Answer to Question #243092 in Physics for Mary may Oseh

Question

If three forces of 12N, 15N and 20N respectively that are concurrent are acting at a point and in equilibrium, calculate the angle between them

Answer 1

Align the 20-N force along the y-axis, then, to keep the body in equilibrium, the horizontal and vertical components of the two other forces must be equal:

$20-12\cos\theta_1-15\cos\theta_2=0,\\\space\\ 12\sin\theta_1-15\sin\theta_2=0.\\\space\\ 12^2(1-\cos^2\theta_1)=15^2(1-\cos^2\theta_2),\\\space\\ (15\cos\theta_2)^2-(12\cos\theta_1)^2-81=0.$

So, we have the upper and lower equations that form quite a pretty system of equations. Solve the system (you may use substitution $12\cos\theta_1=x,\space15\cos\theta_2=y$):

$12\cos\theta_1=7.975,\\ 15\cos\theta_2=12.025,\\ \theta_1=48.3°,\\ \theta_2=36.7°.$