What is the Force of Water on a Bathtub Plug?

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Homework Help Overview

The problem involves calculating the force exerted by water on a bathtub plug, given its area and depth below the water surface. The subject area includes fluid mechanics and pressure calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various formulas related to pressure and force, including P=F/A and F=(density)(Area)(Height)(Gravity). Questions arise regarding the computation of pressure and the relationship between pressure and force.

Discussion Status

The discussion has progressed with participants exploring different formulas and clarifying the concepts of pressure and force. Some guidance has been provided regarding the importance of units and the relationship between density, gravity, and height in calculating pressure.

Contextual Notes

Participants express uncertainty about the correct formulas and calculations, and there is a focus on ensuring proper unit conversions and understanding the implications of depth in fluid pressure.

andorei
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Homework Statement



The plug in a bathtub is 10cm2 in area and is 0.8m below the surface of the water. What is the force of the water on the plug?

Given Data: Area: 10cm^2
h: 0.8m


too lazy to post attempts. sorry.
 
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If you are too lazy to post attempts, then we are too lazy to help you.
 
Okayy fine. I get lost on what formula i should use.
 
Let's see the formulas you are contemplating using.
 
Hmm..
Formula for Pressure P=F/A ,
or Pascal's principle?
 
OK, we now have a beginning. How is pressure computed? How is force computed from pressure.
 
By multiplying Area and Pressure.
 
Pressure increases the deeper you go below the surface of a fluid. Does that give you an idea of how to compute pressure?
 
andorei said:
By multiplying Area and Pressure.

Yes, that will give you force. Now you need to compute pressure. How?
 
  • #10
Wait I found another formula, I think this one could be possible for solving the unknown.

F=(density)(Area)(Height)(Gravity)
 
  • #11
andorei said:
Wait I found another formula, I think this one could be possible for solving the unknown.

F=(density)(Area)(Height)(Gravity)

This formula provides the weight of an object.

We need the pressure where the plug is located because then you can determine the force on the plug by multiplying it by area.

With formulas, units are very important. Look at pressure P. Its units are force per unit area. Recalling what I said earlier about the pressure depending on how far one is below the surface, what two things when multiplied together provide you with force per unit area?
 
  • #12
Hint: What is the density of water?
 
  • #13
density, gravity and height? Coz' height cancels out the m^3?

Sorry for messing things up.
 
  • #14
You've got it. Pressure equals depth times distance under surface if the density is expressed in units of weight per unit volume. If density is mass per units volume, then you have to multiply by gravity as you cite above.

So now you can compute the force on the plug.
 
  • #15
9800N/m^3 then what?
 
  • #16
9800N/m2*

Sorry, been following the thread.
 
  • #17
TaxOnFear said:
9800N/m2*

Sorry, been following the thread.

How can that be, sir?

Density * Gravity = 1.0x103kg/m3(9.81m/s2)

= 9800N/m3

What made the cubic meter turn to a meter squared?
 
  • #18
Density x gravity x height, as you stated earlier.

(kg/m3) x (m/s2) x m

= N/m2
 
  • #19
I'm totally aware of that. But I was only at the first two.

With multiplying with the height the answer would be "7840N/m^2"
 
  • #20
The choices that are given in the book is
7.38N
7.62N
7.75N
7.84N
 
  • #21
Well there you go then. How do you get a force (N) from a pressure(N/m2)
 
  • #22
Multiply N by the area which cancels out both m^2 which then yields 7.84N as a result. Thank you so much TaxOnFear and LawrenceC.

I have to sleep now, it's late night here in the Philippines.
Thanks again.
 

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