# For HW 8, problem 12, how can i find what the scale reads at the end of the problem?

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This is problem 12 in physics 1250. Water falls without splashing at a rate of 0.320 L/s from a height of 2.20 m into a 0.760-kg bucket on a scale. If the bucket is originally empty, what does the scale read 4.00 s after water starts to accumulate in it?

recategorized Oct 14, 2013

The scale reads the total force pushing down on it by the bucket which includes two parts:

1) the weight of the water accumulated in the bucket,

2) the average force created by water falling into the bucket and deaccelerating into zero velocity -- here you will have a change of momentum for water: $\Delta P=F_{ avg }\Delta t$.

The weight of the water in the bucket is easy to find out -- just get the volume and multiple density (1 kg/liter) to get mass.

To find the average force needed to change the water's momentum, first you need to find its velocity right before the water stream hits the bucket. You can get this by using the free-fall of water from a height given in the question: $V_{ f }^{ 2 }-V_{ i }^{ 2 }=2gh$.

To find $\Delta P$ of water for some time, you need to find the volume of the water dropped over that time and then the change of the momentum of that amount of water: (note that the water's density is 1kg/liter and define downward being positive)

$$V_{ water }=flowrate\cdot \Delta t \\ \\ M_{ water }=V_{ water }\cdot density=flowrate\cdot density \cdot \Delta t \\ \\ \Delta P=0-flowrate\cdot density\cdot V_{ f }\cdot \Delta t=F_{ avg }\cdot \Delta t\\ \\ F_{ avg }=-flowrate\cdot density\cdot V_{ f }$$

Give it a try and see if it helps.
answered Oct 14, 2013 by (360 points)