Answer to Question #171777 in Physics for Ade

Question

if three forces f1 =20n, 30 degree north east, f2 = 50n along west, f3 =40n, 50 degree north west act on a body. find the resultant in magnitude and direction

Answer 1

Let's find the $x$- and $y$-components of the resultant force:

$$F_{res,x}=F_{1x}+F_{2x}+F_{3x},$$

$F_{res,x}=20\ N\cdot cos30^{\circ}+50\ N\cdot cos180^{\circ}+40\ N\cdot cos(180^{\circ}-50^{\circ})=-58.4\ N,$

$$F_{res,y}=F_{1y}+F_{2y}+F_{3y},$$

$F_{res,y}=20\ N\cdot sin30^{\circ}+50\ N\cdot sin180^{\circ}+40\ N\cdot sin(180^{\circ}-50^{\circ})=40.64\ N.$

We can find the magnitude of the resultant force from the Pythagorean theorem:

$$F_{res}=\sqrt{F_{res,x}^2+F_{res,y}^2},$$ $$F_{res}=\sqrt{(-58.4\ N)^2+(40.64\ N)^2}=71.15\ N.$$

We can find the direction of the resultant force from the geometry:

$$\theta=sin^{-1}(\dfrac{F_{res,y}}{F_{res}}),$$ $$\theta=sin^{-1}(\dfrac{40.64\ N}{71.15\ N})=34.8^{\circ}.$$

The direction of the resultant force is $34.8^{\circ}\ N\ of\ W.$