# A problem on Impulse & Impulsive Forces [closed]

A jet of water leaves a nozzle of $1$ inch diameter at a speed of $50$ ft/sec and impinges on a plate fixed at right angles to its direction. What pressure is exerted on the plate?

If the nozzle be drawn backwards with a velocity of $20$ ft/sec in the line of the jet, what pressure is then exerted on the plane?

Answer: $26.64~ pounds-wt$; $9.59~pounds-wt$

Attempt: Cross-sectional area of the nozzle=$\pi(\frac{1}{24})^2~sq-ft$

Mass of water issued from jet per sec =Vol. of water per sec $\times$ 62.5 =$\pi(\frac{1}{24})^2\times 50 \times 62.5 =\frac{275}{1008}\times 62.5 ~pounds$

We have $Ft=momentum \implies F =\frac{Mass \times velocity}{t}=\frac{\frac{275}{1008}\times 62.5\times 50 }{1}=852.5545635$

I am not getting the desired result for the first part and unable to solve the 2nd part. Please help me to solve. Thanks in advance.

## closed as off-topic by apnorton, Daniel W. Farlow, Dario, Claude Leibovici, BatominovskiAug 12 '15 at 13:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – apnorton, Daniel W. Farlow, Dario, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

• The non-metric units feet and pound-force makes this problem more difficult for us non-physicists, who do not know the appropriate conversions. Also, are we to assume that the initial velocity of the water jet is horizontal and the horizontal component stays unchanged, and the plate is vertical? Or are we just to ignore what happens to the water between the nozzle and the plate? – Rory Daulton Aug 7 '15 at 18:02
• @RoryDaulton Ignore what happens to the water between the nozzle and the plate & plate is vertical. – user1942348 Aug 7 '15 at 18:07
• Why is the unit for pressure be pounds-wt? Shouldn't it be psi (pounds-force per square inch)? Either something is missing or the answer key is wrong, since the correct answers should be $33.69\text{ psi}$ and $12.13\text{ psi}$, respectively. – Batominovski Aug 7 '15 at 18:21
• Ah ha! You must have mistyped something, or your book is a terrible reference. The answers given here would be roughly correct, if the question were to calculate the "forces" and not the "pressures." – Batominovski Aug 7 '15 at 18:26
• @RobertSoupe (and all other closevoters): Please don't migrate this to Physics, as it would get closed there in a heartbeat (see How to ask a Homework Question on Physics). Now, if anyone actually votes to close this (because it truly isn't a mathematical question, yet a majority of voters have cast in favor of migration), it will be migrated, annoy Physics people, get rejected, and land back here. I'm requesting a moderator to close/reopen to clear the closevote count. – apnorton Aug 10 '15 at 3:55

Notice, the force exerted on the fixed plate in the direction of the jet line$$=\text{rate of change of momentum of water striking the plate}$$$$=\text{mass/sec} \times\text{(velocity before striking-velocity after striking)}=(\rho a V)(V-0)=\rho aV^2$$ Where, density of water, $\rho=1000\ kg/m^3$

Speed of water jet, $V=50\times \frac{30.48}{100}=15.24\ m/sec$

area of cross-section of nozzle, $a=\frac{\pi}{4}(2.54)^2\times 10^{-4}=5.0671\times 10^{-4}\ m^2$ Now, setting all the values in the above expression, we get the force (F) exerted by the jet on the fixed plate $$F=1000\times 5.0671\times 10^{-4}\times (15.24)^2$$ $$=\color{blue}{117.6872 \ \text{N (newtons)}}$$ $$=117.6872\times 0.2248$$ $$=\color{blue}{26.4561 \ \text{lbf (pounds-wt)}}$$ We can calculate pressure $$=\frac{\text{force}}{\text{area of plate}}=\frac{F}{A}$$

Now, if the nozzle is drawn backwards then the relative velocity of the water jet w.r.t. fixed plate $$V=\text{velocity of jet}-\text{velocity of nozzle}=50-20=30\ ft/sec$$$$=30\times 0.3048=9.144\ m/sec$$ Now, setting the value of velocity of jet w.r.t. plate in the above expression, we get force exerted on the plate $F$ $$F=1000\times 5.0671\times 10^{-4}\times (9.144)^2$$$$=\color{blue}{42.3674 \ \text{N (newtons)}}$$$$=42.3674\times 0.2248$$$$=\color{blue}{9.5242 \ \ \text{lbf (pounds-wt)}}$$ We can calculate pressure $$=\frac{\text{force}}{\text{area of plate}}=\frac{F}{A}$$

The main source of confusion here is the definition of a "pound".

You say the mass of water that strikes the plate each second is $$(\text{volume of water per second}) \times 62.5\ \text{pounds}.$$

But actually that is the weight of the water. The mass of the water, which you need in order to correctly compute $\frac{\Delta mv}{t}$, is the $m$ in the formula $F = ma$; when discussing the weight of water, the weight is a force given by $F = mg$. Hence $m = \frac{62.5}{g}\ \text{slugs}$.

The exact solution depends on the values one takes for various constants; $62.5$ is slightly high for the density of water in pounds per cubic foot, and the answer you are "supposed" to get seems to depend on converting lbm to slugs using the approximation $g \approx 32$, whereas $32.174$ would be more accurate.

At any rate, if you assume $g = 32 \ \text{ft}/\text{s}^2$, then the answer you calculated would be off by a factor of $32$, having failed to divide $62.5$ by $g$ when calculating mass. Indeed, $852.5545635 / 32 = 26.64$ to four significant digits, so that explains what happened to your calculation.

All this business with lbf vs. lbm vs. slugs is (to me) is a good argument for doing the whole problem in metric, and only converting from N to lbf at the end, much more compelling than the notion that the conversion factors are much easier in metric (all being powers of $10$ rather than other numbers such as $12$ inches per foot).