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AS physics forces question

The rate of flow of water from a tap is 12kg min^-1. The initial velocity of the water is 0.6 m s^-1. Calculate the force exerted by the water on the tap.


I literally have no clue how to answer this question...any help will be appreciated:smile:.

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Reply 1
Original post by Megan_101
The rate of flow of water from a tap is 12kg min^-1. The initial velocity of the water is 0.6 m s^-1. Calculate the force exerted by the water on the tap.


I literally have no clue how to answer this question...any help will be appreciated:smile:.


Use v = u+at to get the acceleration.

Then use F=ma to get the force.
Reply 2
Original post by Kyx
Use v = u+at to get the acceleration.

Then use F=ma to get the force.


How can you use v=u+at? the only thing you know in that equation is the initial velocity?
Reply 3
Original post by Megan_101
How can you use v=u+at? the only thing you know in that equation is the initial velocity?


you have the final speed of the water in kg/min.

The density of water is equal to 1 kg per cubic metre

Therefore the volume is equal to 12 cubic metres.

Hence the final velocity is 12 metres per minute...

(guessing)
Reply 4
Original post by Kyx
you have the final speed of the water in kg/min.



Sorry where did this come from?
Reply 5
Original post by Megan_101
Sorry where did this come from?



Here:

Original post by Megan_101
The rate of flow of water from a tap is 12kg min^-1. The initial velocity of the water is 0.6 m s^-1. Calculate the force exerted by the water on the tap.


I literally have no clue how to answer this question...any help will be appreciated:smile:.
Original post by Megan_101
The rate of flow of water from a tap is 12kg min^-1. The initial velocity of the water is 0.6 m s^-1. Calculate the force exerted by the water on the tap.


I literally have no clue how to answer this question...any help will be appreciated:smile:.


12/60=0.2kg/s assume water is at rest at first then accelerated to a velocity of 0.6m/s

then using F=mvmutF=\frac{mv-mu}{t}
and Newtons 3rd law find the force.
Reply 7
Original post by Duke Glacia
12/60=0.2kg/s assume water is at rest at first then accelerated to a velocity of 0.6m/s

then using F=mvmutF=\frac{mv-mu}{t}
and Newtons 3rd law find the force.


That's the simplified version of what I said! :wink:
Original post by Kyx
That's the simplified version of what I said! :wink:


I am simple :tongue:
Reply 9
Original post by Duke Glacia
12/60=0.2kg/s assume water is at rest at first then accelerated to a velocity of 0.6m/s

then using F=mvmutF=\frac{mv-mu}{t}
and Newtons 3rd law find the force.


I'm probably being incredibly stupid here, but how do you know the mass?
Original post by Megan_101
I'm probably being incredibly stupid here, but how do you know the mass?


think of the 0.2kg/s. which means that 0.2kg of water is flowing in one second and initially this 0.2kg was at rest which is accelerated to 0.6m/s in a time of 1sec. Get it ?
Reply 11
Original post by Duke Glacia
think of the 0.2kg/s. which means that 0.2kg of water is flowing in one second and initially this 0.2kg was at rest which is accelerated to 0.6m/s in a time of 1sec. Get it ?



0.2kg? I don't know
Original post by Megan_101
0.2kg? I don't know


12kg/min =0.2kg/s so think that the tap has only 0.2kg of water which is initially at rest inside the pipe. this water is accelerated(pushed down through the pipe) by some force. So the pipe will feel the same force upwards right ?
then this 0.2kg of water is accelerated to a speed of 0.6m/s in 1 sec(think of the 0.2kg of water starting to accelerate at the end of the tap/pipe)
Reply 13
Original post by Duke Glacia
12kg/min =0.2kg/s so think that the tap has only 0.2kg of water which is initially at rest inside the pipe. this water is accelerated(pushed down through the pipe) by some force. So the pipe will feel the same force upwards right ?
then this 0.2kg of water is accelerated to a speed of 0.6m/s in 1 sec(think of the 0.2kg of water starting to accelerate at the end of the tap/pipe)


I don't know how you can find the mass from this though?😁
Original post by Megan_101
I don't know how you can find the mass from this though?😁


Think simply, it's easy to think it's more complex that it is. Rather than the mass of all of the water flowing through the pipe, think about the mass of water that flows past a certain point in the pipe in a specific time.
12kg per minute,
12/60= 0.2kg per second
In 1 second, 0.2kg of water flows past a point in the pipe.
Also, in 1 second it moved 0.6m (because of it's velocity)
(edited 8 years ago)
If the tap is exerting a force on the water, then its (the water's) momentum changes
The force is the rate of change of this momentum
Consider also Newton's 3rd
Play around with F=d(mv)/dt, note that the water is actually stationary to begin with, and that the water emerges with the speed given

In the end, you're looking to find that F=v*(dm/dt), where dm/dt is the given mass flow rate
(edited 8 years ago)
Reply 16
Original post by Laurasaur
Think simply, it's easy to think it's more complex that it is. Rather than the mass of all of the water flowing through the pipe, think about the mass of water that flows past a certain point in the pipe in a specific time.
12kg per minute,
12/60= 0.2kg per second
In 1 second, 0.2kg of water flows past a point in the pipe.
Also, in 1 second it moved 0.6m (because of it's velocity)


Ahhhhh right. So could you use F= change in momentum/change in time? As the change in time is 1s, F= essentially, change in momentum...so would you have to work out the final velocity of the water?
Reply 17
Original post by DeuteriumPie
If the tap is exerting a force on the water, then its (the water's) momentum changes
The force is the rate of change of this momentum
Consider also Newton's 3rd
Play around with F=d(mv)/dt, note that the water is actually stationary to begin with, and that the water emerges with the speed given

In the end, you're looking to find that F=v*(dm/dt), where dm/dt is the given mass flow rate

Because the water is stationary to start with, the change in momentum is just 0.2kg*0.6ms^-1 surely?
Reply 18
Original post by Megan_101
Because the water is stationary to start with, the change in momentum is just 0.2kg*0.6ms^-1 surely?


That would give momentum, yes
Reply 19
Original post by Kyx
That would give momentum, yes



Is that the answer then? 0.12Nm?

Original post by DeuteriumPie
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