write the expression for the torque due to force ${\overrightarrow{\text{F}}}_1$ .

  ${\overrightarrow{\tau }}_1={\overrightarrow{\text{r}}}_1\times {\overrightarrow{\text{F}}}_1$

Here, ${\overrightarrow{\tau }}_1$ is the torque due to force ${\overrightarrow{\text{F}}}_1$ .

Write the expression for the torque due to force ${\overrightarrow{\text{F}}}_2$ .

  ${\overrightarrow{\tau }}_2={\overrightarrow{\text{r}}}_2\times {\overrightarrow{\text{F}}}_2$

Here, ${\overrightarrow{\tau }}_2$ is the torque due to force ${\overrightarrow{\text{F}}}_2$ .

Write the expression for the torque due to force ${\overrightarrow{\text{F}}}_3$ .

  ${\overrightarrow{\tau }}_3={\overrightarrow{\text{r}}}_3\times {\overrightarrow{\text{F}}}_3$

Here, ${\overrightarrow{\tau }}_3$ is the torque due to force ${\overrightarrow{\text{F}}}_3$ .

Write the expression for the torque due to force ${\overrightarrow{\text{F}}}_4$ .

  ${\overrightarrow{\tau }}_4={\overrightarrow{\text{r}}}_4\times {\overrightarrow{\text{F}}}_4$

Here, ${\overrightarrow{\tau }}_4$ is the torque due to force ${\overrightarrow{\text{F}}}_4$ .

Conclusion:

Substitute $0\text{N}$ for $F_x$, $-60.0\text{N}$ for $F_y$ and $0\text{N}$ for $F_z$, $2.0\text{m}$ for $r_1$, $2.0\text{m}$ for $r_2$ and $0.0\text{m}$ for $r_3$ in (XII) to get ${\overrightarrow{\tau }}_1$ .

  $\begin{array}{c}{\overrightarrow{\tau }}_1=\left(\left(2.0\text{m}\right)\widehat{\text{i}}+\left(2.0\text{m}\right)\widehat{\text{j}}\text{+}\left(0.0\text{m}\right)\widehat{\text{k}}\right)\times \left(\left(0\text{N}\right)\widehat{\text{i}}-\left(60.0\text{N}\right)\widehat{\text{j}}+\left(0\text{N}\right)\widehat{\text{k}}\right)\\ =\left(\left(2.0\text{m}\right)\left(0\text{N}\right)-\left(0.0\text{m}\right)\left(60.0\text{N}\right)\right)\widehat{\text{i}}+\left(\left(0.0\text{m}\right)\left(0\text{N}\right)-\left(2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{j}}+\left(\left(2.0\text{m}\right)\left(60.0\text{N}\right)-\left(2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{k}}\\ =-120\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

Substitute $30\text{N}$ for $F_x$, $0\text{N}$ for $F_y$ and $0\text{N}$ for $F_z$, $-2.0\text{m}$ for $r_1$, $2.0\text{m}$ for $r_2$ and $0.0\text{m}$ for $r_3$ in (XII) to get ${\overrightarrow{\tau }}_2$ .

  $\begin{array}{c}{\overrightarrow{\tau }}_2=\left(\left(-2.0\text{m}\right)\widehat{\text{i}}+\left(2.0\text{m}\right)\widehat{\text{j}}\text{+}\left(0.0\text{m}\right)\widehat{\text{k}}\right)\times \left(\left(30\text{N}\right)\widehat{\text{i}}+\left(0\text{N}\right)\widehat{\text{j}}+\left(0\text{N}\right)\widehat{\text{k}}\right)\\ =\left(\left(2.0\text{m}\right)\left(0\text{N}\right)-\left(0.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{i}}+\left(\left(0.0\text{m}\right)\left(30\text{N}\right)-\left(-2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{j}}+\left(-\left(2.0\text{m}\right)\left(0\text{N}\right)-\left(2.0\text{m}\right)\left(30\text{N}\right)\right)\widehat{\text{k}}\\ =-60\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

Substitute $0\text{N}$ for $F_x$, $20\text{N}$ for $F_y$ and $0\text{N}$ for $F_z$, $-2.0\text{m}$ for $r_1$, $-2.0\text{m}$ for $r_2$ and $0.0\text{m}$ for $r_3$ in (XII) to get ${\overrightarrow{\tau }}_3$ .

  $\begin{array}{c}{\overrightarrow{\tau }}_3=\left(\left(-2.0\text{m}\right)\widehat{\text{i}}+\left(-2.0\text{m}\right)\widehat{\text{j}}\text{+}\left(0.0\text{m}\right)\widehat{\text{k}}\right)\times \left(\left(0\text{N}\right)\widehat{\text{i}}+\left(20.0\text{N}\right)\widehat{\text{j}}+\left(0\text{N}\right)\widehat{\text{k}}\right)\\ =\left(\left(-2.0\text{m}\right)\left(0\text{N}\right)-\left(0.0\text{m}\right)\left(20.0\text{N}\right)\right)\widehat{\text{i}}+\left(\left(0.0\text{m}\right)\left(0\text{N}\right)-\left(-2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{j}}+\left(\left(-2.0\text{m}\right)\left(20.0\text{N}\right)-\left(-2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{k}}\\ =-40\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

Substitute $55.01\text{N}$ for $F_x$, $55.01\text{N}$ for $F_y$ and $0\text{N}$ for $F_z$, $2.0\text{m}$ for $r_1$, $-2.0\text{m}$ for $r_2$ and $0.0\text{m}$ for $r_3$ in (XII) to get ${\overrightarrow{\tau }}_4$ .

$\begin{array}{c}{\overrightarrow{\tau }}_4=\left(\left(2.0\text{m}\right)\widehat{\text{i}}+\left(-2.0\text{m}\right)\widehat{\text{j}}\text{+}\left(0.0\text{m}\right)\widehat{\text{k}}\right)\times \left(\left(55.01\text{N}\right)\widehat{\text{i}}+\left(55.01\text{N}\right)\widehat{\text{j}}+\left(0\text{N}\right)\widehat{\text{k}}\right)\\ =\left(\left(-2.0\text{m}\right)\left(0\text{N}\right)-\left(0.0\text{m}\right)\left(55.01\text{N}\right)\right)\widehat{\text{i}}+\left(\left(0.0\text{m}\right)\left(55.01\text{N}\right)-\left(2.0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{j}}+\left(\left(2.0\text{m}\right)\left(55.01\text{N}\right)-\left(-2.0\text{m}\right)\left(55.01\text{N}\right)\right)\widehat{\text{k}}\\ =\left(220\widehat{\text{k}}\right)\text{N}\cdot \text{m}\end{array}$

Add ${\overrightarrow{\tau }}_1$, ${\overrightarrow{\tau }}_2$, ${\overrightarrow{\tau }}_3$ and ${\overrightarrow{\tau }}_4$ to check whether it is zero not. If sum of torques are zero, the plate is in rotational equilibrium.

Write the expression for net torque.

  $\overrightarrow{\tau }={\overrightarrow{\tau }}_1+{\overrightarrow{\tau }}_2+{\overrightarrow{\tau }}_3+{\overrightarrow{\tau }}_4$

Substitute $-120\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_1$, $-60\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_2$, $-40\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_3$, $220\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_4$ in above equation to get $\overrightarrow{\tau }$ .

  $\begin{array}{c}\overrightarrow{\tau }=-120\widehat{\text{k}}\text{N}\cdot \text{m}+-60\widehat{\text{k}}\text{N}\cdot \text{m}+-40\widehat{\text{k}}\text{N}\cdot \text{m}+220\widehat{\text{k}}\text{N}\cdot \text{m}\\ =0\text{N}\cdot \text{m}\end{array}$

This indicates that the system is in rotational equilibrium.

Therefore, the pate is in equilibrium.