Chapter 3, Problem 79P

A water trough of semicircular cross section of radius 0.6 m consists of two symmetric parts hinged to each other at the bottom, as shown in Fig. P3-72. The two pans are held together by a cable and turnbuckle placed every 3 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.

To determine

The tension in the cable.

Answer to Problem 79P

The tension in the cable is $5297.4\text{N}$.

Explanation of Solution

Given information:

A water trough of semicircular cross section of two symmetric parts hinged at each other at the bottom and held together by a cable.

The figure below shows the free body diagram of the right-hand side part of water trough.

  

  Figure-(1)

Write the Equation for the horizontal hydrostatic force on the part.

  $F_H=\rho gA{h}_c$  ......(I)

Here, density of fluid is $\rho$, acceleration due to gravity is $g$, projected area is A and centroid distance of projected area is $h_c$.

Write the Equation for vertical projected area.

  $A=r\times l$  ...... (II)

Here, radius of the quarter circle is $r$ and length of the trough measured into the paper is $l$.

Substitute $r\times l$ for $A$ in the Equation (I).

  $F_H=\rho g\left(r\times l\right)h_c$  ......(III)

Write the Equation of the vertical force for curved surface.

  $F_v=\rho gV$  ...... (IV)

Here volume of fluid above curved surface is $V$.

Write the Expression for volume of fluid above curved surface.

  $\begin{array}{c}V=\left({{\rm A}}_{quatercircle}\times l\right)\\ =\left(\frac{\pi }{4}{r}^2\times l\right)\end{array}$   ....... (V)

Substitute $\left(\frac{\pi }{4}{r}^2\times l\right)$ for $V$ in the Equation (IV)

  $F_v=\rho g\left(\frac{\pi }{4}{r}^2\times l\right)$   ....... (VI)

Write the Expression for centre of pressure.

  $h_p=h_c+\frac{I}{A{h}_c}$   ...... (VII)

Write the Expression for moment of Inertia.

  $I=\frac{l{r}^3}{12}$   ...... (VIII)

Substitute $r/2$ for $h_c$, $\frac{l{r}^3}{12}$ for $I$ and $r\times l$ for $A$ in the Equation (VII).

  $\begin{array}{c}{h}_p=\left(r/2\right)+\frac{\frac{l{r}^3}{12}}{\left(r\times l\right)\left(r/2\right)}\\ =\left(\frac{r}{2}\right)+\left(\frac{r}{6}\right)\\ =\left(\frac{2r}{3}\right)\end{array}$  ......(IX)

Write the moment Equation about the hinge point.

  $-F_H\left(r-h_p\right)-F_V\left(x_p\right)+T\left(r\right)=0$  ...... (X)

Here, distance between the line of action of vertical force and hinged point is $x_p$, tension in the cable is $T_2$.

Write the Expression for distance between the line of action of vertical force and hinged point.

  $x_p=\frac{4r}{3\pi }$

Substitute $\left(\frac{2r}{3}\right)$ for $h_p$ and $\frac{4r}{3\pi }$ for $x_p$ in the Equation (IX).

  $-F_H\left(r-\left(\frac{2r}{3}\right)\right)-F_V\left(\frac{4r}{3\pi }\right)+T_2\left(r\right)=0$  ...... (XI)

The figure below shows the free body diagram of water trough.

  

  Figure-(2)

Write the expression for horizontal force equilibrium.

  $\begin{array}{l}{T}_1-T_2=0\\ T_1=T_2\end{array}$   ....... (XII)

Figure-(2)

Calculation:

Substitute $0.6\text{m}$ for $r$ and $3\text{m}$ for $l$ in the Equation (II).

  $\begin{array}{c}A=0.6\text{m}\times 3\text{m}\\ \text{=1}\text{.8}{\text{m}}^2\end{array}$

Substitute, $\text{1}\text{.8}{\text{m}}^2$ for A, $1000\text{kg}/{\text{m}}^3$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.3\text{m}$ for $h_c$ in the Equation (III).

  $\begin{array}{c}{F}_H=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{1}\text{.8}{\text{m}}^2\right)\left(0.3\text{m}\right)\\ =5297.4\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =5297.4\text{N}\end{array}$

Substitute $0.6\text{m}$ for $r$ and $3\text{m}$ for $l$ in the Equation (V).

  $\begin{array}{c}V=\left(\frac{\pi }{4}{\left(0.6\text{m}\right)}^2\times \left(3\text{m}\right)\right)\\ =\frac{3.392{\text{m}}^3}{4}\\ =0.8482{\text{m}}^3\end{array}$

Substitute $0.8482{\text{m}}^3$ for $V$, $1000\text{kg}/{\text{m}}^3$ for $\rho$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (IV).

  $\begin{array}{c}{F}_v=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.84823{\text{m}}^3\right)\\ =8321.136\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =8321.136\text{N}\end{array}$

Substitute $0.6\text{m}$ for $r$ in the Equation (IX).

  $\begin{array}{c}{h}_p=\left(\frac{2\left(0.6\text{m}\right)}{3}\right)\\ =0.4\text{m}\end{array}$

Substitute $8321.136\text{N}$ for $F_v$, $5297.4\text{N}$ for $F_h$, and $0.6\text{m}$ for $r$ in the Equation (XI).

  $\begin{array}{l}-\left(5297.4\text{N}\right)\left(0.6\text{m}-\left(\frac{2\left(0.6\text{m}\right)}{3}\right)\right)-\left(8321.136\text{N}\right)\left(\frac{4\times \left(0.6\text{m}\right)}{3\pi }\right)+T\left(0.6\text{m}\right)=0\\ T_2=\frac{\left(5297.4\text{N}\right)\left(0.6\text{m}-\left(\frac{2\left(0.6\text{m}\right)}{3}\right)\right)+\left(8321.136\text{N}\right)\left(\frac{4\times \left(0.6\text{m}\right)}{3\pi }\right)}{0.6\text{m}}\\ T_2=5297.4\text{N}\end{array}$

Substitute $5297.4\text{N}$ for $T_2$ in the Equation (XII).

  $T_1=5297.4\text{N}$

Here the tension value is same

Conclusion:

The tension in the cable is $5297.4\text{N}$.