At the B end of the recessed horizontal bar, forces F1 and F2 of magnitudes 3 KN and 2KN respectively are applied and oriented as shown in the figure. Determine the magnitude of the resulting force and its orientation with respect to the horizontal.

2.98 kN, 69.1° below the horizontal

Explanation:

Sum of forces in the x direction:

Fₓ = 3 kN cos 30° − 2 kN cos 40°

Fₓ ≈ 1.07 kN

Sum of forces in the y direction:

Fᵧ = -3 kN sin 30° − 2 kN sin 40°

Fᵧ = -2.79 kN

Magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F = 2.98 kN

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

θ = -69.1°

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