Chapter 5, Problem 82A

To determine

The magnitude and the direction of the sixth force required to produce the equilibrium.

Answer to Problem 82A

The magnitude and direction of the sixth force is $-34.2\text{N}$ acting at an angle $43.0°$ with the horizontal in order to keep the object in equilibrium.

Explanation of Solution

Given:

The five forces are:

The force $F_{\text{1}}=60\text{N}$ acts at angle ${\theta }_1=90°$ with the positive direction of $+x$ axis.

The force $F_{\text{2}}=40\text{N}$ acts at angle ${\theta }_2=0°$ with the positive direction of $+x$ axis.

The force $F_3=80\text{N}$ acts at angle ${\theta }_3=270°$ with the positive direction of $+x$ axis.

The force $F_4=40\text{N}$ acts at angle ${\theta }_3=180°$ with the positive direction of $+x$ axis.

The force $F_5=50\text{N}$ acts at angle ${\theta }_3=60°$ with the positive direction of $+x$ axis.

Formula used:

Consider a force $F$ is acting on an object at an angle $\theta$ with the horizontal. The magnitude of horizontal component of the force $F$ is,

  $F_x=F\cos \theta$

The magnitude of vertical component of the forceF is,

  $F_y=F\sin \theta$

Calculation:

Consider five forces $F_1,F_2,F_3,F_4,F_5$ acting on the object.

The horizontal and vertical component of the force $F_1$ as follows:

  $\begin{array}{c}$ {(F_x)}_1=F_1\cos {\theta }_1$=\text{(60 N)}\cos 90°\\ =\text{0}\text{.0 N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_1=F_1\sin {\theta }_1$=\text{(60 N)}\sin 90°\\ =\text{60}\text{.0 N}\end{array}$

The horizontal and vertical component of the force $F_2$ as follows:

  $\begin{array}{c}$ {(F_x)}_2=F_2\cos {\theta }_2$=\text{(40 N)}\cos 0°\\ =\text{40}\text{.0 N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_2=F_2\sin {\theta }_2$=\text{(40 N)}\sin 0°\\ =\text{0}\text{.0 N}\end{array}$

The horizontal and vertical component of the force $F_3$ as follows:

  $\begin{array}{c}$ {(F_x)}_3=F_3\cos {\theta }_3$=\text{(80 N)}\cos 270°\\ =\text{0}\text{.0 N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_3=F_3\sin {\theta }_3$=\text{(80 N)}\sin 270°\\ =-8\text{0}\text{.0 N}\end{array}$

The horizontal and vertical component of the force $F_4$ as follows:

  $\begin{array}{c}$ {(F_x)}_4=F_4\cos {\theta }_4$=\text{(40 N)}\cos 180°\\ =-\text{40}\text{.0 N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_4=F_4\sin {\theta }_4$=\text{(40 N)}\sin 180°\\ =\text{0}\text{.0 N}\end{array}$

The horizontal and vertical component of the force $F_5$ as follows:

  $\begin{array}{c}$ {(F_x)}_5=F_5\cos {\theta }_5$=\text{(50 N)}\cos 60°\\ =25.\text{0 N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_5=F_5\sin {\theta }_5$=\text{(50 N)}\sin 60°\\ =43.30\text{ N}\end{array}$

The net horizontal force due to five forces as follows:

  $\begin{array}{c}$ {\text{(}{F}_{\text{x}}\text{)}}_{\text{resultant}}\text{ = (}{F}_{\text{x}}{\text{)}}_1\text{ + (}{F}_{\text{x}}{\text{)}}_2\text{ + (}{F}_{\text{x}}{\text{)}}_3\text{ + (}{F}_{\text{x}}{\text{)}}_4\text{ + (}{F}_{\text{x}}{\text{)}}_5$\text{= 0}\text{.0N + 40}\text{.0N + 0}\text{.0N + (}-\text{40}\text{.0N) + 25}\text{.0N}\\ \text{= 25}\text{.0N}\end{array}$

The net vertical force due to five forces as follows:

  $\begin{array}{c}$ {(F_y)}_{\text{resultant}}={(F_y)}_1+{(F_y)}_2+{(F_y)}_3+{(F_y)}_4+{(F_{\text{y}}\text{)}}_5$\text{= 60}\text{.0 N + 0}\text{.0 N + (}-\text{80}\text{.0 N) + 0}\text{.0 N + 43}\text{.30 N}\\ \text{= 23}\text{.3 N}\end{array}$

The resultant vector of the five forces acting on the objects as follows:

  ${\text{(}\overrightarrow{\text{F}}\text{)}}_{\text{resultant}}\text{ = }{\left(\overrightarrow{F_x}\right)}_{\text{resultant}}\text{+}{\left(\overrightarrow{F_y}\right)}_{\text{resultant}}$

  ${\text{(}\overrightarrow{\text{F}}\text{)}}_{\text{resultant}}=(25.0\text{N) <mover accent="true"><mo>x </mo><mo>^</mo> </mover> + (23}\text{.3 N) <mover accent="true"><mo>y</mo><mo>^</mo> </mover>}$

The magnitude of the resultant force as follows:

  $\left|{\text{(}\overrightarrow{\text{F}}\text{)}}_{\text{resultant}}\right|=\sqrt{{\text{(25}\text{.0 N)}}^{\text{2}}\text{+(23}{\text{.3 N)}}^{\text{2}}}$

  $=\text{34}\text{.17 N}$

  $\approx 34.2\text{N}$

Find the direction of the resultant force as follows:

  $\begin{array}{c}\theta {\text{= tan}}^{\text{-1}}\left(\frac{{\text{(}{F}_y\text{)}}_{\text{resultant}}}{{\text{(}{F}_x\text{)}}_{\text{resultant}}}\right)\\ {\text{= tan}}^{\text{-1}}\left(\frac{\text{23}\text{.30}}{\text{25}\text{.0}}\right)\\ =43.0°\end{array}$

Hence, the net force acting on the object is $34.2\text{N}$ at angle $43.0°$ with the horizontal.

For the Equilibrium of the forces in the object, the sixth force of magnitude $-34.2\text{N}$ should act angle ${43.0}^{\circ }$ with the horizontal in order to make the net force acting on the object as zero.

Conclusion:

Therefore, the magnitude and direction of the sixth force is $-34.2\text{N}$ acting at an angle $43.0°$ with the horizontal in order keep the object in equilibrium.