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A 10-kN load is suspended from the boom at D. Determine the force in the hydraulic cylinder BC and the pin reaction at A.

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Problem Statement

A 10-kN load is suspended from the boom at $$D$$. Determine the force in the hydraulic cylinder $$BC$$ and the pin reaction at $$A$$.

Solution

Equations of Equilibrium

Hydraulic cylinder $$BC$$ is a two-force member. The force $$F_{BC}$$ can be determined directly by writing the moment equation of

equilibrium about point $$A$$.Refer to the Free Body Diagram (FBD) of the boom:Sum of Moments about Point $$A$$ (ΣM$$_A$$ = 0):$F_{BC} \sin 15^\circ \times 1.5 - 10 \cos 30^\circ \times 4.5 = 0$Solving for $$F_{BC}$$:$F_{BC} = \frac{10 \cos 30^\circ \times 4.5}{\sin 15^\circ \times 1.5}$$F_{BC} = \frac{10 \times 0.866 \times 4.5}{0.259 \times 1.5}$$F_{BC} = \frac{38.97}{0.389}$$F_{BC} \approx 100.38 \text{ kN} \approx 100 \text{ kN}$Force Equations along $$x$$ and $$y$$ AxesUsing the result of $$F_{BC}$$, write the force equations of equilibrium:Sum of Forces in $$x$$-direction (ΣF$$_x$$ = 0):$100.38 \cos 45^\circ - A_x = 0$Solving for $$A_x$$:$A_x = 100.38 \cos 45^\circ$$A_x = 100.38 \times 0.707 = 70.98 \text{ kN} \approx 71 \text{ kN}$Sum of Forces in $$y$$-direction (ΣF$$_y$$ = 0):$100.38 \sin 45^\circ - 10 - A_y = 0$Solving for $$A_y$$:$A_y = 100.38 \sin 45^\circ - 10$$A_y = 100.38 \times 0.707 - 10 = 70.98 - 10 = 60.98 \text{ kN} \approx 61 \text{ kN}$

Summary of Answers - Force in the hydraulic cylinder $$BC$$:

$$F_{BC} \approx 100 \text{ kN}$$

- Pin reaction at $$A$$:

- $$A_x \approx 71 \text{ kN}$$

- $$A_y \approx 61 \text{ kN}$$

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