Chapter 5.7, Problem 7FP

The uniform plate has a weight of 500 lb. Determine the tension in each of the supporting cables.

Prob. F5-7

To determine

The tension in each of the supporting cables.

Answer to Problem 7FP

The tension $\left(T_A\right)$ in cable A is $\underset{\_}{350\text{lb}}$, the tension $\left(T_B\right)$ in cable B is $\underset{\_}{250\text{lb}}$, and the tension $\left(T_C\right)$ in cable C is $\underset{\_}{100\text{lb}}$.

Explanation of Solution

Given information:

• The weight (W) of the plate is 500 lb.
• The load (P) acting on the plate is 200 lb.

Explanation:

Show the free body diagram of the uniform plate as in Figure (1).

Use Figure (1).

Apply the equation of equilibrium along the z axis.

$\begin{array}{l}{\displaystyle \sum F_z=0}\\ T_A+T_B+T_C-P-W=0\end{array}$ (I)

Take the moment about the x axis.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ T_A\left(x_1\right)+T_C\left(x_2\right)-W\left(x_3\right)-P\left(x_4\right)=0\end{array}$ (II)

Here, the horizontal distance between cables B and A is $x_1$, the horizontal distance between cables B and C is $x_2$, the center of gravity of the plate about x axis is $x_3$, and the horizontal distance between the load and cable B is $x_4$.

Take the moment about the y axis.

$\begin{array}{l}{\displaystyle \sum M_y=0}\\ -T_B\left(y_1\right)-T_C\left(y_2\right)-W\left(y_3\right)-P\left(y_4\right)=0\end{array}$ (III)

Here, the vertical distance between cable A and B is $y_1$, the vertical distance between A and C is $y_2$, the center of gravity of the plate about the y axis is $y_3$, the vertical distance of the load acting is $y_4$.

Calculations:

Substitute 200 lb for P and 500 lb for W in Equation (I).

$\begin{array}{l}{T}_A+T_B+T_C-200-500=0\\ T_A+T_B+T_C=700\end{array}$ (IV)

Substitute 3.0 ft for $x_1$, 3.0 ft for $x_2$, 500 lb for W, 1.5 ft for $x_3$, 200 lb for P, and 3.0 ft for $x_4$ in Equation (II).

$\begin{array}{l}{T}_A\left(3.0\right)+T_C\left(3.0\right)-500\left(1.5\right)-200\left(3\right)=0\\ 3T_A+3T_C=1,350\end{array}$ (V)

Substitute 4.0 ft for $y_1$, 4.0 ft for $y_2$, 500 lb for W, 2.0 ft for $y_3$, 200 lb for P, and 2.0 ft for $y_4$ in Equation (III).

$\begin{array}{l}-T_B\left(4.0\right)-T_C\left(4.0\right)+500\left(2.0\right)+200\left(2.0\right)=0\\ -4T_B-4T_C=-1,400\\ 4T_B+4T_C=1,400\end{array}$ (VI)

Simply Equation (I), multiplying by 3 and it can be written as follows.

$\begin{array}{l}3\left(T_A+T_B+T_C\right)=3\left(700\right)\\ 3T_A+3T_B+3T_C=2,100\end{array}$ (VIII)

Conclusion:

Solve Equation (VIII) and (V).

$T_B=250\text{lb}$

Thus, the tension in cable B is $\underset{\_}{250\text{lb}}$

Substitute 250 lb for $T_B$ in Equation (VI).

$\begin{array}{l}4\left(250\right)+4T_C=1,400\\ 1,000+4T_C=1,400\\ T_C=\frac{400}{4}\\ T_C=100\text{lb}\end{array}$

Thus, the tension in cable C is $\underset{\_}{100\text{lb}}$

Substitute 250 lb for $T_B$ and 100 lb for $T_C$ in Equation (IV).

$\begin{array}{l}{T}_A+250+100=700\\ T_A=700-350\\ T_A=350\text{lb}\end{array}$

Thus, the tension in cable A is $\underset{\_}{350\text{lb}}$.