What does the scale read after4.22s?

Question

Water ( density = 1000kg/m^3) falls without splashing at a rate of 0.111 L/s from a height of 24.2m into a 1.07 kg bucket on a scale. The accelaration of gravity is 9.8m/s^2.

If the bucket is originally empty, what does the scale read after 4.22 s? Answer in units of N.

Answer 1

The fall time T = sqrt(2*h/g) = 2.22 s. So only 4.22-2.22 = 2 s of water poured is in the pail. That's 2*.111 kg added to 1.07 kg = 1.292 kg. That mass * g = 12.66 N.

Answer 2

You must solve this problem using the principle of conservation of momentum.

Draw a control volume around the bucket. You can identify forces:
* weight: Fg = - m g
* contact with the scale: Fc (which is the answer sought)

And you can identify a flow crossing the boundary of the control volume
* flow: .111 L/s water at speed?

You know that the speed of the water is found from: y = 1/2 a t^2 --> t = sqrt(2 y / a) = 2.222 sec
v = a t (assume zero initial velocity) = 21.78 m/s

The mass of water is 1 kg/L so the momentum flux across the control surface is (0.111 kg/s) (-21.78 m/s) = -2.42 kg m /s^2. (The velocity is in the -y direction.)

The water can be assumed to be at rest inside the bucket, so there is no internal accumulation of momentum.

The external force must balance the weight force and the momentum flux.

Now we see an ambiguity in the problem. When is t=0? When the first water leaves the source? Or when the first water enters the bucket? I will assume the former but you should have enough understanding to solve the problem if it is the latter case.

Since the water requires 2.22 sec to fall, after 4.22 sec the bucket has accumulated 2 sec worth of flow, or 0.222 L or 0.222 kg. Therefore the weight of water + bucket is (1.292 kg * 1 g) or 12.66 N.

Fc balances the weight and the momentum flux:
-12.66 -2.42 + Fc =0 --> Fc = 15.08 N.

Answer 3

I agree with Kirchwey's answer, if the following stipulations are applied;

(1) The transient (oscillatory) effects of the scale due to impulse have died out.
(2) The bucket is large enough to hold the amount of water added to it.
(3) The scale is calibrated according to ASTM standards.

Yes, Kirchwey has the correct answer given all of this. Perhaps a better question would have been "What is the weight of the bucket + the water assuming that no water was spilled or splashed after 4.22 seconds?" or maybe the question could be posed with the previous assumptions.

Answer 4

I am not 100% but I think 20 N this way and 10N if one end to the wall. Reason is that there is a total gravitationally caused force on the scale of 20 N. One from either side of the scale. With the wall, the wall is exerting a force of 10 N against and the weight is exerting a force of 10 N ON the scale. The wall scenario is easy, it is like holding the scale vertically and the mass of the weight will read on the scale as we would expect. BUT ADDING 10 N to the TENSION in the spring, would sum the masses. It would be like an instantaneous reading if you were to hang 10N on the scale and JERK UPWARDS on the scale, the scale WILL move. If you applied EXACTLY 10 N of "JERK" the scale would temporarily/instantaneously - at the top end - be a total of 20N.

Answer 5

42N