Chapter 4, Problem 2P

To determine

The magnitude of force $F_3$ and its angle.

Answer to Problem 2P

The magnitude of force $F_3$ is $78\text{N}$ and the angle is $9.8°$.

Explanation of Solution

The figure 1 shows the free body diagram for the particle and the x and y component of three forces

Write the expression for x component of the $50\text{N}$ force.

  $F_{50\text{N,}x}=F_1\cos {\theta }_1$        (I)

Here, $F_{50\text{N,}x}$ is the x component of the $50\text{N}$ force, ${\theta }_1$ is the angle, and $F_1$ is the force of the particle.

Write the expression for y component of the $50\text{N}$ force.

  $F_{50\text{N }y}=F_1\sin {\theta }_1$        (II)

Here, $F_{50\text{N,}y}$ is the y component of the $50\text{N}$ force.

Write the expression for x component of the $60\text{N}$ force.

  $F_{60\text{N,}x}=F_2\cos {\theta }_2$        (III)

Here, $F_{60\text{N,}x}$ is the x component of the $60\text{N}$ force and $F_2$ is the force of the particle.

Write the expression for y component of the $60\text{N}$ force.

  $F_{60\text{N,}y}=-F_2\sin {\theta }_2$        (IV)

Here, $F_{60\text{N,}y}$ is the y component of the $60\text{N}$ force.

According to static equilibrium, the sum of the x component of the force acting on the particle is,

  $\begin{array}{c}{\displaystyle \sum F_x}=0\\ F_{50\text{N,}x}+F_{60\text{N,}x}+F_{3x}=0\end{array}$        (V)

Here, $F_{3x}$ is the unknown force acting on the particle along x direction.

According to static equilibrium, the sum of the y component of the force acting on the particle is,

  $\begin{array}{c}{\displaystyle \sum F_y}=0\\ F_{50\text{N,}y}+F_{60\text{N,}y}+F_{3y}=0\end{array}$        (VI)

Here, $F_{3y}$ is the unknown force acting on the particle along y direction.

The net force $F_3$ acting on the particle is,

  $F_3=\sqrt{F_{3x}^2+F_{3y}^2}$        (VII)

The angle acting on the force $F_3$ is,

  ${\theta }_3={\tan }^{-1}\left(\frac{F_{3y}}{F_{3x}}\right)$        (VIII)

Here, ${\theta }_3$ is the angle due to the force $F_3$.

Conclusion:

Substitute $50\text{N}$ for $F_1$ and $60°$ for ${\theta }_1$ in equation (I).

  $\begin{array}{c}{F}_{50\text{N,}x}=\left(50\text{N}\right)\cos \left(60°\right)\\ =25\text{N}\end{array}$

Substitute $50\text{N}$ for $F_1$ and $60°$ for ${\theta }_1$ in equation (II).

  $\begin{array}{c}{F}_{50\text{N,}y}=\left(50\text{N}\right)\sin \left(60°\right)\\ =43.3\text{N}\end{array}$

Substitute $60\text{N}$ for $F_2$ and $30°$ for ${\theta }_2$ in equation (III).

  $\begin{array}{c}{F}_{60\text{N,}x}=\left(60\text{N}\right)\cos \left(30°\right)\\ =52\text{N}\end{array}$

Substitute $60\text{N}$ for $F_2$ and $30°$ for ${\theta }_2$ in equation (IV).

  $\begin{array}{c}{F}_{60\text{N,}y}=-\left(60\text{N}\right)\sin \left(30°\right)\\ =-30\text{N}\end{array}$

Substitute $25\text{N}$ for $F_{50\text{N,}x}$ and $52\text{N}$ for $F_{60\text{N,}x}$ in equation (V) to find $F_{3x}$.

  $\begin{array}{c}25\text{N}+52\text{N}+F_{3x}=0\\ 77\text{N}+F_{3x}=0\\ F_{3x}=-77\text{N}\end{array}$

Substitute $43.3\text{N}$ for $F_{50\text{N,}y}$ and $-30\text{N}$ for $F_{60\text{N,}y}$ in equation (VI) to find $F_{3y}$.

  $\begin{array}{c}43.3\text{N}-30\text{N}+F_{3y}=0\\ 13.3\text{N}+F_{3y}=0\\ F_{3y}=-13.3\text{N}\end{array}$

Substitute $-77\text{N}$ for $F_{3x}$ and $-13.3\text{N}$ for $F_{3y}$ in equation (VII) to find $F_3$.

  $\begin{array}{c}{F}_3=\sqrt{{\left(-77\text{N}\right)}^2+{\left(-13.3\text{N}\right)}^2}\\ =78\text{N}\end{array}$

Substitute $-77\text{N}$ for $F_{3x}$ and $-13.3\text{N}$ for $F_{3y}$ in equation (VIII) to find ${\theta }_3$.

  $\begin{array}{c}{\theta }_3={\tan }^{-1}\left(\frac{-13.3\text{N}}{-77\text{N}}\right)\\ =9.8°\end{array}$

Therefore, the magnitude of force $F_3$ is $78\text{N}$ and the angle is $9.8°$.