Question

Three horizontal forces are applied to a medal ring, as shown. One of the forces is 10N, and the second force is 12N at an angle of 100* from the first. Find the third forces such the ring is in equilibrium (i.e,. the sum of the three forces is zero)

Answer 1

Step-by-step explanation:

Horizontal components:

10 N + 12 cos (-100) = + 7.916 N

Vertical components

0 + 12 sin (-100) = - 11.818 N

To have equilibrium, the third force must be of the same MAGNITUDE and OPPOSITE direction

so Horizontal = - 7.916 N

Vertical = + 11.818 N

Magnitude = sqrt ( -7.916^2 + 11.818^2 ) = 14.22 N

at angle arctan (- 11.818/7.916 ) = 123.8 degrees