Chapter 14, Problem 18PQ

(a)

To determine

The torque exerted by the force $\overrightarrow{\text{F}}=\left(65.8\widehat{\text{i}}+96.3\widehat{\text{j}}\right)\text{N}$ on an object at a point whose position from the axis of rotation is $\overrightarrow{\text{r}}=-5.80\widehat{\text{k}}\text{m}$ .

Answer to Problem 18PQ

The torque exerted by the force $\overrightarrow{\text{F}}=\left(65.8\widehat{\text{i}}+96.3\widehat{\text{j}}\right)\text{N}$ is $\left(559\widehat{\text{i}}-382\widehat{\text{j}}\right)\text{N}\cdot \text{m}$ .

Explanation of Solution

Write the expression for torque.

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

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

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

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 units vector along $x$, $y$, $z$ direction.

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

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

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 equations (II) and (III) in equation (I) to get $\overrightarrow{\tau }$ .

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

Conclusion:

Given that $\overrightarrow{\text{F}}=\left(65.8\widehat{\text{i}}+96.3\widehat{\text{j}}\right)\text{N}$ and $\overrightarrow{\text{r}}=-5.80\widehat{\text{k}}\text{m}$. That is $F_x=65.8\text{N}$, $F_y=96.3\text{N}$, $r_x=0\text{m}$, $r_y=0\text{m}$ and $r_z=-5.80\widehat{\text{k}}\text{m}$.

Substitute $65.8\text{N}$ for $F_x$, $96.3\text{N}$ for $F_y$, $0\text{N}$ for $F_z$, $0\text{m}$ for $r_x$, $0\text{m}$ for $r_y$ and $-5.80\text{m}$ for $r_z$ in equation (IV) to get $\overrightarrow{\tau }$ .

  $\begin{array}{c}\overrightarrow{\tau }=\left(\left(0\text{m}\right)\left(0\text{N}\right)-\left(-5.80\text{m}\right)\left(96.3\text{N}\right)\right)\widehat{\text{i}}+\left(\left(-5.80\text{m}\right)\left(65.8\text{N}\right)-\left(0\text{m}\right)\left(0\text{N}\right)\right)\widehat{\text{j}}+\left(\left(0\text{m}\right)\left(96.3\text{N}\right)-\left(0\text{m}\right)\left(65.8\text{N}\right)\right)\widehat{\text{k}}\\ \text{=}-\left(-5.80\widehat{\text{k}}\text{m}\right)\left(96.3\text{N}\right)\widehat{\text{i}}+\left(-5.80\widehat{\text{k}}\text{m}\right)\left(65.8\text{N}\right)\widehat{\text{j}}\\ =\left(559\widehat{\text{i}}-382\widehat{\text{j}}\right)\text{N}\cdot \text{m}\end{array}$

Therefore, the torque exerted by the force $\overrightarrow{\text{F}}=\left(65.8\widehat{\text{i}}+96.3\widehat{\text{j}}\right)\text{N}$ is $\left(559\widehat{\text{i}}-382\widehat{\text{j}}\right)\text{N}\cdot \text{m}$ .

(b)

To determine

The change in answer in part (a) if the force’s $x$ component increases while the $y$ component remains constant.

Answer to Problem 18PQ

The $y$ component of torque increases in magnitude with increases in the $x$ component of force.

Explanation of Solution

Rewrite equation (IV) to get relation between torque and component of forces.

  $\overrightarrow{\tau }=\left(r_y{F}_z-r_z{F}_y\right)\widehat{\text{i}}+\left(r_z{F}_x-r_x{F}_z\right)\widehat{\text{j}}+\left(r_x{F}_y-r_y{F}_x\right)\widehat{\text{k}}$

The $y$ component of torque is function of $F_x$ and $F_z$. The $z$ component of torque is a function of $F_x$ and $F_y$ . The $y$ component of torque is given by $\left(r_z{F}_x-r_x{F}_z\right)$ and $z$ component of torque is given by $\left(r_x{F}_y-r_y{F}_x\right)$. Since all components of $\overrightarrow{\text{r}}$ except $z$ component are zero, $z$ component of torque is zero and magnitude of $y$ component equal to $r_z{F}_x$.

It is given that $F_y$ remains the same and $F_z$ is zero. Then above equation implies if the force’s $x$ component increases, $y$ component of torque increases with magnitude equal to $r_z{F}_x$.

Conclusion:

Therefore, the $y$ component of torque increases in magnitude with increases in the $x$ component of force.

(c)

To determine

The change in answer in part (a) if the force’s $y$ component increases while the $x$ component remains constant.

Answer to Problem 18PQ

If the force’s $y$ component increases while the $x$ component remains constant, $x$ component of torque increases in magnitude.

Explanation of Solution

Rewrite equation (IV) to get relation between torque and component of forces.

  $\overrightarrow{\tau }=\left(r_y{F}_z-r_z{F}_y\right)\widehat{\text{i}}+\left(r_z{F}_x-r_x{F}_z\right)\widehat{\text{j}}+\left(r_x{F}_y-r_y{F}_x\right)\widehat{\text{k}}$

The $x$ component of torque is function of $F_y$ and $F_z$. The $z$ component of torque is a function of $F_x$ and $F_y$ . The $x$ component of torque is given by $\left(r_y{F}_z-r_z{F}_y\right)$ and $z$ component of torque is given by $\left(r_x{F}_y-r_y{F}_x\right)$. Since all other components of $\overrightarrow{\text{r}}$ except $z$ component are zero, $z$ component of torque is zero and magnitude of $x$ component equal to $r_z{F}_y$.

It is given that $F_y$ remains the same and $F_z$ is zero. Then above equation implies, if the force’s $y$ component increases, $x$ component of torque increases with magnitude equal to $r_z{F}_y$.

Conclusion:

Therefore, if the force’s $y$ component increases while the $x$ component remains constant, $x$ component of torque increases in magntidue.