Chapter 9.4, Problem 75AYU

To determine

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

Answer to Problem 75AYU

  $F=80.27N;F_y=-{16.2}^{\circ }$ .

Explanation of Solution

Given:

Two force of magnitudes $40N\text{ and 6}0N$ act on object at angles of ${30}^{\circ }\text{ and }-{45}^{\circ }$ with the positive x-axis.

  

Calculation:

As it is known that,

Components of force vector are

  $\begin{array}{l}{F}_{1x}=40\cos {30}^{\circ }=40\left(\frac{\sqrt{3}}{2}\right)=20\sqrt{3}\{\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\}\\ F_{1y}=40\sin {30}^{\circ }=40\left(\frac{1}{2}\right)=20\{sin{30}^{\circ }=\frac{1}{2}\}\\ F_{2x}=60\cos {\left(-45\right)}^{\circ }=60\cos {\left(45\right)}^{\circ }=60\left(\frac{1}{\sqrt{2}}\right)=30\sqrt{2}\{\cos {\left(-45\right)}^{\circ }=\frac{1}{\sqrt{2}}\}\\ F_{2y}=60\sin {\left(-45\right)}^{\circ }=-60\sin {\left(45\right)}^{\circ }=60\left(\frac{1}{\sqrt{2}}\right)=-30\sqrt{2}\{sin{30}^{\circ }=\frac{1}{2}\}\end{array}$ .

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

  $\begin{array}{l}{F}_x=F_{1x}+F_{2x}=77.07N\\ F_y=F_{1y}+F_{2y}=-22.43N\end{array}$

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

  $\begin{array}{l}\left|F\right|=\sqrt{F_x^2+F_y^2}=80.27N\\ F_y=\arctan \frac{F_y}{F_x}=-{16.2}^{\circ }\end{array}$

Hence, the direction and magnitude of the resultant force is $F=80.27N;F_y=-{16.2}^{\circ }$ .