Chapter 14, Problem 14.112RP

To determine

The force couple system that must be applied at A to hold the vane in place.

Answer to Problem 14.112RP

The x component of force $\left(A_x\right)$ applied at A to hold the vane in place is $\underset{\_}{55.5\text{lb}(\to )}$.

The y component of force $\left(A_y\right)$ applied at A to hold the vane in place is $\underset{\_}{20.2\text{lb}(\downarrow )}$.

The couple $\left(M_A\right)$ applied at A to hold the vane in place is $\underset{\_}{41.4\text{lb}\cdot \text{ft}\left(\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The nozzle discharges (Q) water at the rate of $200\text{gal/min}$.

The stream of the water moves with a velocity (v) of magnitude is $100\text{ft/s}$.

Assume unit weight of the water $\left(\gamma \right)$ as $62.4{\text{lb/ft}}^{\text{3}}$.

Assume the acceleration due to gravity (g) as $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Convert the gallons per minutes to cubic feet per second.

$\begin{array}{c}Q=200\text{gal/min}\times \frac{1{\text{ft}}^{\text{3}}}{\text{7}\text{.48gal}}\times \frac{1\min }{60\sec }\\ =0.44563{\text{ft}}^{\text{3}}\text{/s}\end{array}$

Calculate the rate of the water discharge $\left(\frac{dm}{dt}\right)$ per unit length along BC using the relation:

$\left(\frac{dm}{dt}\right)=\frac{\gamma Q}{g}$

Substitute $62.4{\text{lb/ft}}^{\text{3}}$ for $\gamma$, $32.2{\text{ft/s}}^{\text{2}}$ for g and $0.44563{\text{ft}}^{\text{3}}\text{/s}$ for Q.

$\begin{array}{c}\left(\frac{dm}{dt}\right)=\frac{62.4\times 0.44563}{32.2}\\ =0.8636\text{lb}\cdot \text{s}/\text{ft}\end{array}$

Express the velocity of water $\left(v_B\right)$ at point B in terms of vector:

$v_B=\left(100\text{ft/s}\right)j$

Express the velocity of water $\left(v_C\right)$ at point C in terms of vector:

$v_C=\left(100\text{ft/s}\right)\left(\sin 40°i+\cos 40°j\right)$

Consider right side and upward as positive.

Show the impulse momentum equation diagram as in Figure (1).

Calculate the x component of force $\left(A_x\right)$ applied at A to hold the vane using Figure (1):

$0+A_x\left(\Delta t\right)=\left(\Delta m\right)\left(v_C\right)i$

Substitute $\left(100\sin 40°\right)$ for $\left(v_C\right)i$.

$\begin{array}{c}{A}_x=\frac{\left(\Delta m\right)}{\left(\Delta t\right)}\left(100\sin 40°\right)\\ =\frac{\left(dm\right)}{\left(dt\right)}\left(100\sin 40°\right)\end{array}$

Substitute $0.8636\text{lb}\cdot \text{s}/\text{ft}$ for $\left(\frac{dm}{dt}\right)$.

$\begin{array}{c}{A}_x=0.8636\times \left(100\sin 40°\right)\\ =55.5\text{lb}(\to )\end{array}$

Calculate the y component of force $\left(A_y\right)$ applied at A to hold the vane using Figure (1):

$\left(\Delta m\right)v_B+A_y\left(\Delta t\right)=\left(\Delta m\right)\left(v_C\right)j$

Substitute $\left(100\cos 40°\right)$ for $\left(v_C\right)j$ and $\left(100\text{ft/s}\right)j$ for $v_B$.

$\begin{array}{l}\left(\Delta m\right)\left(100\text{ft/s}\right)+A_y\left(\Delta t\right)=\left(\Delta m\right)\left(100\cos 40°\right)\\ A_y=\frac{\left(\Delta m\right)\left(100\cos 40°\right)-\left(\Delta m\right)\left(100\text{ft/s}\right)}{\left(\Delta t\right)}\\ A_y=\frac{\left(\Delta m\right)}{\left(\Delta t\right)}\left(100\right)\left(\cos 40°-1\right)\\ A_y=\frac{\left(dm\right)}{\left(dt\right)}\left(100\right)\left(\cos 40°-1\right)\end{array}$

Substitute $0.8636\text{lb}\cdot \text{s}/\text{ft}$ for $\left(\frac{dm}{dt}\right)$.

$\begin{array}{c}{A}_y=\left(0.8636\right)\left(100\right)\left(\cos 40°-1\right)\\ =-20.2\text{lb}\\ =20.2\text{lb}(\downarrow )\end{array}$

Show the given nozzle system as in Figure (2).

Consider clockwise as negative.

Calculate the couple applied $\left(M_A\right)$ at the point A to hold the vane using Figure (1) and Figure (2).

$\left(3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)v_B+M_A\left(\Delta t\right)=\left[\begin{array}{l}\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)\left(\left(v_C\right)j\right)\\ -\left(15\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)\left(\left(v_C\right)i\right)\end{array}\right]$

Substitute $\left(100\cos 40°\right)$ for $\left(v_C\right)j$,$\left(100\sin 40°\right)$ for $\left(v_C\right)i$ and $\left(100\text{ft/s}\right)j$ for $v_B$.

$\begin{array}{l}\left(3\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)\left(100\text{ft/s}\right)+M_A\left(\Delta t\right)=\left[\begin{array}{l}\left(9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)\left(\left(100\cos 40°\right)\right)\\ -\left(15\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(\Delta m\right)\left(\left(100\sin 40°\right)\right)\end{array}\right]\\ 25\left(\Delta m\right)+M_A\left(\Delta t\right)=57.453\left(\Delta m\right)-80.348\left(\Delta m\right)\\ M_A\left(\Delta t\right)=57.453\left(\Delta m\right)-80.348\left(\Delta m\right)-25\left(\Delta m\right)\\ M_A=-47.895\frac{\left(\Delta m\right)}{\left(\Delta t\right)}\\ M_A=-47.895\frac{\left(dm\right)}{\left(dt\right)}\end{array}$

Substitute $0.8636\text{lb}\cdot \text{s}/\text{ft}$ for $\left(\frac{dm}{dt}\right)$.

$\begin{array}{c}{M}_A=-47.895\left(0.8636\right)\\ =-41.362\text{lb}\cdot \text{ft}\approx -\text{41}\text{.4}\text{lb}\cdot \text{ft}\\ \text{=}\text{41}\text{.4}\text{lb}\cdot \text{ft}\left(\text{clockwise}\right)\end{array}$

Therefore, the x component of force $\left(A_x\right)$ applied at A to hold the vane in place is $\underset{\_}{55.5\text{lb}(\to )}$.

Therefore, the y component of force $\left(A_y\right)$ applied at A to hold the vane in place is $\underset{\_}{20.2\text{lb}(\downarrow )}$.

Therefore, the couple $\left(M_A\right)$ applied at A to hold the vane in place is $\underset{\_}{41.4\text{lb}\cdot \text{ft}\left(\text{clockwise}\right)}$.