Chapter 3.3, Problem 11P

Determine the maximum weight W of the block that can be suspended in the position shown if cords AB and CAD can each support a maximum tension of 80 lb. Also, what is the angle θ for equilibrium?

To determine

The maximum weight of the block and the angle $\theta$ for equilibrium.

Answer to Problem 11P

The maximum weight of the block (W) is $\underset{\_}{42.6\text{lb}}$ and the angle for equilibrium $\left(\theta \right)$ is $\underset{\_}{40°}$.

Explanation of Solution

Given information:

• The cords AB and CAD can each support a maximum tension of 80 lb.

Assumption:

• For equilibrium, the tension along the length of cord CAD is constant.
• The tension in cord AB $\left(T_{AB}\right)$ reaches the limit first.

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

Determine the maximum weight of the block and the angle $\theta$ by applying the equation of equilibrium.

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ W\sin \theta -T_{AB}\sin 20°=0\\ W\sin \theta =T_{AB}\sin 20°\\ W=\frac{T_{AB}\sin 20°}{\sin \theta }\end{array}$ (I)

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ T_{AB}\cos 20°-W\cos \theta -W=0\\ W\left(1+\cos \theta \right)=T_{AB}\cos 20°\\ W=\frac{T_{AB}\cos 20°}{\left(1+\cos \theta \right)}\end{array}$ (II)

Formula used:

$\sin \theta \cos \varphi -\cos \theta \sin \varphi =\sin \left(\theta -\varphi \right)$

Conclusion:

Substitute $80\text{lb}$ for $T_{AB}$ in Equation (I).

$W=\frac{80\sin 20°}{\sin \theta }$ (III)

Substitute $80\text{lb}$ for $T_{AB}$ in Equation (II).

$W=\frac{80\cos 20°}{\left(1+\cos \theta \right)}$ (IV)

Substitute $\frac{80\sin 20°}{\sin \theta }$ for W in Equation (IV).

$\begin{array}{l}\frac{80\sin 20°}{\sin \theta }=\frac{80\cos 20°}{\left(1+\cos \theta \right)}\\ \sin 20°\left(1+\cos \theta \right)=\sin \theta \cos 20°\\ \sin 20°+\sin 20°\cos \theta =\sin \theta \cos 20°\\ \sin \theta \cos 20°-\sin 20°\cos \theta =\sin 20°\end{array}$ (V)

Substitute $\sin \left(\theta -20°\right)$ for $\left(\sin \theta \cos 20°-\cos \theta \sin 20°\right)$ in Equation (V).

$\begin{array}{l}\sin \left(\theta -20°\right)=\sin 20°\\ \theta -20°=20°\\ \theta =40°\end{array}$

Substitute $40°$ for $\theta$ in Equation (III).

$\begin{array}{c}W=\frac{80\sin 20°}{\sin 40°}\\ =42.6\text{lb<80}\text{lb}\end{array}$

Thus, the maximum weight of the block is $\underset{\_}{42.6\text{lb}}$.

Thus, the angle $\left(\theta \right)$ for equilibrium is $\underset{\_}{40°}$.