Chapter 3.3, Problem 10P

The block has a weight of 20 lb and is being hoisted at uniform velocity. Determine the angle θ for equilibrium and the force in cord AB.

To determine

The angle $\theta$ for equilibrium and the force in cord AB.

Answer to Problem 10P

The angle $\left(\theta \right)$ is $\underset{\_}{40°}$ and the force in cord AB $\left(T_{AB}\right)$ is $\underset{\_}{37.6\text{lb}}$.

Explanation of Solution

Given information:

• The weight of block is 20 lb.

Assumption:

• The tension along the length of cord CAD is constant.

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

Determine the angle and force in cord AB by applying the Equation of equilibrium.

Along horizontal direction:

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

Along vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ T_{AB}\cos 20°-20\cos \theta -20=0\end{array}$ (II)

Formula used:

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

Substitute $\frac{20\sin \theta }{\sin 20°}$ for $T_{AB}$ in Equation (II).

$\begin{array}{l}{T}_{AB}\cos 20°-20\cos \theta -20=0\\ \frac{20\sin \theta }{\sin 20°}\cos 20°-20\cos \theta -20=0\\ 20\sin \theta \cos 20°-20\cos \theta \sin 20°=20\sin 20°\\ 20\left(\sin \theta \cos 20°-\cos \theta \sin 20°\right)=20\sin 20°\end{array}$ (III)

Conclusion:

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

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

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

$\begin{array}{c}{T}_{AB}=\frac{20\sin 40°}{\sin 20°}\\ =37.6\text{lb}\end{array}$

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

Thus, the force in cord AB is $\underset{\_}{37.6\text{lb}}$.