Chapter 6.1, Problem 49E

Solution

To find: The direction and magnitude of the resultant force.

The direction is equal to ${89.42}^{\circ }$ and the magnitude is equal to $100.33$ .

Given information:

A force of $50$ lb acts on an object at an angle of ${45}^{\circ }$ . A second force of $75$ lb acts on the object at an angle of $-{30}^{\circ }$ .

Concept used:

The component form of a vector is $\overrightarrow{a}=〈\left|a\right|\cos \theta ,\left|a\right|\sin \theta 〉$ .

The magnitude of the vector is $\left|a\right|=\sqrt{x^2+y^2}$ .

The direction of the vector is $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ .

The resultant force is equal to $F_R=F_1+F_2$ .

Calculation:

Given,

  $F_1=50,{\theta }_1={45}^{\circ },F_2=75,{\theta }_2=-{30}^{\circ }$

Apply the above concept and find the magnitude and direction of the resultant force below.

So, the resultant force will be

  $\begin{array}{c}{F}_R=F_1+F_2\\ =〈50\cos {45}^{\circ },50\sin {45}^{\circ }〉+〈75\cos \left(-{30}^{\circ }\right),75\sin \left(-{30}^{\circ }\right)〉\\ =〈\frac{50}{\sqrt{2}},\frac{50}{\sqrt{2}}〉+〈\frac{75\sqrt{3}}{2},-\frac{75}{2}〉\\ =〈\frac{50}{\sqrt{2}}+\frac{75\sqrt{3}}{2},\frac{50}{\sqrt{2}}-\frac{75}{2}〉\\ =〈\frac{100+75\sqrt{6}}{2\sqrt{2}},\frac{100-75\sqrt{2}}{2\sqrt{2}}〉\end{array}$

Now, calculate the direction of the resultant force below:

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\frac{100-75\sqrt{2}}{2\sqrt{2}}}{\frac{100+75\sqrt{6}}{2\sqrt{2}}}\right)\\ ={\tan }^{-1}\left(\frac{100-75\sqrt{2}}{100+75\sqrt{6}}\right)\\ \approx {89.42}^{\circ }\end{array}$

Calculate the magnitude of the resultant force below:

  $\begin{array}{c}\left|F_R\right|=\sqrt{{\left(\frac{100+75\sqrt{6}}{2\sqrt{2}}\right)}^2+{\left(\frac{100-75\sqrt{2}}{2\sqrt{2}}\right)}^2}\\ =\sqrt{\frac{10000+33750+15000\sqrt{6}+10000+11250-15000\sqrt{2}}{8}}\\ =\sqrt{\frac{65000+15000\sqrt{6}-15000\sqrt{2}}{8}}\\ \approx 100.33\end{array}$