Calculating Force Exerted by a Water Fountain on a Vertical Board

[Biker]

Homework Statement

ِA fountain of water the pushes water from a hole its radius = 2.5 cm with a speed of 30 m/s. This water hits a flat board that was put vertically in respect to the water and it is moving away of water with a speed of 6 meters
1) calculate the force that the water exerts on the board
2) Calculate this force if the board was moving to the water with a speed of 6 m/s and the water density is 1000 kg/m3

Homework Equations

Density = mass/volume
Momentum = mv

The Attempt at a Solution

Aren't I missing something here? The question didn't give me the mass of the board.. Neither it told me the final speed of the water. How I am supposed to answer this?

 

[mfb]

How would you use the mass of the board?

What do you expect for the final speed of the water?

 

[Biker]

Maybe if I have the mass of the board I could find through that the velocity of the water so I can basically get the force from that.

Final speed of water: There is nothing here can really tell me the speed of water is there?

 

[mfb]

Maybe if I have the mass of the board I could assume that they both move with the same velocity so I can basically get the force from that.

How would you use the mass of the board to do so? In which formula would it appear?

Final speed of water: There is nothing here can really tell me the speed of water is there?

Well, do you expect it to bounce off and move backwards away from the board? Do you expect it to stay faster than the board? If not, what is left as option?

 

[Biker]

How would you use the mass of the board to do so? In which formula would it appear?
Well, do you expect it to bounce off and move backwards away from the board? Do you expect it to stay faster than the board? If not, what is left as option?

as probably
m(water)v = m(board)*v + m(water)*v
Or if the questions means that the board was moving intially with a speed of 6 m/s then
m(water)v + m(board)v = m(board)v + m(water)v
but now I have to unknowns, Those are v of the board and v of water
I can get through that the velocity of the water after the collision and use that to determine the force

Move backwards.

 

[haruspex]

as probably
m(water)v = m(board)*v + m(water)*v
Or if the questions means that the board was moving intially with a speed of 6 m/s then
m(water)v + m(board)v = m(board)v + m(water)v
but now I have to unknowns, Those are v of the board and v of water
I can get through that the velocity of the water after the collision and use that to determine the force

Move backwards.

Don't worry about the board, just consider what is happening to the water. In a short time period t, how much water hits the board? What is the change in its velocity?

 

[Biker]

Don't worry about the board, just consider what is happening to the water. In a short time period t, how much water hits the board? What is the change in its velocity?

Lets just calculate what happens in one sec.
You can get through that the mass of the water
It will be around 58 kg every second with a speed of 30 m/s

I can't say that the speed of the water is 6 m/s just as the board because it gives me wrong answer and I guess the question meant that the board is moving at 6 m/s initially.

 

[mfb]

The speed of the board is constant, the water does not accelerate it.

I can't say that the speed of the water is 6 m/s just as the board because it gives me wrong answer and I guess the question meant that the board is moving at 6 m/s initially.

What do you get as answer, and how do you know it is wrong?

 

[Biker]

The speed of the board is constant, the water does not accelerate it.
What do you get as answer, and how do you know it is wrong?

If I use 6 m/s as the final velocity of the water then
F * 1 = 58.9(6-30)
F = -1413 N

However the answer in my book is 83 N

 

[haruspex]

If I use 6 m/s as the final velocity of the water then
F * 1 = 58.9(6-30)
F = -1413 N

However the answer in my book is 83 N

When you calculate the the rate at which water reaches the board (your 58.9 kg/s) you need to take into account the movement of the board.
(Edit: But I still get much more than 83N.)

 

[Biker]

When you calculate the the rate at which water reaches the board (your 58.9 kg/s) you need to take into account the movement of the board.
(Edit: But I still get much more than 83N.)

Oh yes I didn't put that concept in my calculations. Maybe there is the effect of gravity in the problem? because it said vertically.
I will try to do some calculations and see whether I get the right answer or no.

 

[haruspex]

Oh yes I didn't put that concept in my calculations. Maybe there is the effect of gravity in the problem? because it said vertically.
I will try to do some calculations and see whether I get the right answer or no.

It is not clear, but my interpretation is that the jet is horizontal. Yes, it will fall in an arc, but the horizontal velocity won't change, so you can ignore gravity.
If the jet were vertical then the board would be horizontal and you would need to know how high up it is.

 

[Biker]

It is not clear, but my interpretation is that the jet is horizontal. Yes, it will fall in an arc, but the horizontal velocity won't change, so you can ignore gravity.
If the jet were vertical then the board would be horizontal and you would need to know how high up it is.

Ehh yea.

This question is a bit hard to grasp. I solved the rest of the questions easily it is from a first year college book (I am still in high school)
I have a bit of problem with this. So it says that the board is moving away from the water with a velocity of 6 m/s is that before the collision or after the collision.
Whether it is horizontal or vertical we can try each one of them and check the answer but I just want to know the statement above so I can at least find a way to get the velocity of the water after the collision and the time that the force affected the board in it

 

[haruspex]

it says that the board is moving away from the water with a velocity of 6 m/s is that before the collision or after the collision.

Well, it's 6m/s away from the source of the water. What do you think the speed of the water is after it hits the board? (I think you got this right in your previous calculation.)

 

[Biker]

Well, it's 6m/s away from the source of the water. What do you think the speed of the water is after it hits the board? (I think you got this right in your previous calculation.)

I know I was just asking if it was moving with a speed of 6 m/s before even the water fountains starts pushing water. That is what bothers me a bit.
Anyway I will check the calculations after few hours with both ways and I will post them so you can check it. I will also try to find the rate of water flowing into the board in respect of the board's speed.
Edit: Could you draw your perspective of the question? I used to think that both were vertical

 

[haruspex]

I know I was just asking if it was moving with a speed of 6 m/s before even the water fountains starts pushing water.

The way I read the question, it is 6m/s at some instant, and you are to find the force at that instant. Maybe it is constant 6m/s, it doesn't matter.

I used to think that both were vertical

No, I'm pretty sure the jet is horizontal. Mind you, I get a much higher force than the answer you give, so it could be intended to be at some angle to the horizontal. But you would need to be told that angle. It cannot be vertical, since it would then not hit the board.

 

[mfb]

I see no way to get a force as low as 83N.
I also imagined the stream as horizontal. If the board is very close to the source of the stream in a vertical orientation it does not matter, however.