5
$\begingroup$

If a ball under the influence of gravity falls straight down from a height $h$, collides elastically with the floor, at the instant of collision, what forces does it experience?

Shouldn't the floor exert a force equal to the weight of the ball, on the ball itself? Then won't the forces cancel out?(since the only to forces on the ball are it's weight and the force of the floor, which are equal and Opposite)

I know that the better way to find resulting velocity, etc. Would be to use work-energy or kinematics, but I would like to reconcile those with the force treatment

$\endgroup$
  • 1
    $\begingroup$ Please see my answer to an earlier question about bouncing balls for some insight: physics.stackexchange.com/a/117141/26969 $\endgroup$ – Floris Jun 27 '14 at 12:38
  • $\begingroup$ Note $F=m g$ ONLY applies to a static case. Here as the ball experiences acceleration it is dynamic with $F-m g = m a$ $\endgroup$ – John Alexiou Jul 28 '14 at 5:52
2
$\begingroup$

The floor provides a normal force on the ball. That force is a constraint. It prevents the ball from continuing downward. Constraint forces provide whatever force is required to maintain the constraint. For example: a book on a table. Regardless of the weight of the book, the table is able to provide exactly the right force to counter gravity. The mechanism for the normal force is the elasticity of the tabletop (or floor). Think of the atoms as balls and the bonds between them as springs. The top layer of springs compress just enough to provide the needed force.

In the case of the dropped ball, the normal force has to accelerate the ball in the upward direction. That is, it provides a force greater than the weight of the ball.

How do those springs "know" how much force to provide? The springs are compressed by the action of the KE of the ball. Absent losses (elastic) all of that KE is converted to the PE of the springs, and then returned back to the ball, so it rebounds with the same speed with which it hit the floor.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

There is a force from the floor on the ball - so the ball experiences a force. There is no net force on the "ball plus floor" system - that is where the "forces cancel out".

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If a ball under the influence of gravity falls straight down from a height h, collides elastically with the floor, at the instant of collision, what forces does it experience?

Impact forces during short duration collision events can be very, very high. Typical values are hundreds or even thousand times the gravitational force (Cross 1999).

Shouldn't the floor exert a force equal to the weight of the ball, on the ball itself?

The contact forces have to be much greater than the weight of the ball. The ball would penetrate into the surface and continue at a constant velocity if the force was only that of the weight of the ball. That's not what happens. The ball's velocity changes from downward to upward in a few milliseconds or less. That means there are huge transient (short-lived) forces involved in the collision.

Then won't the forces cancel out?

You have a misunderstanding of the forces in Newton's second and third laws. The force in Newton's second law $\vec F = m \vec a$ is the net force acting on some object. The equal but opposite forces in Newton's third law act on two different objects.


Reference:
Rod Cross, "The bounce of a ball," Am. J. Phys. 67 (3), March 1999

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The weight of the ball cancels out the constraint force exerted by the Floor.

But the ball does have some momentum due to its velocity which makes the ball bounce again.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let us consider the idealized case of a point particle with mass $m$, and let the time of collision be $t=0$ for simplicity. We choose upward (downward) as a positive (negative) direction, respectively. Let the speed just before and after the elastic collision be

$$v_0~:=~\lim_{t\to 0^{+}} v(t)~=~ -\lim_{t\to 0^{-}} v(t)~>~0. $$

Then the time profiles of velocity, momentum, acceleration and force are

$$v(t) ~=~-gt +v_0 ~{\rm sgn}(t) , \qquad p(t)~=~m~v(t), $$ $$a(t) ~=~-g +2v_0 ~\delta(t) , \qquad F(t)~=~m~a(t)~=~F_g+F_n(t), $$

respectively. In other words, the gravity force is

$$F_g~=~-mg,$$

while the normal force from the floor is given by a Dirac delta function

$$F_n(t)~=~ 2mv_0 ~\delta(t).$$

Note in particular, that the two forces $F_g$ and $F_n$ at no instant $t$ cancel out.

| cite | improve this answer | |
$\endgroup$

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.