0
$\begingroup$

This bowling ball is floating in water.enter image description here The force of gravity attracting the ball to the weights is very weak, about 2 ten millionths of a newton. Is the water impeding the ball's movement? I know the water is pushing on the ball, but each force has a counter force on the other side of the ball. So is the net force on the ball zero? It may look like the sides of the bucket are squeezing the ball, but they are not. Any ideas on what I could do to allow the feeble fraction of a newton to move the ball to the weights? Thanks in advance for any insights you offer.

$\endgroup$
  • $\begingroup$ In these cases the culprit is almost always static friction. Read up on the Cavendish experiment to see the pains people have gone to to measure gravity between two normal sized objects. $\endgroup$ – Javier Oct 8 '16 at 16:39
  • $\begingroup$ I agree the Cavendish experiment will drive you mad. Getting the bar to come to rest takes forever, and then any movement in the room will cause it to move again. That's what I like about the floating ball, very little wait time, and it tolerates movements in the room. $\endgroup$ – Lambda Oct 8 '16 at 17:25
0
$\begingroup$

If your bowling ball is floating and experiences your estimated extremely small gravitational force, it will extremely slowly move towards the weights slowed only by the viscous resistance force of the water. You can easily calculate the velocity of the ball as a function of time given the force, the mass and the frictional resistance force on the ball by the viscosity in water (Stokes Law). However, you will probably have to wait eons until a movement will be visible. And much earlier your ball will get stuck because the water has evaporated.

$\endgroup$
  • $\begingroup$ That sounds like and interesting calculation. I know the force and distance. I'll look up stokes law. What would the equation look like? $\endgroup$ – Lambda Oct 8 '16 at 17:21
  • $\begingroup$ The equation of motion would approximately be M·x''= FG - 6·π·𝜇·R·x' , where x is the ball coordinate, M and R are the mass and radius of your ball, 𝜇 is the dynamic viscosity of water and x' is the velocity of the ball. Have fun! I would be interested how long it would take for the ball to move one cm. $\endgroup$ – freecharly Oct 8 '16 at 17:44
  • $\begingroup$ FG is the gravitational force of your weights on the ball. $\endgroup$ – freecharly Oct 8 '16 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.