Chapter 16.1, Problem 16.61P

To determine

Tension in each cable.

Answer to Problem 16.61P

The tension in cable $A$ is $202.056\text{lb}$.

The tension in cable $B$ is $431.01\text{lb}$.

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 $20{\text{ft}/\text{s}}^2$, the deceleration of cable $B$ is $2{\text{ft}/\text{s}}^2$ and distance between cable is $12\text{ft}$.

The below figure represents the kinematics 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.

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

Here, the acceleration of cable $A$ is $a_A$, the distance between cable is $r$ and the angular acceleration of the beam is $\alpha$.

Write the expression of total acceleration on the beam.

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

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

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 distance between mid of beam and the cable $B$ as shown in Figure.

$h=\frac{L}{2}-\left(L-r\right)$   ............... (VI)

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

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

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

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

$\sum M_B=T_{A}r-Wh$   ............... (VIII)

Here, the tension in the cable $A$ is $T_A$ and distance between mid of beam and the cable $B$ is $h$.

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

${\sum \left(M_B\right)}_{\text{effective}}=T+F_{\text{total}}h$   ............... (IX)

Here, the angular acceleration of the rod is $\alpha$, the total acceleration on the beam is $a$ and the total force is $F_{\text{total}}$.

Write the expression of generated torque in the beam.

$T=I\alpha$   ............... (X)

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

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

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

Substitute $T_{A}r-Wh$ for $\sum M_B$ and $I\alpha +F_{\text{total}}h$ for ${\sum \left(M_B\right)}_{\text{effective}}$ in Equation (VII).

$\begin{array}{l}{T}_{A}r-Wh=I\alpha +F_{\text{total}}h\\ T_{A}r=I\alpha +F_{\text{total}}h+Wh\\ T_A=\frac{1}{r}\left(I\alpha +F_{\text{total}}h+Wh\right)\end{array}$   ............... (XII)

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

$T_A=\frac{1}{r}\left(I\alpha +mah+Wh\right)$   ............... (XIII)

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

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

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

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

$T_B=ma+W-T_A$   ............... (XV)

Calculation:

Substitute $2{\text{ft}/\text{s}}^2$ for $a_B$, $20{\text{ft}/\text{s}}^2$ for $a_A$ and $12\text{ft}$ for $r$ in Equation (II).

$\begin{array}{l}2{\text{ft}/\text{s}}^2=20{\text{ft}/\text{s}}^2+\left(12\text{ft}\right)\alpha \\ \left(2{\text{ft}/\text{s}}^2\right)-\left(20{\text{ft}/\text{s}}^2\right)=\left(12\text{ft}\right)\alpha \\ \alpha =\frac{-18{\text{ft}/\text{s}}^2}{12\text{ft}}\\ \alpha =-1.5{\text{rad}/\text{s}}^2\end{array}$

Substitute $20{\text{ft}/\text{s}}^2$ for $a_A$, $15\text{ft}$ for $L$ and $-1.5{\text{rad}/\text{s}}^2$ for $\alpha$ in Equation (III).

$\begin{array}{c}a=\left(20{\text{ft}/\text{s}}^2\right)+\frac{\left(15\text{ft}\right)}{2}\left(-1.5{\text{rad}/\text{s}}^2\right)\\ =\left(20{\text{ft}/\text{s}}^2\right)-\left(7.5\text{ft}\right)\left(1.5{\text{rad}/\text{s}}^2\right)\\ =\left(20{\text{ft}/\text{s}}^2\right)-\left(11.25{\text{ft}/\text{s}}^2\right)\\ =8.75{\text{ft}/\text{s}}^2\end{array}$

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

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

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

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

Substitute $15\text{ft}$ for $L$ and $12\text{ft}$ for $r$ in Equation (VI).

$\begin{array}{c}h=\frac{\left(15\text{ft}\right)}{2}-\left(\left(15\text{ft}\right)-\left(12\text{ft}\right)\right)\\ =\left(7.5\text{ft}\right)-\left(3\text{ft}\right)\\ =4.5\text{ft}\end{array}$

Substitute $12\text{ft}$ for $r$, $291.13\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$ for $I$, $-1.5{\text{rad}/\text{s}}^2$ for $\alpha$, $15.527\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $4.5\text{ft}$ for $h$, $500\text{lb}$ for $W$ and $8.75{\text{ft}/\text{s}}^2$ for $a$ in Equation (XIII).

$\begin{array}{c}{T}_A=\frac{1}{\left(12\text{ft}\right)}\left\{\begin{array}{l}\left(291.13\text{lb}\cdot \text{ft}\cdot {\text{s}}^2\right)\times \left(-1.5{\text{rad}/\text{s}}^2\right)\\ +\left(15.527\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times \left(8.75{\text{ft}/\text{s}}^2\right)\times \left(4.5\text{ft}\right)+\left(500\text{lb}\right)\times \left(4.5\text{ft}\right)\end{array}\right\}\\ =\frac{1}{\left(12\text{ft}\right)}\left\{-\left(436.695\text{lb}\cdot \text{ft}\right)+\left(611.376\text{lb}\cdot \text{ft}\right)+\left(2250\text{lb}\cdot \text{ft}\right)\right\}\\ =\frac{2424.68\text{lb}\cdot \text{ft}}{\left(12\text{ft}\right)}\\ =202.056\text{lb}\end{array}$

Substitute $15.527\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $500\text{lb}$ for $W$, $8.75{\text{ft}/\text{s}}^2$ for $a$ and $202.056\text{lb}$ for $T_A$ in Equation (XV).

$\begin{array}{c}{T}_B=\left(15.527\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\times \left(8.75{\text{ft}/\text{s}}^2\right)+\left(500\text{lb}\right)-202.056\text{lb}\\ \text{=}\left(\text{133}\text{.066}\text{lb}\right)+\left(500\text{lb}\right)-\left(202.056\text{lb}\right)\\ =431.01\text{lb}\end{array}$

Conclusion:

The tension in cable $A$ is $202.056\text{lb}$.

The tension in cable $B$ is $431.01\text{lb}$.