Question

A 4.30 kg sign hangs from two wires. The
left wire exerts a 31.0 N force at 122°.
What is the magnitude of the force exerted
by the second wire?
F₁"
*W
magnitude (N)
F2
Enter

Answer 1

The magnitude of the force exerted by the second wire is 16.5 N.

Resultant of the forces

Apply cosine rule to determine the force on the second wire;

W² = F₁² + F₂² - 2F₁F₂ cos(θ)

where;

• W is the suspended weight
• θ is the angle between the two wires

(4.3 x 9.8)² = 31²  +  F₂²  - 2(31) F₂(cos 122)

1775.78 = 961  +  F₂²  + 32.85F₂

814.78 = F₂²  + 32.85F₂

F₂²  + 32.85F₂ - 814.78 = 0

Solve the quadratic equation using formula method;

F₂ = 16.5 N

Thus, the magnitude of the force exerted by the second wire is 16.5 N.

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