Chapter 3, Problem 1FP

The crate has a weight of 550 lb. Determine the force in each supporting cable.

To determine

The force in each supporting cable.

Answer to Problem 1FP

The force in cable AB $\left(F_{AB}\right)$ is $\underset{\_}{478\text{lb}}$.

The force in cable AC $\left(F_{AC}\right)$ is $\underset{\_}{518\text{lb}}$.

Explanation of Solution

Given information:

The weight of a crate is (W) is 550 lb.

Show the free body diagram of the cable as in Figure 1.

Determine the force of the members AB and AC by applying the Equation of equilibrium.

Along horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{AC}\cos {\theta }_{AC}-F_{AB}\cos {\theta }_{AB}=0\end{array}$ (I)

Here, the angle of force AC is ${\theta }_{AC}$ and the angle of force AB is ${\theta }_{AB}$.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{AC}\sin {\theta }_{AC}+F_{AB}\sin {\theta }_{AB}-W=0\end{array}$ (II)

Conclusion:

Substitute $\left(\frac{4}{5}\right)$ for $\cos {\theta }_{AC}$ and $30°$ for ${\theta }_{AB}$ in Equation (I).

$\begin{array}{l}{F}_{AC}\left(\frac{4}{5}\right)-F_{AB}\cos 30°=0\\ F_{AC}\left(\frac{4}{5}\right)=0.866F_{AB}\\ F_{AC}=\left(0.866\times \frac{5}{4}\right)F_{AB}\\ F_{AC}=1.0825F_{AB}\end{array}$ (III)

Substitute $1.0825F_{AB}$ for $F_{AC}$, $\left(\frac{3}{5}\right)$ for $\sin {\theta }_{AC}$ and $30°$ for ${\theta }_{AB}$, 550 lb for W in Equation (II).

$\begin{array}{l}\left(1.0825F_{AB}\right)\left(\frac{3}{5}\right)+F_{AB}\sin 30°-550=0\\ 0.649F_{AB}+0.5F_{AB}=550\\ 1.149F_{AB}=550\\ F_{AB}=\frac{550}{1.149}\end{array}$

$F_{AB}=478\text{lb}$

Thus, the force in cable AB $\left(F_{AB}\right)$ is $\underset{\_}{478\text{lb}}$.

Substitute 478 lb for $F_{AB}$ in Equation (III).

$\begin{array}{c}{F}_{AC}=1.0825\times 478\\ =518\text{lb}\end{array}$

Thus, the force in cable AC $\left(F_{AC}\right)$ is $\underset{\_}{518\text{lb}}$.