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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 \( A \):       

   - \( A_x \approx 71 \text{ kN} \)   

   - \( A_y \approx 61 \text{ kN} \)

by (115k points)

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