Water falling into bucket, Force question

AI Thread Summary
Water falls at a rate of 271 g/s from a height of 56.2 m into a bucket weighing 776 g. After 3.14 seconds, the total mass of water collected is approximately 0.851 kg, leading to a combined weight of 1.63 kg. The force calculation initially yields 15.9 N, but it does not account for the additional force needed to stop the water's momentum. The distance is relevant for determining the water's speed upon impact, calculated as 17.9 m/s. The discussion highlights the importance of correctly interpreting the time and speed of the falling water to solve the problem accurately.
Leeoku
Messages
18
Reaction score
0

Homework Statement


13. Water falls at the rate of 271 g/s from a height of 56.2 m into a 776g bucket on a scale (without splashing). If the bucket is originally empty, what does the scale read after 3.14 s?
Answer: 2.50e+01 N



Homework Equations


Density, estimate 1kg/m^3
F = ma (a = gravity)
1kg = 1L


The Attempt at a Solution


In 3 seconds, .271*3 = .851 L dropped->.851 KG
Add weight of bucket, .851+.776 = 1.63 kg
F =ma
= 15.9 N
 
Physics news on Phys.org
You found the weight of the water + bucket. Good! But an additional force is required to stop the momentum of the flowing water.
 
i actually have no clue. what I am puzzled at is why they gave us distance. the first thing that comes to mind is work, but i don't think that helps..
 
Leeoku said:
i actually have no clue. what I am puzzled at is why they gave us distance. the first thing that comes to mind is work, but i don't think that helps..
They gave you the distance so you can figure out the speed of the water as it hits the bucket. Hint: Use that speed to find the rate of change of the momentum of the water as it hits the bucket.
 
ok using v=d/t, 56.2/3.14 = 17.9 m/s
Impulse = delta p = (sum F)delta T
Delta p = 15.9*3.14
= 50 N.s
Using 50 as momentum, p = mv
50 = 17.89 M
m = 2.8 kg

Weird part is answer is given in Newtons, so mass*g = 27.4 N. Still not right, close though..
 
Leeoku said:
ok using v=d/t, 56.2/3.14 = 17.9 m/s
Two problems:
(1) The 3.14 s time given is not the time it takes for the water to fall from the given height.
(2) The speed of the water is not constant as it falls.

To find the speed of the water as it hits the scale, treat it as a falling body (falling from rest through a height of 56.2 m).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top