Tag Archives: components along non-perpendicular axes

Hibbeler Statics 14E P2.4 – Components of a Force Along Two Non-Perpendicular Axes

The vertical force F\textbf{F} acts downward at A on the two-membered frame. Determine the magnitudes of the two components of F\textbf{F} directed along the axes of AB and AC. Set F\textbf{F} = 500 N.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-4

Solution:

Draw the components of the force using the parallelogram law. Then the triangulation rule.

Parallelogram Law
Triangulation Rule

Solving for FAC using sine law.

FACsin⁡ 45∘=500 Nsin⁡ 75∘FAC=500 N sin⁡45∘sin⁡ 75∘FAC=366.0254 NFAC≈366 N\begin{align*} \frac{\text{F}_\text{AC}}{\sin \ 45^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\ \text{F}_\text{AC} & = \frac{500 \ \text{N} \ \sin45^\circ }{\sin\ 75^\circ }\\ \text{F}_\text{AC} & =366.0254 \ \text{N}\\ \text{F}_\text{AC} &\approx 366 \ \text{N} \end{align*}

Solve for FAB using sine law.

FABsin⁡ 60∘=500 Nsin⁡ 75∘FAB=500 N sin⁡60∘sin⁡ 75∘FAB=448.2877 NFAB≈448 N\begin{align*} \frac{\text{F}_\text{AB}}{\sin \ 60^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\ \text{F}_\text{AB} & = \frac{500 \ \text{N} \ \sin60^\circ }{\sin\ 75^\circ }\\ \text{F}_\text{AB} & =448.2877 \ \text{N}\\ \text{F}_\text{AB} &\approx 448 \ \text{N} \end{align*}
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