Chapter 3.3, Problem 13P

A nuclear-reactor vessel has a weight of 500(10³) lb. Determine the horizontal compressive force that the spreader bar AB exerts on point A and the force that each cable segment CA and AD exert on this point white the vessel is hoisted upward at constant velocity.

Prob. 3–13

To determine

The horizontal compressive force that the spreader bar AB exerts on point A.

The force that each cable segment CA and AD exerts on point A.

Answer to Problem 13P

The horizontal compressive force that the spreader bar AB exerts on point A $\left(F_{AB}\right)$ is $\underset{\_}{433\times {10}^3\text{lb}}$.

The force that each cable segment CA and AD exert on point A, $\underset{\_}{F_{CA}=500\times {10}^3\text{lb}}$ $\underset{\_}{F_{AD}=250\times {10}^3\text{lb}}$.

Explanation of Solution

Given information:

The weight of a nuclear reactor vessel (W) is $500\times {10}^3\text{lb}$.

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

Show the free body diagram of the cable about point A as in Figure 2.

Determine the horizontal compressive force of each cable segment CA and AD by applying the equation of equilibrium.

At point C

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_{CB}\cos \theta -F_{CA}\cos \theta =0\\ F_{CB}\cos \theta =F_{CA}\cos \theta \\ F_{CB}=F_{CA}\end{array}$

Here, the force acting on cable CB is $F_{CB}$ and the force acting on cable CA is $F_{CA}$.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ W-F_{CA}\sin \theta -F_{CB}\sin \theta =0\end{array}$ (I)

At point A

Along the horizontal direction:

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

Here, the force acting on cable AB is $F_{AB}$.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ W\sin \theta -F_{AD}=0\\ F_{AD}=W\sin \theta \end{array}$ (III)

Here, the force acting on cable AD is $F_{AD}$.

Conclusion:

Substitute $30°$ for $\theta$ and $F_{CB}$ for $F_{CA}$ and $500\times {10}^3\text{lb}$ for W in Equation (I).

$\begin{array}{l}500\times {10}^3-F_{CA}\sin 30°-F_{CA}\sin 30°=0\\ 2F_{CA}\sin 30°=500\times {10}^3\\ F_{CA}=\frac{500\times {10}^3}{2\sin 30°}\\ F_{CA}=500\times {10}^3\text{lb}\end{array}$

Substitute $30°$ for $\theta$ and $500\times {10}^3\text{lb}$ for W in Equation (II).

$\begin{array}{c}{F}_{AB}=500\times {10}^3\cos 30°\\ =433\times {10}^3\text{lb}\end{array}$

Substitute $30°$ for $\theta$ and $500\times {10}^3\text{lb}$ for W in Equation (III).

$\begin{array}{c}{F}_{AD}=500\times {10}^3\sin 30°\\ =250\times {10}^3\text{lb}\end{array}$

Thus, the horizontal compressive force that the spreader bar AB exerts on point A is $\underset{\_}{433\times {10}^3\text{lb}}$.

The force that each cable segment CA and AD exerts on point A is $\underset{\_}{500\times {10}^3\text{lb}}$ and $\underset{\_}{250\times {10}^3\text{lb}}$.