A 4.30 kg sign hangs from two wires. The left wire experts a 31.0 N force at 122 degrees. What is the direction of the force exerted by a second wire?

Answer 1

we have the equations:

F₁ sin α + F₂ sin β = F₀ = mg (Eq.1)

F₁ cos α + F₂ cos β =0 (Eq.2)

solve the equations we can get the answer β.

From Eq.2, we have cos β = - F₁ cos α / F₂, so , put it into Eq.1, we have:

,

with the known values of m, g, F₁ and α, we can obtain F₂ = 22.83N, and β = 43.98°.