Find the force acting per unit width on the vertical wall

In summary: In that case, the force per unit width would be the pressure force.My understanding is that they are asking about the force on...the vertical wall.
  • #1
Kaushik
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17
Homework Statement
A container of height 3m and width 5m is filled with water upto 2 m. It is accelerated such that the water is at the brim (it is about to spill). When it moves with this acceleration find the force acting per unit width on the vertical wall.
Relevant Equations
I got ##a = 4 \frac{m}{s^2}##
A container of height 3m and width 5m is filled with water upto 2 m. It is accelerated such that the water is at the brim (it is about to spill). When it moves with this acceleration find the force acting per unit width on the vertical wall.
I found the acceleration for the condition provided but how to find the force per unit width?
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  • #2
Kaushik said:
I got a = 4
Units?

how to find the force per unit width?
Is there some relation between force and acceleration?
 
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  • #3
TSny said:
Units?
I added the units. Sorry!

TSny said:
Is there some relation between force and acceleration?
##F = ma ##
## F = ρAba## ; where ##A## is Area, ##b## is width, ##a## is acceleration and ##ρ## is density.
 
  • #4
Kaushik said:
##F = ma ##
## F = ρAba## ; where ##A## is Area, ##b## is width, ##a## is acceleration and ##ρ## is density.
Great! You're almost home.
 
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  • #5
TSny said:
Great! You're almost home.
Oh! Now?
 
  • #6
Can you express "force per unit width" in terms of F and b?
 
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  • #7
TSny said:
Can you express "force per unit width" in terms of F and b?
Dividing by ##b## both sides of the equation.
 
  • #8
Kaushik said:
Dividing by ##b## both sides of the equation.
I think you're implying that "Force per unit width" is F/b. If so, good!
 
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  • #9
TSny said:
I think you're implying that "Force per unit width" is F/b. If so, good!
Yes, I actually meant that.
 
  • #10
Now what how to proceed further?
 
  • #11
Kaushik said:
Now what how to proceed further?
So what expression did you get for force per unit width? What number do you get when you plug in the data?
 
  • #12
haruspex said:
ber do you get when yo
I think I was wrong before. We will be getting the following expression : ##F = ρALa ## where ##A= h.b## is the area, ##L=5## is the length, ##a## is the acceleration.
## \frac{F}{b} = ρ.h.L.a = 6 * 10^4##
 
  • #13
Kaushik said:
I think I was wrong before.
Yes, I hadn't checked post #3 because TSny confirmed it. He must have thought your A was the base area.
Kaushik said:
F=ρALa
Ok, but what is A here? Think about the volume of the water.
 
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  • #14
haruspex said:
Yes, I hadn't checked post #3 because TSny confirmed it. He must have thought your A was the base area.

Ok, but what is A here? Think about the volume of the water.
Oh , the volume of water is different. Should I consider the ##A## of trapezium?
In the above expression I considered area of the wall (left) . I think it would be better if I consider area of the trapezium and multiply it by width to get Volume.

So ##V = A_{trapezium} * b = 10b##
## F = ρVa = 10ρba \implies \frac{F}{b} = 4 * 10^4 N/m##
 
  • #15
Kaushik said:
Oh , the volume of water is different. Should I consider the ##A## of trapezium?
In the above expression I considered area of the wall (left) . I think it would be better if I consider area of the trapezium and multiply it by width to get Volume.

So ##V = A_{trapezium} * b = 10b##
## F = ρVa = 10ρba \implies \frac{F}{b} = 4 * 10^4 ##
That is the right answer, but you did not need to use the trapezium formula. The volume is not changed by the acceleration.
 
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  • #16
haruspex said:
That is the right answer, but you did not need to use the trapezium formula. The volume is not changed by the acceleration.
What about the Force due to pressure due to g? (sorry for not writing the units once again)
 
  • #17
Kaushik said:
What about the Force due to pressure due to g? (sorry for not writing the units once again)
Whoops! You are right. All you have found is the additional force due to the acceleration. You need to add in the force when there is no acceleration.
 
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  • #18
I think I totally misinterpreted the problem statement. I thought it was asking about the force per unit width required to accelerate the system if you neglect the mass of the container and assume no external friction acting on the system. I thought of "width" here as referring to the dimension of the container perpendicular to the diagram.

I'm still not sure about the meaning of "the force per unit width acting on the vertical wall". Are they asking about the force caused by the fluid pressure on the 3 m x 5 m wall while the system is accelerating? In that case, it seems odd to me to ask for the force per unit width rather than just the force.

Anyway, sorry for the confusion.
 
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  • #19
TSny said:
I'm still not sure about the meaning of "the force per unit width acting on the vertical wall". Are they asking about the force caused by the fluid pressure on the 3 m x 5 m wall while the system is accelerating? In that case, it seems odd to me to ask for the force per unit width rather than just the force.
My understanding is that they are asking about the force on the left hand vertical wall against which the water is piling up. Expressed as force per unit depth-into-the-paper-drawing. It seems to me that there is an insight which can reduce much work and allow the answer to be readily extracted with almost no calculation required.

Let me try to give a huge hint at the relevant insight by asking some easy questions. What is the pressure at the top of the fluid piled up at the left wall? What would the vertical pressure gradient be if the fluid were exposed to a vertical gravitational field with magnitude g with no horizontal acceleration? How does that vertical pressure gradient change when the tank is exposed to a horizontal acceleration as well?
 
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  • #20
TSny said:
it seems odd to me to ask for the force per unit width rather than just the force.
Yes, it would have made more sense if the width had not been given.
 
  • #21
jbriggs444 said:
My understanding is that they are asking about the force on the left hand vertical wall against which the water is piling up. Expressed as force per unit depth-into-the-paper-drawing. It seems to me that there is an insight which can reduce much work and allow the answer to be readily extracted with almost no calculation required.

Let me try to give a huge hint at the relevant insight by asking some easy questions. What is the pressure at the top of the fluid piled up at the left wall? What would the vertical pressure gradient be if the fluid were exposed to a vertical gravitational field with magnitude g with no horizontal acceleration? How does that vertical pressure gradient change when the tank is exposed to a horizontal acceleration as well?
I got the vertical pressure gradient when there is no horizontal acceleration. But when there is horizontal acceleration won't that add up to the force that is already present?
 
  • #22
Kaushik said:
I got the vertical pressure gradient when there is no horizontal acceleration. But when there is horizontal acceleration won't that add up to the force that is already present?
The [vertical component of the] pressure gradient is unchanged by the presence of the horizontal acceleration, yes.

That does not mean that the total force on the left hand wall is unchanged when more water piles up on it. But it does mean that you can easily calculate the total force on the left hand wall based on the depth of the water against the wall.
 
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Related to Find the force acting per unit width on the vertical wall

What is the meaning of "force acting per unit width"?

"Force acting per unit width" refers to the amount of force that is applied to a specific width or length of a surface. It is a measure of the intensity of force over a certain area.

Why is it important to find the force acting per unit width on a vertical wall?

It is important to find the force acting per unit width on a vertical wall in order to understand the distribution of force on the wall and ensure that it is able to withstand the applied force without collapsing or failing.

How is the force acting per unit width calculated?

The force acting per unit width can be calculated by dividing the total force applied to the wall by the width of the wall. This will give the amount of force exerted on each unit of width.

What factors can affect the force acting per unit width on a vertical wall?

The force acting per unit width can be affected by various factors such as the magnitude and direction of the applied force, the material and construction of the wall, and the stability of the surrounding environment.

What are some real-life applications of finding the force acting per unit width on a vertical wall?

Finding the force acting per unit width on a vertical wall is important in engineering and construction, as it helps determine the strength and stability of structures. It is also useful in fields such as geology and environmental science, where understanding the distribution of force on natural structures like cliffs or dams is important for predicting potential failures.

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