Calculating Average Force from Water Flow Rate and Height

[mooshasta]

Hey.. I was looking over a physics team test, and I can't figure out how to solve this problem. I've asked my teachers and they can't help me either.


Water runs out of a horizontal drainpipe at a rate of 120 kg per minute. It falls 3.2 m to the ground. Assuming the water doesn't splash up, what average force does it exert on the ground?


I'm not sure where to start with this- I believe that if the diameter of the pipe was given, you could find the velocity of the water, but without it I'm not sure where to begin. Just a little help would be greatly appriciated.

 

[einstone]

The change in the momentum of a particle of fluid in the vertical direction equals the negative of (its mass)* sqrt(2* acceleration due to gravity* 3.2m). This momentum is transferred to the ground ( assuming, as you said, the water doesn't splash around). Since the particles weighing 120kg fall on the ground per minute, the average force is
- (120)*sqrt( 2*acc. due to gravity* 3.2) Newton per min.
A fluid particle may have a velocity in the horizontal direction, but it remains unchanged throughout the fall & the fluid continues to flow along the ground. I assumed, based on the few drainpipes I've seen in my life ,that initially the vertical component of the velocity was negligible.
I am, with great respect,
Einstone.

 

[xanthym]

The change in the momentum of a particle of fluid in the vertical direction equals the negative of (its mass)* sqrt(2* acceleration due to gravity* 3.2m). This momentum is transferred to the ground ( assuming, as you said, the water doesn't splash around). Since the particles weighing 120kg fall on the ground per minute, the average force is
- (120)*sqrt( 2*acc. due to gravity* 3.2) Newton per min.

To clarify Einstone's solution, we have for the momentum of fluid mass "m":
p = m*v
Δp = m*Δv
{Average Force During Time ΔT} = Δp/ΔT = m*Δv/ΔT

The kinetic energy change of fluid mass "m" is related to the work performed by gravity over distance "d" on "m" by:
Δ{(1/2)*m*(v^2)} = (1/2)*m*{(v_f)^2 - (v_i)^2} = F*d
Or in this case, since v_i=(0):
(1/2)*m*(v_f)^2 = m*g*d
v_f= sqrt{2*g*d}
Δv = sqrt{2*g*d} - 0 = sqrt{2*g*d}

{Average Force On Ground During Time ΔT} = -Δp/ΔT =
= -m*Δv/ΔT = -m*sqrt{2*g*d}/ΔT =
= -(120 kg)*sqrt{2*(9.8 m/sec^2)*(3.2 m)}/(60 sec)
= (-15.84 N)


~~

 

[mooshasta]

Thank you so much, 16 N was the answer given. :)