Finding normal force with momentum

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vu10758
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Water falls without splashing at a X kg/s from a height H into a bucket of mass M. The bucket sits on a sacle. Determine the reading of the sacle as a function of time.

I know the the sum of all forces is equal to the derivative of momentum with respect to time.

Mg is the force at time = 0.

The correct answer is mg + xtg + x*SQRT(2gH)

I know

F = dp / dt
dp = F dt

but I don't know what to do.
 
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Since p = mv, the expression F = dp/dt can be expanded for situations involving changing masses: F = m dv/dt + v dm/dt.
 
vu10758 said:
Water falls without splashing at a X kg/s from a height H into a bucket of mass M. The bucket sits on a sacle. Determine the reading of the sacle as a function of time.

I know the the sum of all forces is equal to the derivative of momentum with respect to time.

Mg is the force at time = 0.

The correct answer is mg + xtg + x*SQRT(2gH)

I know

F = dp / dt
dp = F dt

but I don't know what to do.
The scale is going to measure the downward force. There are two things that contribute the downard force. What are they?

AM
 
The total force measured by the bucket is the weight of the bucket + weight of water + force of collision.

So, I understand mg + xtg which gives me force. But where did the x * SQRT(2gH) come from. SQRT(2gH) is equivalent to time, meaing that x * SQRT(2gH) is a mass. Why is a mass included in an equation for force?
 
Last edited:
vu10758 said:
The total force measured by the bucket is the weight of the bucket + weight of water + force of collision.

So, I understand mg + xtg which gives me force. But where did the x * SQRT(2gH) come from. SQRT(2gH) is equivalent to time, meaing that x * SQRT(2gH) is a mass. Why is a mass included in an equation for force?
The rate of change of momentum is the rate of mass flow x the speed of the water when it hits the bucket.

[tex]dp/dt = vdm/dt[/tex]

In order to determine the speed, use the fact that potential energy is converted to kinetic energy. So, for an element of mass, [itex]\Delta m[/itex]:

[tex]\Delta mgh = xtgh = \frac{1}{2}\Delta mv^2 = \frac{1}{2}xtv^2[/tex]

AM