Chapter 12, Problem 53PQ

A square plate with sides 2.0 m in length can rotate around an axle passing through its center of mass (CM) and perpendicular to its surface (Fig. P12.53). There are four forces acting on the plate at different points. The rotational inertia of the plate is 24 kg · m². Use the values given in the figure to answer the following questions. a. What is the net torque acting on the plate? b. What is the angular acceleration of the plate?

FIGURE P12.53

Problems 53 and 54.

To determine

The net torque acting on the plate.

Answer to Problem 53PQ

The net torque acting on the plate is $\underset{\_}{-53\widehat{\text{k}}\text{N}\cdot \text{m}}$.

Explanation of Solution

Write the expression for the net torque on the plate.

    ${\overrightarrow{\tau }}_{\text{tot}}={\overrightarrow{\tau }}_1+{\overrightarrow{\tau }}_2+{\overrightarrow{\tau }}_3+{\overrightarrow{\tau }}_4$                                                                                               (I)

Here, ${\overrightarrow{\tau }}_{\text{tot}}$ is the net torque on the plate, ${\overrightarrow{\tau }}_1$ is the torque due to first force, ${\overrightarrow{\tau }}_2$ is the torque due to second force, ${\overrightarrow{\tau }}_3$ is the torque due to third force,${\overrightarrow{\tau }}_4$ is the torque due to fourth force.

Write the expression for the torque acting on an object in the cross product form.

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

Here, $\overrightarrow{r}$ is the displacement vector, $\overrightarrow{F}$ is the force acting on the object.

Equation (II) can be solved as,

    $\tau =rF\sin \theta$                                                                                                            (III)

Conclusion:

For each forces $60\text{N}$ for $F_1$ , $30\text{N}$ for $F_2$, $20\text{N}$ for $F_3$ and the moment arm is $1.0\text{m}$ and using the right hand rule, the torque points into the page in the negative $z$ direction.

    $\begin{array}{c}{\overrightarrow{\tau }}_1=-\left(60\text{N}\right)\left(1.0\text{m}\right)\widehat{\text{k}}\\ =-60\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

    $\begin{array}{c}{\overrightarrow{\tau }}_2=-\left(30\text{N}\right)\left(1.0\text{m}\right)\widehat{\text{k}}\\ =-30\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

    $\begin{array}{c}{\overrightarrow{\tau }}_3=-\left(20\text{N}\right)\left(1.0\text{m}\right)\widehat{\text{k}}\\ \text{=}-20\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

The diagonal of the square is the square root of the sum of the squares of the two sides of the square which can be found that,

    $\begin{array}{c}d=\sqrt{{\left(2.00\text{m}\right)}^2+{\left(2.00\text{m}\right)}^2}\\ =2\sqrt{2}\text{m}\end{array}$

For the torque due to force $F_4$,$40\text{N}$ the moment arm is half of the diagonal of the square,

  $\begin{array}{c}r=\frac{1}{2}\left(2\sqrt{2}\text{m}\right)\\ =\sqrt{2}\text{m}\end{array}$

The angle between the moment arm and the force is $90°$, the fourth torque will be,

    $\begin{array}{c}{\tau }_4=\left(40\right)\sqrt{2}\widehat{\text{k}}\\ \text{=57}\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

Substitute $-60\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_1$, $-30\widehat{\text{k}}\text{Nm}$ for ${\overrightarrow{\tau }}_2$, $-20\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_3$ and $57\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_4$ in equation (I) to find ${\overrightarrow{\tau }}_{\text{total}}$.

    $\begin{array}{c}{\tau }_{\text{total}}=\left(-60\widehat{\text{k}}\text{N}\cdot \text{m}\right)+\left(-30\widehat{\text{k}}\text{Nm}\right)+\left(-20\widehat{\text{k}}\text{N}\cdot \text{m}\right)+\left(57\widehat{\text{k}}\text{N}\cdot \text{m}\right)\\ =-53\widehat{\text{k}}\text{N}\cdot \text{m}\end{array}$

Therefore, the net torque acting on the plate is $\underset{\_}{-53\widehat{\text{k}}\text{N}\cdot \text{m}}$.

To determine

The angular acceleration of the plate.

Answer to Problem 53PQ

The angular acceleration of the plate is $\underset{\_}{-2.2\widehat{\text{k}}\text{rad}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the total torque in terms of rotational inertia.

    ${\overrightarrow{\tau }}_{\text{tot}}=I\overrightarrow{\alpha }$                                                                                                                  (IV)

Here, $I$ is the moment of inertia, $\overrightarrow{\alpha }$ is the angular acceleration.

Rearrange equation (IV),

    $\overrightarrow{\alpha }=\frac{{\overrightarrow{\tau }}_{\text{tot}}}{I}$                                                                                                                    (V)

Conclusion:

Substitute $-53\widehat{\text{k}}\text{N}\cdot \text{m}$ for ${\overrightarrow{\tau }}_{\text{tot}}$ and $24\text{kg}\cdot {\text{m}}^2$ for $I$ in equation (V),

    $\begin{array}{c}\overrightarrow{\alpha }=\frac{-53\widehat{\text{k}}\text{N}\cdot \text{m}}{24\text{kg}\cdot {\text{m}}^2}\\ =-2.2\widehat{\text{k}}\text{rad}/{\text{s}}^2\end{array}$

Therefore, the angular acceleration of the plate is $\underset{\_}{-2.2\widehat{\text{k}}\text{rad}/{\text{s}}^2}$.