Calculating Force on a Simple Water Squirter Plunger: A1, V1, A2, V2

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The discussion centers on calculating the force applied to a plunger in a water squirter using fluid dynamics principles. Given the areas and velocity at the nozzle, the user calculated the plunger's velocity and applied Bernoulli's equation to find the pressure at the plunger. The density of water was assumed to be 1, and the gauge pressure at the nozzle was considered zero. Ultimately, the force exerted on the plunger was determined to be approximately 15.94 N, without accounting for fluid friction or viscosity. The calculations illustrate the application of fluid dynamics equations in a practical scenario.
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Homework Statement



Area at the end with the plunger (A1) = 8cm^2
Area at the nozzle (A2) = 0.5cm^2
Velocity at the nozzle (V2) = 2m/s

What is the force being applied to the plunger?

Homework Equations



A1*V1=A2*V2
P + (1/2)pV^2 + pgy = P + (1/2)pV^2 + pgy
P=F/A

The Attempt at a Solution



I used the first equation listed above to find the velocity of the plunger (V1) to be 0.125m/s. I'm having quite a difficult time finding any of the other variables (Pressure, Force, mass, density) for either side.
 
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How about changing the velocity (speed), which is acceleration, of the water?

I presume one is ignoring the friction/viscous forces.
 
I figured out that the density (p) is just 1 (because it's a water squirter) and the gauge pressure at the nozzle (P2) is 0. And the pgy terms are also 0 because the squirter is just horizontal. So using equation 2 from above (bernoulli's eqn) I got:

P1 + 1/2(1)(V1^2) + 0 = 0 + 1/2(1)(V2^2) + 0
P1=1/2(V2^2-V1^2)
F1=P1*A1
F1=1/2(2^2-.125^2)(8)=15.9375N

And yes, I have no idea about fluid friction or viscosity at this point, hehe.
 
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