View Full Version : Finding normal force with momentum
vu10758
Oct29-06, 03:37 PM
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.
Since p = mv, the expression F = dp/dt can be expanded for situations involving changing masses: F = m dv/dt + v dm/dt.
Andrew Mason
Oct29-06, 05:44 PM
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
vu10758
Oct29-06, 06:04 PM
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?
Andrew Mason
Oct29-06, 10:18 PM
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.
dp/dt = vdm/dt
In order to determine the speed, use the fact that potential energy is converted to kinetic energy. So, for an element of mass, \Delta m:
\Delta mgh = xtgh = \frac{1}{2}\Delta mv^2 = \frac{1}{2}xtv^2
AM
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