Question

Two forces, 80 N and 100 N, acting at an angle of 60 with each other, pull on an object. (a) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

Answer 1

(a) Find a single force that would replace the two forces algebraically. First, find the x- and y-components of the forces. We measure all angles as if the 100N-force were the x-axis:

$$F_{1x}=80\cos60°=40\text{ N}.\\ F_{2x}=100\cos0=100\text{ N}.\\ F_{1y}=80\sin60°=69.3\text{ N}.\\ F_{2y}=100\sin0=0\text{ N}.$$

The components of the resultant:

$$F_x=F_{1x}+F_{2x}=140\text{ N},\\ F_y=F_{1y}+F_{2y}=69.3\text{ N}.\\ F=\sqrt{F_x^2+F_y^2}=156\text{ N}.\\\space\\ \theta=\arctan \frac{F_y}{F_x}=26.3°.$$

This angle is measured above the axis on which the 100-N force lies.

(b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

The force that would balance must be opposite in direction yet equal in magnitude to the force we have found above:

$$F'=F=156\text{ N},\\ \theta'=\theta+180°=206.3°.$$