Answer to Question #202574 in Molecular Physics | Thermodynamics for Tejay Mawal

Question

Four forces act on a point 150 gm SE, 300 gm due S 60° W . 280 gm due W 

and 180 gm 60° S of E. (a) What single force will replace the four forces? (b) What 

single force (called the equilibrant) would balance the four forces? Solve algebraically

Answer 1

(a) Let x-axis along east and y-axis along the north direction. Breaking the forces along the x and y-axis.

150 gm SE component, $\vec{F_1} = 150cos(45)\hat{i} - 150sin(45)\hat{j}$

300 gm due S 60° W, $\vec{F_2} =- 200sin(60)\hat{i} - 200cos(60)\hat{j}$

280 gm due W , $\vec{F_3} = -280\hat{i}$

180 gm 60° S of E, $\vec{F_4} = 180cos(60)\hat{i} - 180sin(60)\hat{j}$

Resultant Force:

$\vec{F}=\vec{F_1} +\vec{F_2}+\vec{F_3}+\vec{F_4}$

$\vec{F} = 150cos(45)\hat{i} - 150sin(45)\hat{j} - 200sin(60)\hat{i} - 200cos(60)\hat{j} -280\hat{i} +180cos(60)\hat{i} - 180sin(60)\hat{j}$

$\vec{F} = (-257.14)\hat{i}+(-361.95)\hat{i}$

Magnitude of the force, $F = \sqrt{(257.14)^2+(361.95)^2} = 443.99 gm$

Direction, $\theta = tan^{-1}\frac{-361.95}{-257.14} = 54.61 ^0$ south of west.

(b) The single force that can balance it will be of magnitude 443.99 gm in 54.61 degree north of east.