Chapter 5, Problem 64A

To determine

The net force acting on the ring.

Answer to Problem 64A

The net force acting on the ring is $79.4\text{N}$ at angle $53.7°$ with the horizontal.

Explanation of Solution

Given:

The free body diagram of forces acting on the ring is shown below.

  

Formula used:

The magnitude of horizontal component of the force F , that makes an angle $\theta$ with the horizontal is,

  $F_x=F\cos \theta$

The magnitude of vertical component of the force $F$ is,

  $F_y=F\sin \theta$

Calculation:

Consider the given free body diagram. Three forces $F_1,F_2,F_3$ acting on the ring such as,

The force $F_{\text{1}}=64\text{N}$ acts at angle ${\theta }_1=0°$ with the positive direction of $+x$ axis.The force $F_{\text{2}}=128\text{N}$ acts at angle ${\theta }_2=30°$ with the positive direction of $+x$ axis.The force $F_3=128\text{N}$ acts at angle ${\theta }_3=180°$ with the positive direction of $+x$ axis.

The horizontal and vertical component of the force $F_1$ is,

  $\begin{array}{c}$ {(F_x)}_1=F_1\cos {\theta }_1$=\text{(64 N)}\cos 0°\\ =64\text{ N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_1=F_1\sin {\theta }_1$=\text{(64 N)}\sin 0°\\ =\text{0 N}\end{array}$

The horizontal and vertical component of the force $F_2$ is,

  $\begin{array}{c}$ {(F_x)}_2=F_2\cos {\theta }_2$=\text{(128 N)}\cos 30°\\ =111\text{ N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_2=F_2\sin {\theta }_2$=\text{(128 N)}\sin 30°\\ =64\text{ N}\end{array}$

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

  $\begin{array}{c}$ {(F_x)}_3=F_3\cos {\theta }_3$=\text{(128 N)}\cos 180°\\ =-128\text{ N}\end{array}$

  $\begin{array}{c}$ {(F_y)}_3=F_3\sin {\theta }_3$=\text{(128 N)}\sin 180°\\ =\text{0 N}\end{array}$

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$\text{= 64}\text{N + 111}\text{N }-128\text{N}\\ \text{= 47}\text{N}\end{array}$

The net vertical force due to five forces as follows:

  $\begin{array}{c}$ {\text{(}{F}_{\text{y}}\text{)}}_{\text{resultant}}\text{ = (}{F}_{\text{y}}{\text{)}}_1\text{ + (}{F}_{\text{y}}{\text{)}}_2\text{ + (}{F}_{\text{y}}{\text{)}}_3$\text{= 0 N + 64 N + 0 N}\\ \text{= 64}\text{N}\end{array}$

The magnitude of the resultant force as follows:

  $\begin{array}{c}\left|{\left(\overrightarrow{F}\right)}_{\text{resultant}}\right|=\sqrt{{\left(F_y\right)}_{\text{resultant}}^2+{\left(F_y\right)}_{\text{resultant}}^2}\\ =\sqrt{{\text{(47 N)}}^{\text{2}}{\text{+(64 N)}}^{\text{2}}}\\ =79.4\text{ N}\end{array}$

Find the direction of the resultant force as follows:

  $\begin{array}{c}{\text{α = tan}}^{\text{-1}}\left(\frac{{\left(F_y\right)}_{\text{resultant}}}{{\left(F_x\right)}_{\text{resultant}}}\right)\\ {\text{= tan}}^{\text{-1}}\left(\frac{64}{47}\right)\\ =53.7°\end{array}$

Conclusion:

Therefore, the net force acting on the ring is $79.4\text{N}$ at angle $53.7°$ with the horizontal.