# Simple Water Squirter

1. Nov 26, 2007

### Genza

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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.

2. Nov 26, 2007

### Staff: Mentor

How about changing the velocity (speed), which is acceleration, of the water?

I presume one is ignoring the friction/viscous forces.

3. Nov 26, 2007

### Genza

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.