Chapter 6, Problem 78P

Indiana Jones needs So ascend a 10-m-high building. There is a large hose filled with pressurized water hanging down from the building top. He builds a square platform and mounts four 4-cm-diameter nozzles pointing down at each corner. By connecting hose branches, a water jet with I Sni s velocity can be produced from each nozzle. Jones, the platform, and the nozzles have a combined mass of 150 kg. Determine (a) the minimum water jet velocity needed to raise the system. (b) how long it takes for the system to rise 10 m when the water jet velocity is 18 m/s and the velocity of the platform at that moment, and (c) how much higher will the momentum raise Jones if he shuts off the water at the moment the platform reaches 10 in above the ground. How much time does he have to jump from the platform to the roof?

Answers: (a) 17.1 m/s. (b) 4.37 s, 4.57 m/s. (C) 1.07 m, 0.933 S

To determine

(a)

The minimum velocity needed to raise the platform.

Answer to Problem 78P

The minimum velocity of the water jet is $17.1\text{m}/\text{s}$.

Explanation of Solution

Given Information:

The combined mass of the system is $150\text{kg}$ and the velocity of flow through each nozzle is $18\text{m}/\text{s}$.

Write the expression to calculate the total weight of the platform.

  $W=mg$...... (I)

Here, the combined mass is $m$ and the acceleration due to gravity is $g$.

Write the moment equation.

  $W={\dot{m}}_{out}{V}_{out}$...... (II)

Here, the mass flow rate at outlet is ${\dot{m}}_{out}$, the velocity at the outlet is $V_{out}$.

Write the expression for the mass flow rate from the four nozzles.

  $\dot{m}=4\rho A{V}_{\min }$...... (III)

Here, the density of the fluid is $\rho$, the area of nozzle is $A$ and the velocity of flow is $V_{\min }$.

Write the expression to calculate the area.

  $A=\frac{\pi D^2}{4}$

Here, the diameter of the nozzles is $D$.

Calculation:

Substitute $\text{150}\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(150\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =1471.5\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =1471.5\text{N}\end{array}$

Substitute $\frac{\pi D^2}{4}$ for $A$ in Equation (III).

  $\begin{array}{c}\dot{m}=4\rho \left(\frac{\pi D^2}{4}\right)V_{\min }\\ =\rho \left(\pi D^2\right)V_{\min }\end{array}$...... (IV)

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $4\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}\dot{m}=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(\pi {\left(4\text{cm}\right)}^2\right)V_{\min }\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(16\pi {\text{cm}}^{\text{2}}\left(\frac{1{\text{m}}^2}{{10}^{-4}\text{cm}}\right)\right)V_{\min }\\ =5.027V_{\min }\text{kg}/\text{m}\end{array}$...... (V)

Substitute $1471.5\text{N}$ for $W$, $V_{\min }$ for $V_{out}$ and $5.026V_{\min }\text{kg}/{\text{m}}^{}$ for $\dot{m}$ in Equation (II).

  $\begin{array}{l}1471.5\text{N}=\left(5.026V_{\min }\text{kg}/{\text{m}}^{}\right)V_{\min }\\ V_{\min }=\sqrt{\frac{1471.5\text{N}}{5.027\text{kg}/{\text{m}}^{}}}\\ V_{\min }=17.1\text{m}/\text{s}\end{array}$

Conclusion:

The minimum velocity of the water jet is $17.1\text{m}/\text{s}$

To determine

(b)

The time taken by the system to rise $\text{10}\text{m}$ and the velocity at that moment.

Answer to Problem 78P

The time taken to cover a distance of $10\text{m}$ is $5.12\text{s}$.

The velocity at that moment is $\text{3}\text{.9}\text{m}/\text{s}$.

Explanation of Solution

Write the moment equation in the vertical direction.

  $F_{R2}=W-\dot{m}V$...... (VI)

Here, the vertical reaction on the platform is $F_{R2}$ and the velocity of the platform is $V$.

Write the expression for vertical reaction force.

  $F_{R2}=ma$...... (VII)

Here, the acceleration of the system is $a$.

Write the expression of third law of motion.

  $s=ut+\frac{1}{2}a{t}^2$...... (VIII)

Here, the distance travelled is $s$, the initial velocity is $u$ and the time taken us $t$.

Write the expression for first equation of motion.

  $V_f=u+at$...... (IX)

Here, the final velocity is $V_f$.

Calculation:

Substitute $15\text{m}/\text{s}$ for $V_{out}$ in Equation (V).

  $\begin{array}{c}\dot{m}=\left(5.027\text{kg}/\text{m}\right)\left(15\text{m}/\text{s}\right)\\ =75.405\text{kg}/\text{s}\end{array}$

Substitute $75.405\text{kg}/s$ for $\dot{m}$, $18\text{m}/\text{s}$ for $V$ and $1471.5\text{N}$ for $W$ in Equation (VI).

  $\begin{array}{c}{F}_{R2}=1471.5\text{N}-\left(18\text{m}/\text{s}\right)\left(75.405\text{kg}/s\right)\\ =1471.5\text{N}-1357.29\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =114.21\text{N}\end{array}$

Substitute $150\text{kg}$ for $m$ and $114.21\text{N}$ for $F_{R2}$ in Equation (VII).

  $\begin{array}{l}114.21\text{N}=150a\text{kg}\\ a=\frac{114.21\text{N}}{150\text{kg}}\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ a=0.761\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $\text{10}\text{m}$ for $s$, $0$ for $u$, $\text{0}\text{.761}\text{m}/{\text{s}}^{\text{2}}$ for $a$ in Equation (VIII).

  $\begin{array}{l}10\text{m}=0+\frac{1}{2}\times \text{0}\text{.761}\text{m}/{\text{s}}^{\text{2}}\times t^2\\ t^2=26.28\text{s}\\ t=\text{5}\text{.12}\text{s}\end{array}$

Substitute $5.12\text{s}$ for $t$, $0$ for $u$, $\text{0}\text{.761}\text{m}/{\text{s}}^{\text{2}}$ for $a$ in Equation (IX).

  $\begin{array}{c}{V}_f=0+\left(0.761\text{m}/{\text{s}}^2\right)\left(5.12\text{s}\right)\\ =3.9\text{m}/\text{s}\end{array}$

Conclusion:

The time taken to cover a distance of $10\text{m}$ is $5.12\text{s}$ and the velocity at that moment is $\text{3}\text{.9}\text{m}/\text{s}$.

To determine

(c)

The momentum raised.

The time taken to jump on the roof.

Answer to Problem 78P

The momentum raised is $0.802\text{m}$.

The time for getting off the platform is $0.794\text{s}$.

Explanation of Solution

Write the expression for momentum rise.

  $s^{\prime }=V_f{t}_1-\frac{1}{2}g{t}_1^2$...... (X)

Here, the time taken to reach the ground is $t_1$, the momentum rise is $s^{\prime }$ and the acceleration due to gravity is $g$.

Write the expression for velocity when platform is descending.

  $V_f=g{t}_1$...... (XI)

Write the expression for time taken to jump off the platform.

  $t^{\prime }=2t_1$...... (XII)

Calculation:

Substitute $3.9\text{m}/\text{s}$ for $V_f$, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XI).

  $\begin{array}{l}3.9\text{m}/\text{s}=\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)t_1\\ t_1=0.397\text{s}\end{array}$

Substitute $3.9\text{m}/\text{s}$ for $V_f$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.397\text{s}$ for $t_1$ in Equation (X).

  $\begin{array}{c}{s}^{\prime }=\left(3.9\text{m}/\text{s}\right)\left(0.397\text{s}\right)-\left(\frac{1}{2}\left(9.81\text{m}/{\text{s}}^2\right)\left(0.397\text{s}\right)\right)\\ =1.5483\text{m}-0.746\\ =0.802\text{m}\end{array}$

Substitute $0.802\text{m}$ for $s^{\prime }$ in Equation (XII).

  $\begin{array}{c}{t}^{\prime }=2\times 0.397\\ =0.794\text{s}\end{array}$

Conclusion:

The momentum raised is $0.802\text{m}$.

The time for getting off the platform is $0.794\text{s}$.