Question

1. 3 forces 47 N, 36 N and 20 N are oriented with x axis with angles of 45 degree, 135 degree and 225 degree respectively. What is their resultant?

2. A 20 N force is oriented at a certain angle with x axis and you measured y component = 7.3310 N. What is the angle of 20 N with x axis?

Instruction - Take sine of the answer

Answer 1

1)

Fnetx = F1x + F2x + F3x

= 47*cos(45) + 36*cos(135) + 20*cos(225)

= -6.36 N

Fnety = F1y + F2y + F3y

= 47*sin(45) + 36*sin(135) + 20*sin(225)

= 44.5 N

magnitude of resulstant force,
Fnet = sqrt(Fnetx^2 + Fnety^2)

= sqrt(6.36^2 + 44.5^2)

= 45 N <<<<<<<<<-----------------------Answer

2) given F = 20 N
Fy = 7.3310 N
Fx = ?

let theta is the angle made by the Force with +x axis.

use, sin(theta) = Fy/F

sin(theta) = 7.331/20

theta = sin^-1(7.331/20)

= 21.50 degrees

now use, cos(theta) = Fx/F

Fx = F*cos(theta)

= 20*cos(21.50)

= 18.610 N <<<<<<<<<<<<<<--------------Answer