Chapter 1, Problem 67P

(a)

To determine

The equivalent force between the two pulling forces and its direction.

Answer to Problem 67P

The equivalent force between the two pulling forces is $185\text{N}$ and its direction is $77.8°$.

Explanation of Solution

Write the expression for person on the right pull with a force ${\overrightarrow{F}}_1$.

  ${\overrightarrow{F}}_1=F_{1x}\cos \theta \widehat{i}+F_{1y}\sin \theta \widehat{j}$        (I)

Here, ${\overrightarrow{F}}_1$ is the person on the right pull with a force, $\theta$ is the direction of the right pull force, $F_{1y}$ is right force vector y component, and $F_{1x}$ is the right force vector along x component.

Write the expression for person on the left pull with a force ${\overrightarrow{F}}_2$.

  ${\overrightarrow{F}}_2=-F_{2x}\cos \theta \widehat{i}+F_{2y}\sin \theta \widehat{j}$        (II)

Here, ${\overrightarrow{F}}_2$ is the person on the left pull with a force, $\theta$ is the direction of the left pull force, $F_{2y}$ is left force vector y component, and $F_{2x}$ is the left force vector along x component.

Write the expression for the net force is acting between the two forces.

  $\overrightarrow{F}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$        (III)

Here, $\overrightarrow{F}$ is the net equivalent force.

Write the expression for the magnitude of net force is acting between the two forces.

  $\left|\overrightarrow{F}\right|=\sqrt{F_x^2+F_y^2}$        (IV)

Here, $\left|\overrightarrow{F}\right|$ is the magnitude of net force is acting between the two forces.

Write the expression for the angle of equivalent force.

  $\varphi ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$        (V)

Here, $\varphi$ is the angle of equivalent force.

Conclusion:

Substitute $120\text{N}$ for $F_{1x}$, $60.0°$ for $\theta$, and $120\text{N}$ for $F_{1y}$ in the equation (I) to find ${\overrightarrow{F}}_1$.

  $\begin{array}{c}{\overrightarrow{F}}_1=\left(120\text{N}\right)\cos \left(60.0°\right)\widehat{i}+\left(120\text{N}\right)\sin \left(60.0°\right)\widehat{j}\\ =\left(60\widehat{i}+104\widehat{j}\right)\text{N}\end{array}$

Substitute $80.0\text{N}$ for $F_{2x}$, $75.0°$ for $\theta$, and $80.0\text{N}$ for $F_{2y}$ in the equation (II) to find ${\overrightarrow{F}}_2$.

  $\begin{array}{c}{\overrightarrow{F}}_2=-\left(80.0\text{N}\right)\cos \left(75.0°\right)\widehat{i}+\left(80.0\text{N}\right)\sin \left(75.0°\right)\widehat{j}\\ =\left(-20.7\widehat{i}+77.3\widehat{j}\right)\text{N}\end{array}$

Substitute $\left(60\widehat{i}+104\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_1$ and $\left(-20.7\widehat{i}+77.3\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_2$ in equation (III) to find $\overrightarrow{F}$.

  $\begin{array}{c}\overrightarrow{F}=\left(60\widehat{i}+104\widehat{j}\right)\text{N}+\left(-20.7\widehat{i}+77.3\widehat{j}\right)\text{N}\\ =\left(39.3\widehat{i}+181\widehat{j}\right)\text{N}\end{array}$

Substitute $39.3\text{N}$ for $F_x$ and $181\text{N}$ for $F_y$ in equation (IV) to find $\left|\overrightarrow{F}\right|$.

  $\begin{array}{c}\left|\overrightarrow{F}\right|=\sqrt{{\left(39.3\text{N}\right)}^2+{\left(181\text{N}\right)}^2}\\ =185\text{N}\end{array}$

Substitute $39.3\text{N}$ for $F_x$ and $181\text{N}$ for $F_y$ in equation (V) to find $\varphi$.

  $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{181\text{N}}{39.3\text{N}}\right)\\ =77.8°\end{array}$

Therefore, the equivalent force between the two pulling forces is $185\text{N}$ and its direction is $77.8°$

(b)

To determine

The force of the third person exerted on the mule to make the resultant force is equal to zero.

Answer to Problem 67P

The force of the third person is $\left(-39.3\widehat{i}-181\widehat{j}\right)\text{N}$ exerted on the mule to make the resultant force is equal to zero.

Explanation of Solution

Write the expression for the third person exerted on the mule.

  ${\overrightarrow{F}}_3=-\overrightarrow{F}$        (VI)

Here, ${\overrightarrow{F}}_3$ is the force of the third person exerted on the mule to make the resultant force is equal to zero.

Conclusion:

Substitute $\left(39.3\widehat{i}+181\widehat{j}\right)\text{N}$ for $\overrightarrow{F}$ in the equation (VI) to find ${\overrightarrow{F}}_3$.

  $\begin{array}{c}{\overrightarrow{F}}_3=-\left(39.3\widehat{i}+181\widehat{j}\right)\text{N}\\ =\left(-39.3\widehat{i}-181\widehat{j}\right)\text{N}\end{array}$

Therefore, the force of the third person is $\left(-39.3\widehat{i}-181\widehat{j}\right)\text{N}$ exerted on the mule to make the resultant force is equal to zero.