Equilibrium of a particle: two suspended loads

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Question 1

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Two loads $W_1 = \var{w1}$ , and $W_2 = \var{w2}$ are suspended from a system of cables as shown. Determine the tensions in the cables.

First Particle

Draw a free body diagram of point A and apply the equations of equilibrium to determine the tensions in cables AB and AC.

$T_{AB}$ = Expected answer:

$T_{AC}$ = Expected answer:

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Second Particle

Draw a free body diagram of point C and apply the equations of equilibrium to determine the tensions in cables CE and CD.

$T_{CE}$ = Expected answer:

$T_{CD}$ = Expected answer:

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Advice

Begin by drawing a free body diagram of particle A, then apply the equations of equilibrium.

$\begin{align}\\ \Sigma F_x = 0 \\-{T_B}_x + {T_C}_x = 0 \end{align} \\ T_B \cos(\var{a1}°) = T_C \sin(\var{90-a2}°) \qquad(1)$

$\begin{align}\\ \Sigma F_y = 0 \\{T_B}_y + {T_C}_y - W_1 = 0 \end{align} \\ T_B \sin(\var{a1}°) + T_C \cos(\var{90-a2})° = W_1 \qquad(2)$

Solve (1) and (2) simultaneously for $T_B$ and $T_C$ .

$T_B = \var{siground(TB,4)}, \quad T_C = \var{siground(TC,4)}$

With $T_C$ known, use a similar approach on particle $C$ to find $T_D$ and $T_E$ .

Since cable CD is horizontal $T_D$ has no y-component, so the $\Sigma F_y = 0$ equation only has one unknown and simultaneous equations are unnecessary in this case.

$T_D = \var{siground(TD,4)}, \quad T_E = \var{siground(TE,4)}$

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Equilibrium of a particle: two suspended loads

First Particle

Second Particle

Advice