Chapter 6, Problem 1FP

To determine

The resultant force the elbow exerts on the water and the direction of the resultant force.

Explanation of Solution

Draw the free body diagram of the pipe as in Figure (1).

Cross section of the both the end of the pipe are same $\left(d_A=d_B\right)$.

Calculate the area of pipe.

  $\begin{array}{c}A=\frac{\pi }{4}{\left(d\right)}^2\\ A=\frac{\pi }{4}{\left(40\text{mm}\right)}^2\\ =\frac{\pi }{4}{\left(40\text{mm}\times \times \frac{1\text{ m}}{1000\text{ mm}}\right)}^2\\ =0.001257{\text{m}}^{\text{2}}\end{array}$

Calculate the velocity of pipe $\left(V\right)$ using the discharge equation.

  $\begin{array}{l}{Q}_{A/B}=A_{A/B}{V}_{A/B}\\ 0.012{\text{m}}^{\text{3}}/\text{s}=0.001257{\text{m}}^2\times V_{A/B}\\ V_{A/B}=\frac{0.012}{0.001257}\\ V_{A/B}=9.547\text{m}/\text{s}\end{array}$

Write the expression for the Equation for linear momentum:

  ${\displaystyle \sum F=\frac{\partial }{\partial t}{\displaystyle \underset{cv}{\int }V\rho d\sout{V}}}+{\displaystyle \underset{cs}{\int }V\rho V\cdot dA}$

Here, F is the force, cv is the control volume, V is the velocity, $\rho$ is the density of water, cs is the control surface, $d\sout{V}$ is the differential volume, and $dA$ is the differential area.

Apply the Equation of equilibrium to obtain the horizontal force $F_x$ the elbow exerts on the water.

  $\begin{array}{l}{\displaystyle \sum F_x=0}+V_B{\rho }_w{V}_B{A}_B\\ F_x=V_B{\rho }_w{V}_B{A}_B\\ F_x=\left(9.547\text{m}/\text{s}\right)\left(1,000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.547\text{m}/\text{s}\times 0.001257{\text{m}}^2\right)\\ =114.57\text{N}\end{array}$

Apply the equation of equilibrium to obtain the vertical force $F_y$ the elbow exerts on the water.

  $\begin{array}{l}{\displaystyle \sum F_y=0}+V_A\left(-{\rho }_w{V}_A{A}_A\right)\\ \left(p_A\times A_A\right)-F_y=V_A\left(-{\rho }_w{V}_A{A}_A\right)\end{array}$

  $\begin{array}{l}\left(160\times {10}^3\text{N}/{\text{m}}^{\text{2}}\times 0.001257{\text{m}}^{\text{2}}\right)-F_y=9.547\text{m}/\text{s}\left(-1,000\text{kg}/{\text{m}}^{\text{3}}\times 9.547\text{m}/\text{s}\times 0.001257{\text{m}}^{\text{2}}\right)\\ -F_y=-114.57-201.12\\ F_y=315.69\text{N}\end{array}$

Calculate resultant force the elbow exerts on the water $\left(R\right)$.

  $\begin{array}{c}R=\sqrt{F_x^2+F_y^2}\\ R=\sqrt{{\left(114.57\text{N}\right)}^2+{\left(315.69\text{N}\right)}^2}\\ =336\text{N}\end{array}$

Thus, the resultant force the elbow exerts on the water $\left(R\right)$ is $\underset{\_}{336\text{N}}$.

Calculate the direction of the resultant force $\left(\theta \right)$ .

  $\begin{array}{l}\tan \theta =\left(\frac{F_y}{F_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{315.69\text{N}}{114.57\text{ N}}\right)\\ \theta =70.0°\end{array}$

Thus, the magnitude $\theta$ of the resultant force is $\underset{\_}{70.0°}$.