Chapter 16.1, Problem 16.60P

A 15-ft beam weighing 500 Ib is lowered by mean of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration of cable A is 20 ft/s² and the deceleration of cable B is 2 ft/s², determine the tension in each cable.

To determine

The tension in each cable.

Answer to Problem 16.60P

The tension in cable $A$ is $312.108\text{lb}$.

The tension in cable $B$ is $358.689\text{lb}$.

Explanation of Solution

Given information:

The length of the beam is $15\text{ft}$, the weight of the beam is $500\text{lb}$, the deceleration of cable $A$ is $20{\text{ft}/\text{s}}^2$ and the deceleration of cable $B$ is $2{\text{ft}/\text{s}}^2$.

The below figure represent the kinetics of the beam.

Figure-(1)

Write the expression of total force applied on the beam.

$F_{\text{total}}=ma$ ...... (I)

Here, the mass of the beam is $m$ and the acceleration is $a$.

Consider the tension in the cable $A$ is $T_A$ and the tension in the cable $B$ is $T_B$.

Write the expression of deceleration of cable $B$ by using kinematics of the beam.

$a_B=a_A+L\alpha$ ...... (II)

Here, the acceleration of cable $A$ is $a_A$, the length of the beam is $L$ and the angular acceleration of the beam is $\alpha$.

Write the expression of total acceleration on the beam.

$a=\frac{a_A+a_B}{2}$ ...... (III)

Here, the acceleration of the cable $B$ is $a_B$.

Write the expression of moment of inertia of beam.

$I=\frac{m{L}^2}{12}$ ...... (IV)

Here, the mass of the slender rod is $m$ and the length of the rod is $L$.

Write the expression of mass of the rod.

$m=\frac{W}{g}$ ...... (V)

Here, the weight of the beam is $W$ and the acceleration due to gravity is $g$.

Write the expression of total moment about point $B$ by applying equation of motion as shown in Figure-(1)

$\sum M_B=\sum {\left(M_B\right)}_{\text{effective}}$ ...... (VI)

Here, the effective moment is ${\left(M_G\right)}_{\text{effective}}$.

Write the expression of moment about point $B$ as shown in Figure-(1).

$\sum M_B=T_{A}L-\frac{WL}{2}$ ...... (VII)

Here, the tension in the cable $A$ is $T_A$

Write the expression of effective moment about point $B$ as shown in Figure-(1).

${\sum \left(M_B\right)}_{\text{effective}}=T+\frac{F_{\text{total}}L}{2}$ ...... (VIII)

Here, the angular acceleration of the rod is $\alpha$ and total acceleration on the beam is $a$.

Write the expression of generated torque in the beam.

$T=I\alpha$ ...... (IX)

Here, the of inertia of beam is $I$.

Substitute $I\alpha$ for $T$ in Equation (IX).

${\sum \left(M_B\right)}_{\text{effective}}=I\alpha +F_{\text{total}}h$ ...... (X)

Substitute $ma$ for $F_{\text{total}}$ in Equation (VIII).

${\sum \left(M_B\right)}_{\text{effective}}=I\alpha +\frac{maL}{2}$ ...... (XI)

Substitute $I\alpha +\frac{maL}{2}$ for ${\sum \left(M_B\right)}_{\text{effective}}$ and $T_{A}L-\frac{WL}{2}$ for $\sum M_B$ in Equation (V).

$\begin{array}{l}{T}_{A}L-\frac{WL}{2}=I\alpha +\frac{maL}{2}\\ T_{A}L=I\alpha +\frac{maL}{2}+\frac{WL}{2}\\ T_A=\frac{1}{L}\left(I\alpha +\frac{maL}{2}+\frac{WL}{2}\right)\\ T_A=\frac{I\alpha }{L}+\frac{ma}{2}+\frac{W}{2}\end{array}$ ...... (XII)

Write the expression of total force applied on the beam by equilibrium of the beam as shown in Figure-(1).

$\begin{array}{l}{T}_A+T_B-W=F_{\text{total}}\\ T_B=F_{\text{total}}+W-T_A\end{array}$ ...... (XIII)

Substitute $ma$ for $F_{\text{total}}$ in Equation (XI)

$T_B=ma+W-T_A$ ...... (XIV)

Calculation:

Substitute $2{\text{ft}/\text{s}}^2$ for $a_B$, $20{\text{ft}/\text{s}}^2$ for $a_A$ and $15\text{ft}$ for $L$ in Equation (II).

$\begin{array}{l}2{\text{ft}/\text{s}}^2=20{\text{ft}/\text{s}}^2+\left(15\text{ft}\right)\alpha \\ \left(2{\text{ft}/\text{s}}^2\right)-\left(20{\text{ft}/\text{s}}^2\right)=\left(15\text{ft}\right)\alpha \\ \alpha =\frac{-18{\text{ft}/\text{s}}^2}{15\text{ft}}\\ \alpha =-1.2{\text{rad}/\text{s}}^2\end{array}$

Substitute $20{\text{ft}/\text{s}}^2$ for $a_A$ and $2{\text{ft}/\text{s}}^2$ for $a_B$ in Equation (II).

$\begin{array}{c}a=\frac{\left(20{\text{ft}/\text{s}}^2\right)+\left(2{\text{ft}/\text{s}}^2\right)}{2}\\ =\frac{22{\text{ft}/\text{s}}^2}{2}\\ =11{\text{ft}/\text{s}}^2\end{array}$

Substitute $500\text{lb}$ for $W$ and $32.2{\text{ft}/\text{s}}^2$ for $g$ in Equation (IV).

$\begin{array}{c}m=\frac{\left(500\text{lb}\right)}{\left(32.2{\text{ft}/\text{s}}^2\right)}\\ =15.527\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Substitute $15.527\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$ and $15\text{ft}$ for $L$ in Equation (III).

$\begin{array}{c}I=\frac{\left(15.527\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times {\left(15\text{ft}\right)}^2}{12}\\ =\frac{3493.575\text{lb}\cdot \text{ft}\cdot {\text{s}}^2}{12}\\ =291.13\text{lb}\cdot \text{ft}\cdot {\text{s}}^2\end{array}$

Substitute $291.13\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$ for $I$, $-1.2{\text{rad}/\text{s}}^2$ for $\alpha$, $15\text{ft}$ for $L$, $15.527\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $11{\text{ft}/\text{s}}^2$ for $a$ and $500\text{lb}$ for $W$ in Equation (XII).

$\begin{array}{c}{T}_A=\frac{\left(291.13\text{lb}\cdot \text{ft}\cdot {\text{s}}^2\right)\times \left(-1.2{\text{rad}/\text{s}}^2\right)}{\left(15\text{ft}\right)}+\frac{\left(15.527\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times \left(11{\text{ft}/\text{s}}^2\right)}{2}+\frac{\left(500\text{lb}\right)}{2}\\ =\frac{-\left(349.356\text{lb}\cdot \text{ft}\right)}{\left(15\text{ft}\right)}+\frac{\left(170.797\text{lb}\right)}{2}+\frac{\left(500\text{lb}\right)}{2}\\ =-\left(23.29\text{lb}\right)+\left(85.398\text{lb}\right)+\left(250\text{lb}\right)\\ =312.108\text{lb}\end{array}$

Substitute $312.108\text{lb}$ for $T_A$, $15.527\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $11{\text{ft}/\text{s}}^2$ for $a$ and $500\text{lb}$ for $W$ in Equation (XIV).

Conclusion:

The tension in cable $A$ is $312.108\text{lb}$.

The tension in cable $B$ is $358.689\text{lb}$.