How Much Force is Needed to Hold a Hose Emitting Water at 20 m/s?

[Priyadarshini]

Homework Statement

Water is pumped through a hose-pipe at a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with a speed of 20 m/s.What is the force is required from a person holding the hose-pipe to prevent it moving backwards?

Homework Equations

F=ma--?
Ft=mv-mu

The Attempt at a Solution

90kg/min=0.025kg/s
This is m/t
So if we assume u=0m/s
Then, F= 0.025 x 20 (using Ft=mv-mu)
F= 0.5N
The answer is 30N
Can we assume that u is 0m/s?

 

[SteamKing]

Homework Statement

Water is pumped through a hose-pipe at a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with a speed of 20 m/s.What is the force is required from a person holding the hose-pipe to prevent it moving backwards?

Homework Equations

F=ma--?
Ft=mv-mu

The Attempt at a Solution

90kg/min=0.025kg/s

{You might want to check this calculation again!}

 

[Priyadarshini]

{You might want to check this calculation again!}

Oh!
Even then,
90kg/min=0.0004167kg/s
then the answer becomes,
0.0004167 x 20
= 0.0083 N

 

[SteamKing]

Oh!
Even then,
90kg/min=0.0004167kg/s
then the answer becomes,
0.0004167 x 20
= 0.0083 N

I repeat, check your mass flow calculation again. How many seconds are in 1 minute?

 

[CWatters]

Can I recommend you work out the number of kg/s using a pencil and paper rather than a calculator.

 

[SteamKing]

Can I recommend you work out the number of kg/s using a pencil and paper rather than a calculator.

90 kg/min you should be able to work out in your head.

 

[Adoniram]

Homework Statement

Water is pumped through a hose-pipe at a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with a speed of 20 m/s.What is the force is required from a person holding the hose-pipe to prevent it moving backwards?

Homework Equations

F=ma--?
Ft=mv-mu

The Attempt at a Solution

90kg/min=0.025kg/s
This is m/t
So if we assume u=0m/s
Then, F= 0.025 x 20 (using Ft=mv-mu)
F= 0.5N
The answer is 30N
Can we assume that u is 0m/s?

What is Ft exactly?

Try applying basic principles:
F = ma is what we usually use when applying the 2nd Law, but a more appropriate form might be:
F = d/dt (m*v) = v*(d/dt m) + m*(d/dt v)

In the case of constant mass, this reduces to F = ma. In the case of constant velocity (this problem), this can be reduced to F = v*(d/dt m)

You are given the rate of water being pumped, which is in units of kg/min (can be converted to kg/s). This looks a lot like... dm/dt, right?

So, you can calculate the force being pressed back against the person holding the hose. But you are asked how much force it takes to keep the hose from moving. If that's what you want, what is true about the total acceleration of the hose? Now apply that to the 2nd Law, and what do you get?

 

[Priyadarshini]

I repeat, check your mass flow calculation again. How many seconds are in 1 minute?

Whoops! It's 1.5kg/s. It works now. Thank you!

 

[Priyadarshini]

What is Ft exactly?

Try applying basic principles:
F = ma is what we usually use when applying the 2nd Law, but a more appropriate form might be:
F = d/dt (m*v) = v*(d/dt m) + m*(d/dt v)

In the case of constant mass, this reduces to F = ma. In the case of constant velocity (this problem), this can be reduced to F = v*(d/dt m)

You are given the rate of water being pumped, which is in units of kg/min (can be converted to kg/s). This looks a lot like... dm/dt, right?

So, you can calculate the force being pressed back against the person holding the hose. But you are asked how much force it takes to keep the hose from moving. If that's what you want, what is true about the total acceleration of the hose? Now apply that to the 2nd Law, and what do you get?

Oh,thanks!