Calculating force of water coming out

[Raghav Gupta]

Homework Statement

The side wall of a wide vertical cylindrical vessel of height , h = 75 cm has a narrow vertical slit running all the way to the bottom of the vessel. The length of slit is l = 50 cm and width b = 0.1 cm. With the slit closed, the vessel is filled with water. What is the resultant force of reaction of the water flowing out of the vessel immediately after the slit is opened?

A. 4.9 N
B. 0.49 N
C. 49 N
D. 490 N

Homework Equations

P = P₀ + ρgh
F = PA
g = 9.8 m/s²
ρ = 10³kg/m³
b = 10^-3m
h = 3/4 m
l = 1/2 m

The Attempt at a Solution

The P₀ will be canceled as outside pressure is also P₀
$$ F = \int_0^l bρg(h-l)dl $$

$F = bρg [hl - \frac{l^2}{2} ]$

⇒ F= 9.8 /4 = 2.45 N

Where is the flaw?
Why I am getting a wrong answer?

 

[haruspex]

I get the same answer. Do you know what it is supposed to be?

 

[Raghav Gupta]

I get the same answer. Do you know what it is supposed to be?

Yes, it is supposed to be 4.9 N.
They have given solution showing Toricelli law.
Is this approach wrong?

 

[haruspex]

Yes, it is supposed to be 4.9 N.
They have given solution showing Toricelli law.
Is this approach wrong?

Can you post the whole supposed solution?

 

[haruspex]

But... it clearly says "immediately after the slit is opened". It will take time to reach a steady velocity, so I do not see how Torricelli's law can be applied here.

 

[Raghav Gupta]

But... it clearly says "immediately after the slit is opened". It will take time to reach a steady velocity, so I do not see how Torricelli's law can be applied here.

So if they would not have written "immediately" would then my answer have been wrong and their's correct?

 

[Chestermiller]

They added the word immediately to let you know that you are to use the starting level of fluid in the tank to calculate the velocity out of the slot. As far as the time required to establish a steady velocity distribution is concerned, they intended you to assume that this happens instantaneously, and can be neglected. The time required would be on the order of the depth divided by the speed of sound. The level in the tank would not change significantly during that time interval.

What they are doing here is a horizontal momentum balance on the fluid. They are finding the force that the tank has to exert on the fluid to accelerate it out of the slot at the indicated velocity. This is equal and opposite to the horizontal force that the fluid exerts on the tank.

Their analysis is correct.

Chet

 

[haruspex]

They added the word immediately to let you know that you are to use the starting level of fluid in the tank to calculate the velocity out of the slot. As far as the time required to establish a steady velocity distribution is concerned, they intended you to assume that this happens instantaneously, and can be neglected. The time required would be on the order of the depth divided by the speed of sound. The level in the tank would not change significantly during that time interval.

What they are doing here is a horizontal momentum balance on the fluid. They are finding the force that the tank has to exert on the fluid to accelerate it out of the slot at the indicated velocity. This is equal and opposite to the horizontal force that the fluid exerts on the tank.

Their analysis is correct.

Chet

Chet, can you explain what is wrong with the analysis adopted by both Raghav and me? Prior to removal of the strip, the water exerted a force on the strip of 2.45N. The instant the strip is removed, that force is no longer exerted, but all the other forces on the container remain. So the reaction force should be 2.45N.
To make the Torricelli solution correct, it should have asked for the force when the flow is established but the level has barely fallen.

 

[Chestermiller]

Chet, can you explain what is wrong with the analysis adopted by both Raghav and me? Prior to removal of the strip, the water exerted a force on the strip of 2.45N. The instant the strip is removed, that force is no longer exerted, but all the other forces on the container remain. So the reaction force should be 2.45N.
To make the Torricelli solution correct, it should have asked for the force when the flow is established but the level has barely fallen.

I agree. The problem is poorly posed.

Chet

 

[Raghav Gupta]

Thanks guys.
I posted this problem since I was amazed that why my attempt was giving a solution not in options and I tried and tried but still same 2.45 N, but I see the solution which they have given is interpreting something else.

 

[Parthc]

Let us consider an infinetisimal portion of length $dx$ of the slit at a depth $x$ below water level.
Reaction force due to this portion is given by:
$dF = a(v^2)p = bdx(2gx)p$
Total reaction force is Integration of this expression from:
$(h-l)→ (h)$.
Here $b$ is the Breadth, $p$ is the Density, $a$ is the Area.

 

[Chestermiller]

Let us consider an infinetisimal portion of length $dx$ of the slit at a depth $x$ below water level.
Reaction force due to this portion is given by:
$dF = a(v^2)p = bdx(2gx)p$
Total reaction force is Integration of this expression from:
$(h-l)→ (h)$.
Here $b$ is the Breadth, $p$ is the Density, $a$ is the Area.

Did you realize that this thread is over 3 1/2 years old and that the OP has not been seen since early July?