# Force Applied by a Water Jet

1. Feb 11, 2014

### wahaj

To experiment with conservation of linear momentum I did lab where a jet of water is shot at a flat plate and a hemispherical cup. After simplification the final equation for the theoretical force came out to be
$$F_t= \dot{m}V(1-cos \beta)$$
where $\dot{m}$ is the mass flow rate of water and V is the velocity of water hitting the surface. $\beta$ is the angle at which the water deflects off of the surface. For the flat plate is was 90 and for the cup it was 180. I have a hard time physically interpreting how the angle of deflection determines the force of the water. In both cases the same amount of water is hitting both surfaces with the same velocity at the same angle (which would be vertically upwards in this case). So why does the angle of deflection determine the force applied by the water?

2. Feb 11, 2014

### Travis_King

Take a look at the structure of that equation. What happens when Beta is 90 degrees? What happens when it is 180?

Think of this equation as a sum of forces (as that's what it is). The plate resists the force imparted by the incoming water stream and the water is deflected 90 degrees (no horizontal force, since there's no horizontal component to the stream). So, after the water is deflected, there's no additional force for the plate to resist.

With the cup, the surface must first resist the stream (m_dot*V_initial) and then experiences an equal and opposite reaction from the horizontal component of redirected water stream.

3. Feb 11, 2014

### jim hardy

When the math and intuition agree you are probably getting someplace. At least it's easier to remember the equation when they agree... Nice job there T_K.

Simply put - the plate just stops the water but the cup throws it back.
Imagine a similar experiment measurement with a baseball. Try it yourself in a rocking chair.

4. Feb 11, 2014

### wahaj

Excellent explanation, both of you. Thanks you very much for clearing this up for me.