Question

Force x has a magnitude of 1210 ​pounds, and force y has a magnitude of 1540 pounds. they act on a single point at an angle of 15 degrees to each other. find the magnitude of the equilibrant and the angle between the equilibrant and the 1210 ​-pound force.

Answer 1

Two forces are given at 15 degree and the equilibrant is required between them

The formula to find the equilibrant is given by

$F = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 cos\theta}

[tex]F = \sqrt{1210^2 + 1540^2 + 2*1210*1540*cos15]

[tex]F = 2726.8 lb$

So the magnitude of equilibrant is 2726.8 lb

now in order to find the angle with respect to 1210 lb we can say

$\phi = tan^{-1}\frac{1540sin\theta}{1210 + 1540cos\theta}$

$\phi = tan^{-1}0.147$

$\phi = 8.4 degree$