Pushing ball underwater

In summary, the problem involves calculating the work done when a ball of radius 15 inches is pushed halfway under water, taking into account the weight of the ball and the lift provided by the water. The difficult part of the problem is calculating the average volume of the ball that is underwater, which cannot be done as a sum and would require an integral. As this concept has not been taught yet, the easiest way to solve the problem is to approximate the average volume.
  • #1
2
0

Homework Statement



A ball of which radius is 15 inches is put into water and it sinks 2.5 inches. How much is the work done when it's pushed halfway under water?

Homework Equations



Air pressure doesn't have to be taken into account (not sure about that but that one I can do myself anyways).

Lift by the water (the weight of the ball can be calculated with this):

[tex]N=\rho Vg[/tex]

Average V of the part of the ball that is underwater is (this one I made up myself):

[tex]V_{Av}=\frac{ \sum^{\infty}_{n=0} V_{n}}{\infty}[/tex]

Where[tex] V_{n}[/tex] is the volume of the ball underwater when the ball is up to its [tex]\frac{n}{\infty}*r[/tex] underwater (where r is the radius of the ball).

Work done:

[tex]W= \rho V_{Av}g * 15 inches - mg * 15 inches[/tex] (Where m is the mass of the ball and W is the final answer. )

The Attempt at a Solution



I can do everything else in this exercise except I can't calculate [tex]V_{Av}=\frac{\sum ^{\infty}_{n=0} V_{n}}{\infty}[/tex] (the hard part is getting the [tex]\infty[/tex] out of the expression) but I'm not even sure if it's supposed to be calculated this way as we haven't been taught the concept of [tex] \sum [/tex].

This exercise is one of the preliminary exercises that came with my physics book so it's not homework or something like that and therefore I'm not interested in the exact answer but rather the theory and easiest way to solve it.

Thanks in advance for any kind of help.
 
Last edited:
Physics news on Phys.org
  • #2
Drektjal said:

Homework Statement



A ball of which radius is 15 inches is put into water and it sinks 2.5 inches. How much is the work done when it's pushed halfway under water?

Homework Equations



Air pressure doesn't have to be taken into account (not sure about that but that one I can do myself anyways).
Since it says you don't have to worry about that I imagine you can!

Lift by the water (the weight of the ball can be calculated with this):

[tex]N=\rho Vg[/tex]

Average V of the part of the ball that is underwater is (this one I made up myself):

[tex]V_{Av}=\frac{ \sum^{\infty}_{n=0} V_{n}}{\infty}[/tex]

Where[tex] V_{n}[/tex] is the volume of the ball underwater when the ball is up to its [tex]\frac{n}{\infty}*r[/tex] underwater (where r is the radius of the ball).

The Attempt at a Solution



I can do everything else in this exercise except I can't calculate [tex]V_{Av}=\frac{\sum ^{\infty}_{n=0} V_{n}}{\infty}[/tex] (the hard part is getting the [tex]\infty[/tex] out of the expression) but I'm not even sure if it's supposed to be calculated this way as we haven't been taught the concept of [tex] \sum [/tex].

This exercise is one of the preliminary exercises that came with my physics book so it's not homework or something like that and therefore I'm not interested in the exact answer but rather the theory and easiest way to solve it.

Thanks in advance for any kind of help.
You can't do it as a sum and you don't have "infinity" in it- if you want to do it exactly you will need to use an integral, not a sum.
 
  • #3
HallsofIvy said:
You can't do it as a sum and you don't have "infinity" in it- if you want to do it exactly you will need to use an integral, not a sum.

-So you'd recon there's no other way around this than that as integral is also something that hasn't been taught to us (if I remember right I don't have the integral-class in maths until next semester)?

Thanks for the quick answer though!
 

1. What is the concept behind pushing a ball underwater?

The concept behind pushing a ball underwater is based on the principles of buoyancy and fluid dynamics. When an object, such as a ball, is placed in water, it experiences an upward force known as buoyancy, which is equal to the weight of the water displaced by the object. When the ball is pushed underwater, it must displace a larger volume of water, resulting in a greater upward force. This creates a situation where the force pushing the ball upward is greater than the force of gravity pushing it downward, allowing the ball to be pushed underwater.

2. Why does the ball float back to the surface after being pushed underwater?

The ball floats back to the surface because of the principle of buoyancy. As mentioned before, when an object is placed in water, it experiences an upward force equal to the weight of the water displaced. When the ball is pushed underwater, it displaces a larger volume of water, creating a greater upward force. However, as the ball rises back to the surface, the volume of water it displaces decreases, resulting in a decrease in the upward force. When the upward force is less than the weight of the ball, it will sink back to the bottom.

3. How does the depth at which the ball is pushed affect its behavior?

The depth at which the ball is pushed underwater will affect its behavior because the pressure and density of the water increases with depth. As the ball is pushed deeper, it will experience a greater upward force due to the increased pressure and density of the water. This means that the ball will be easier to push underwater at greater depths, and will also float back to the surface more quickly.

4. Can the ball be pushed underwater indefinitely?

No, the ball cannot be pushed underwater indefinitely. Eventually, the ball will reach a depth where the upward force is equal to the weight of the ball, and it will stop sinking. This is known as the equilibrium point. If the ball is pushed below this point, the upward force will be greater than the weight of the ball, causing it to rise back to the surface.

5. How does the size and weight of the ball affect its behavior when pushed underwater?

The size and weight of the ball will affect its behavior when pushed underwater because it determines the amount of water the ball will displace and the force required to push it underwater. A larger and heavier ball will displace more water and require a greater force to push it underwater, while a smaller and lighter ball will displace less water and require less force to push it underwater. This means that a larger and heavier ball will sink deeper and faster, while a smaller and lighter ball will not sink as deep and will resurface more slowly.

Suggested for: Pushing ball underwater

Replies
12
Views
668
Replies
3
Views
740
Replies
25
Views
1K
Replies
10
Views
705
Replies
4
Views
159
Replies
10
Views
985
Replies
5
Views
850
Back
Top