Chapter 9.4, Problem 76AYU

To determine

To find: the direction and magnitude of the resultant force ${\text{F}}_1+{\text{F}}_2$ .

Answer to Problem 76AYU

  $F=82.98N;F_y={99.57}^{\circ }$ .

Explanation of Solution

Given:

Two force of magnitudes $30N\text{ and 7}0N$ act on object at angles of ${45}^{\circ }\text{ and }{120}^{\circ }$ with the positive x-axis.

  

Calculation:

As it is known that,

Components of force vector are

  $\begin{array}{l}{F}_{1x}=30\cos {45}^{\circ }=30\left(\frac{1}{\sqrt{2}}\right)=15\sqrt{2}\{\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\}\\ F_{1y}=30\sin {45}^{\circ }=30\left(\frac{1}{\sqrt{2}}\right)=15\sqrt{2}\{sin{30}^{\circ }=\frac{1}{2}\}\\ F_{2x}=70\cos {120}^{\circ }=-70\cos {\left(60\right)}^{\circ }=-70\left(\frac{1}{2}\right)=-35\{\cos {\left(-45\right)}^{\circ }=\frac{1}{\sqrt{2}}\}\\ F_{2y}=70\sin {120}^{\circ }=70\sin {\left(60\right)}^{\circ }=70\left(\frac{\sqrt{3}}{2}\right)=35\sqrt{3}\{sin{30}^{\circ }=\frac{1}{2}\}\end{array}$ .

  $x\text{ and }y$ components of the resultant vector F are just sum of corresponding components ${\text{F}}_1\text{and}{\text{F}}_2$

  $\begin{array}{l}{F}_x=F_{1x}+F_{2x}=-13.79N\\ F_y=F_{1y}+F_{2y}=81.83N\end{array}$

Finally, magnitude $\left|\text{F}\right|$ and direction $\theta$ of the resulting force are

  $\begin{array}{l}\left|F\right|=\sqrt{F_x^2+F_y^2}=82.98N\\ F_y=\arctan \frac{F_y}{F_x}={99.57}^{\circ }\end{array}$

Hence, the direction and magnitude of the resultant force is $F=82.98N;F_y={99.57}^{\circ }$ .