Answer to Question #81912 in Physics for cyril earl jay remedio

Question

Three forces; f1=40 newstons and horizontal to the left, f2 =20 newstons and vertically up, and f3 =20 newstons of 45 degree below positive x-axis, act at common point in the body. Find the fourth force of the equilibrant which is required to make the body in equilibrium.

Answer 1

f_1=(-40,0) N;

f_2=(0,20) N;

f_3=(20 cos⁡45,-20 sin⁡45 ) N.

The fourth force of the equilibrant which is required to make the body in equilibrium:

f_4=-(f_1+f_2+f_3)

f_4=-((-40,0)+(0,20)+(20 cos⁡45,-20 sin⁡45 ))=(40-20 cos⁡45,20 sin⁡45-20)

f_4=(25.9,-5.86) N.

The fourth force is 27 N of 13 degree below positive x-axis.