Integration buoyancy problem

In summary, the conversation revolves around finding the amount of work needed to push a 4 cm diameter cylinder 10 cm deeper into water. The attempt at a solution involves using the integral of (.02m)^2 * pi * (1000kg/m^3) * (9.80) * x dx over the range of 0 to 0.1 m. However, the calculated answer of .0615 J does not match the answer in the book of .615 J. The discrepancy may be due to an error in the book or a mistake in the problem data.
  • #1
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Homework Statement


A 4 cm diameter cylinder floats in the water. How much work must be done to push the cylinder 10 cm deeper into the water?


Homework Equations





The Attempt at a Solution



I did the integral from 0 to .1 m of (.02m)^2 * pi * (1000kg/m^3) * (9.80) * x dx

= (12.3 x^2)/2 evaluated from 0 to .1 = .0615 J. However, the back of my book says .615 J. Somehow I ended up a decimal place off. Could someone help me?
 
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  • #2
Anyone have any thoughts on this one?
 
  • #3
Isn't the radius=0.2 instead of 0.02 ?
 
  • #4
no, I'm pretty sure the radius is correct

4 cm/2 = radius of 2 cm 2cm * 1m/100cm = .02 m
 
  • #5
Any thoughts would be greatly appreciated
 
  • #6
I think that given the values you are working with that your answer is correct.
 
  • #7
I am concerned because the back of my book says it's .615 J and I can't figure out what's wrong with mine
 
  • #8
Can somebody find my mistake?
 
  • #9
I need some help with this one pretty soon because my homework is due tomorrow. Thanks!
 
  • #10
Sometimes the answer in the book is wrong...or the problem data has been misread. In either case it looks like to me that your answer is correct, If not, I would also like to know why and maybe another reader can set us both strait.
 

1. What is integration buoyancy problem?

The integration buoyancy problem is a phenomenon that occurs when an object is submerged in a liquid and experiences a buoyant force. This force is caused by the pressure difference between the top and bottom of the object, and it can either help or hinder the object's movement through the liquid.

2. How does buoyancy affect integration in fluids?

Buoyancy plays a crucial role in the integration of objects in fluids. The buoyant force acts in an upward direction, opposing the force of gravity on the object. This force helps to keep the object afloat or suspended in the fluid, making it easier to integrate it into the surrounding environment.

3. What factors influence the integration buoyancy problem?

The integration buoyancy problem is influenced by several factors, including the density and volume of the object, the density of the fluid, and the depth at which the object is submerged. The shape and orientation of the object also play a role in determining the magnitude and direction of the buoyant force.

4. How is the buoyant force calculated?

The buoyant force can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. This can be represented by the equation Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

5. How can the integration buoyancy problem be utilized in scientific research?

The integration buoyancy problem is an important concept in many scientific fields, such as fluid dynamics and oceanography. It is used to study the behavior of objects in fluids and to understand how different factors affect the buoyant force. It is also used in designing and testing various structures, such as ships and submarines, to ensure they have proper integration and stability in water.

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