Having problems trying to solve part b of this problem.

Rocket Propulsion Elements 8th - Chapter 2

  1. A jet of fluid hits a stationary flat plate in the manner shown below.

(a) If there is 50 kg of fluid flowing per minute at an absolute velocity of 200 m/sec, what will be the force on the plate?

(b) What will this force be when the plate moves in the direction of flow at u = 50 km/h? Explain the methodology.

Part A just involves plugging in mass flow rate, changing units to seconds, multiplying it by characteristic velocity to get force.

Part B, I have zero idea what's going on. I thought it was a conservation of momentum problem, so I set up the problem as F = dm * v2, where v is 200-13.8 (13.8 being the change in speed in m/s as the plate moves away). The problem is, I'm getting 155 N, whereas the correct answer is 144 N.

  • $\begingroup$ Not an answer: $F = dp/dt$. If $p = mv$ then $F= d(mv)/dt$, if $v$ ($dx/dt$) is constant then it's $F = v \ dm/dt$ Step 1 is figure out your flow in kg/m ($dm/dx$, kilograms per meter) in the initial case, then choose your velocity carefully, then use $dm/dt = dm/dx \times dx/dt$. hmm... I get (200 - 13.899) * 0.8333 = 155.08 N also! I get 144 only if the block moves at 100 km/hr. $\endgroup$
    – uhoh
    Commented Nov 11, 2018 at 18:32
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    $\begingroup$ The answer might be misprinted as well. Check with your classmates solution as well to verify the answer. $\endgroup$
    – Amar
    Commented Nov 11, 2018 at 20:05
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    $\begingroup$ Just thinking of it another way: if the plate was moving 200 m/s the force would be zero N, when it's moving 0 m/s the force is 167 N. I see nothing to indicate the variation isn't linear so 155 N at a speed of 13.8 m/s makes sense to me. Incidentally in my tattered 4th edition it's 50 lb/s, 200 ft/s, and 50 mph. Plus they don't give the answers :) $\endgroup$ Commented Nov 11, 2018 at 22:07
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    $\begingroup$ @amar actually not reading this book for a class: My school doesn't have an aeronautics or astronautics program, so I'm going through the process of self-teaching myself. Good to know I wasn't completely crazy in my calculation, thanks for the help everyone! $\endgroup$
    – Colin Warn
    Commented Nov 17, 2018 at 16:25

3 Answers 3


The thing is that in the second case, the mass of the fluid hitting the plate in one second is not the same as in the first case. If the fluid flows at the rate $0.8333$ kg/s and speed $200$ m/s, then the linear density of the jet is $\frac{0.8333~ \text{kg/s}}{200~\text{m/s}}\approx 4.167~\text{g/m}$ (i.e., 1 meter of the jet contains 4.167 g of fluid). Relative to the plate, the fluid moves at approximately $186.1$ m/s, so the rate at which the fluid hits the plate is $4.167~\text{g/m}\cdot 186.1~\text{m/s} \approx 0.7755~\text{kg/s}$. Multiplying this rate by the velocity change of the fluid, we get the answer $0.7755~\text{kg/s}\cdot 186.1~\text{m/s}\approx 144.3~\text{N}$.


I am getting the same answer as well (155N). I am using mass flow rate (0.83kg/s) multiplied by u (13.8m/s) to yield a 12N force and subtracting that from my original 167N force to yield 155N.


Here's the answer to this problem;

m= 50kg/60sec = 0.833kg/sec

F= mass flow rate × v2 (vehicle velocity)

F= 0.833 × 200m/sec

= 166.6N ( convert it to the nearest unit)

= 167N

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    $\begingroup$ The question the OP, and answerers, are primarily concerned with is part B, not part A. $\endgroup$
    – Erin Anne
    Commented Apr 7, 2023 at 17:33

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