Chapter 6.1, Problem 50E

Solution

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

The direction is equal to ${35.65}^{\circ }$ and the magnitude is equal to $113.81$ .

Given information:

Three forces with magnitudes $100,50,80$ lb act on an object at angle of ${50}^{\circ },{160}^{\circ },-{20}^{\circ }$ respectively

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 $F_R=F_1+F_2+F_3$ .

Calculation:

Given,

  $F_1=100,{\theta }_1={50}^{\circ },F_2=50,{\theta }_2={160}^{\circ },F_3=80,{\theta }_3=-{20}^{\circ }$

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

So, the resultant force will be

  $\begin{array}{c}{F}_R=F_1+F_2+F_3\\ =〈100\cos {50}^{\circ },100\sin {50}^{\circ }〉+〈50\cos {160}^{\circ },50\sin {160}^{\circ }〉+〈80\cos \left(-{20}^{\circ }\right),80\sin \left(-{20}^{\circ }\right)〉\\ =〈64.28,76.60〉+〈-46.98,17.10〉+〈75.18,-27.36〉\\ =〈64.28-46.98+75.18,76.60+17.10-27.36〉\\ =〈92.48,66.34〉\end{array}$

Now, calculate the direction of the resultant force below:

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{66.34}{92.48}\right)\\ \approx {35.65}^{\circ }\end{array}$

Calculate the magnitude of the resultant force below:

  $\begin{array}{c}\left|F_R\right|=\sqrt{{\left(92.48\right)}^2+{\left(66.34\right)}^2}\\ \approx 113.81\end{array}$