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February 22, 2026, 07:53:27 am

Author Topic: Reaction force in bouncing balls  (Read 833 times)  Share 

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taqi

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Reaction force in bouncing balls
« on: April 29, 2011, 04:20:48 pm »
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Ive seen in my book, that when a ball is bouncing, and is at rest (at max compression hitting the floor, just before it starts going up again), Fn >> Fg, so the normal force is much greater than the weight force. But how can this be? if the ball is at rest, then aren't all the forces equal, so no movement?

moekamo

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Re: Reaction force in bouncing balls
« Reply #1 on: April 29, 2011, 04:27:21 pm »
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it can be at rest, but there is still an acceleration, think of throwing a ball in the air, when it is at the top of its flight it is stationary, but gravity is still acting on the ball, so there is still a force and an acceleration.
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xZero

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Re: Reaction force in bouncing balls
« Reply #2 on: April 29, 2011, 04:29:13 pm »
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The velocity is 0 but there can still be acceleration.( the gradient of the velocity graph at 0 can be anything) hence net force doesn't have to be 0
« Last Edit: April 29, 2011, 04:33:24 pm by xZero »
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kushwoho44

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Re: Reaction force in bouncing balls
« Reply #3 on: April 29, 2011, 04:31:50 pm »
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The velocity is 0 but the can still be acceleration.( the gradient of the velocity graph at 0 can be anything) hence net force doesn't have to be 0

Yeh, same as when you're throwing a ball in the air. It's acceleration is a constant 10 metres per second per second towards the earth yet at the top of it's flight it's velocity is zero.

taqi

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Re: Reaction force in bouncing balls
« Reply #4 on: April 29, 2011, 08:25:36 pm »
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ah yes, that makes sense. thanks

VivaTequila

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Re: Reaction force in bouncing balls
« Reply #5 on: May 10, 2011, 10:09:03 pm »
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this is something i've always struggled to understand conceptually, but meh. just come to accept it.

i just think of it using methods, if you have a quadratic, at it's turning point the gradient equals zero, but if you sketch the derivative of the quadratic you'll find that the gradient function is on a slope.

i find imagining the slope of the derivative helps, but that's just me, don't know if it has any relevance but it's a helpful analogy