3 forces, 5N, 6N and 7N act on a particle which remains in a state of equilibrium. What is the angle between the 5N and 6N force?

Question

how do I do this using vectors?

Answer 1

Let the 5 Newton force be in the positive x-direction.

Let the 6 Newton force be at angle θ counterclockwise, and the 7 Newton force have angle φ clockwise from the 5 Newton force.

Then for the zero net force, we get two equations.

5 + 6 cos θ + 7 cos φ = 0

6 sin θ - 7 sin φ = 0

We can solve this algebraically to get cos θ = -1/5 and cos φ = -19/35.

So the angle between the 5 Newton and 6 Newton forces is θ = arccos(-1/5) ≈ 101.537°.