If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of $g = 9.81\,{\rm m/s^2}$. (I don't know what terminal velocity is that well as I'm only in physics 1 in high school, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)
Also, if a ball is dropped high enough to reach terminal velocity, then it accelerates at $0\,{\rm m/s^2}$, so it has a force of ZERO when it hits the ground?
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9 Answers
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I think the other two answers may have overlooked the source of your confusion, which is quite simple.
The $F$ in $F=ma$ is the force being exerted on the object of mass $m$ to give it the acceleration $a$, not the force that that object will exert when it hits something.
In the case of your example, the force of gravity on the basketball is independent of the height. The force that the basketball exerts on the ground is an entirely different matter.
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$\begingroup$ +1 Great explanation. And to further clear out a typical misconception: A force is not something an object "has" or "carries around" until impact. It is something acting in the moment. $\endgroup$– SteevenCommented Apr 9, 2017 at 20:00
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While the ball is in free fall, yes, the net force it exerts on the ground is roughly independent of its height above the ground, and is independent of its speed.
However, you are interested in the force between the basketball and the ground while the ball is colliding with the ground, and that is entirely a different matter.
When the ball is colliding with the ground, its speed drops from a very high speed going down to a speed of zero, or to a speed going up if it bounces. That means that the acceleration of the ball during the collision, points up. While the ball is falling, its acceleration is down, and that acceleration changes directions during the impact.
Further, since the duration of the impact is very short, the acceleration is extremely high. If you drop the ball from a high height, it will be going very fast when it strikes the ground, and this also means the acceleration is very great. Thus, a ball dropped from higher exerts the same force on the Earth while falling, but a higher force during its impact because its acceleration is higher during impact.
If a ball falls at terminal velocity, there are two significant forces on it: gravity and air drag. The force from gravity is unchanged. Gravity still pulls the ball down and the ball still exerts and equal and opposite upward force on the Earth. However, the force from air drag pushes the ball up, and the equal and opposite force from the ball on the air is down. Overall, there is zero net force on the ball. If we consider the atmosphere and Earth excluding the ball as one system, there is zero net force from the ball on that system as well.
When a ball moving at terminal velocity hits the ground, it will suddenly pick up a huge acceleration that it didn't have a moment before. This means there will be a big net force on it that wasn't there a moment before. The fact that the net force was zero before the impact does not mean the impact itself occurs with zero forces at play.
answered Dec 17, 2011 at 18:44
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The problem is that $9.81~\text{m/s^2}$ is the acceleration during the fall, not the impact.
Let's suppose the impact lasts $0.1~\text{s}$ when a ball hits the ground at $10~\text{m/s}$, assuming inelastic collision (meaning, the ball doesn't bounce back), average acceleration DURING impact will be $\frac{10}{0.1} = 100~\text{m/s^2}$ rather than $9.81~\text{m/s^2}$ and that's the acceleration you should be using for the formula.
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I believe you are confusing force with energy -- kinetic energy, to be more precise. Remember $$E {\scriptsize _K} = \textstyle \frac{1}{2} m v^2$$
The ball dropped from the higher location falls a greater distance with nothing opposing the force of gravity. Therefore, the ball accelerates to a higher velocity than the ball dropped from 1 ft.
Using the above formula, the velocity squared, times the mass of the ball divided by 2 gives the Kinetic Energy, which will be useful in determining the force with which each impacts the ground.
See http://www.physicsclassroom.com/calcpad/energy for a more detailed explanation.
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$\begingroup$ "Let's play ball" - I think that is the wisest answer to no. 1 of the question. Go for no. 2 which I do not manage to understand, that part of the question. $\endgroup$ Commented Oct 28, 2022 at 19:25
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You're not stupid. You've shown you have intellectual curiosity about extremely complicated topics you've only just been introduced to.
"1.If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of g=9.81 m/s^2. (I don't know what terminal velocity is that well as im only in physics 1 in highschool, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)"
No. Both balls have built up something called kinetic energy and momentum. Each of these two concepts depends upon how long the force has acted upon them. Intuitively, you will realize that since gravity worked longer on the "tower ball" than it did on the "lower ball", it will take more work or "punch" to stop the "tower ball".
"2.Also, if a ball IS dropped high enough to reach terminal velocity, then it accelerates at 0 m/s^2, so it has a force of ZERO when it hits the ground?"
No. Again, a significant force was applied over a period of time and, even though acceleration stopped because of wind resistance, significant momentum and energy had been built up before that happened. A strong "punch" is still required to stop it.
Continue to work very hard on your math and physics and chemistry courses. You'll do well only if you work very hard.
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As Jack M has answered, if an object of mass $m$ is having an acceleration of $a$ this means that the net force $F$ acting on the object is $ma$. In this question the basketball of mass $m$ is having a downward acceleration of $g$ due to earth's gravity. The force acting on the basketball is $mg$. We have assumed here that everything is happening on the surface of the earth where we can safely take $g$ to be constant or independent of the height on the surface of the earth.
Neglecting any other forces, drags, the only force acting on the basketball is $mg$. But from Newton's third law this is also the force that the basketball exerts on the earth, $GMm/R^2=mg=Ma_E$, no matter from what height it is falling. Acceleration of the earth $a_E$ as you can see is very very small as its mass $M$ is very very big. $Ma_E$ is the force exerted on the earth by the basketball which is equal to $mg$. The interesting thing is that when the basketball falls from a height it gives momentum to the earth when it hits the ground because the basketball gains some velocity when it falls from a height. But this does not not imply that the force exerted on the earth by the basketball should be more when it falls from a greater height. The force from Newton's second law is also given by $\text{d}p/\text{d}t$, the rate of change of momentum.
Considering elastic collision, when the basketball hits the ground and rebounds the rate of change of momentum of the basketball is $\text{d}p/\text{d}t=\text{d}(mv)/\text{d}t=m\text{d}(gt)/\text{d}t=mg$ which is also equal to the rate of change of momentum of the earth.
If the basketball is falling with terminal velocity, the net force on the basketball is zero. The air drag exactly counter-balances $mg$. And again this does not imply that the force on the earth by the basketball when it hits the ground is zero because the basketball gives momentum to the earth.
answered Jan 21, 2015 at 6:12
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This is the case of jerk in physics , rightly pointed above, so when the ball hits ground it change in acceleration is 2a in very less time. This is the reason why you get more injured when you fall from high tower.
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You are misunderstanding the equation $F=ma$. It states that if $F$ is the total external force acting on a body of mass $m$, then the mass $m$ accelerates with acceleration $a$. It does not say, as you are interpreting that a body accelerating with acceleration $a$ exerts a force $F$.
That being said, what is the answer to your question - What is the force exerted on the Earth by a bouncing ball? We can answer this as follows. We go back to a more basic definition of force, namely the change the rate of change of momentum
$$ F = \frac{ \Delta p }{ \Delta t}~. $$
Now, consider a ball that is falling down. Suppose that its speed is $-v$ (pointing downwards) just before it hits the Earth. The initial momentum of the ball is $-mv$ (pointing down). We can safely assume that the Earth is much much more massive than the ball. If we further assume that the Earth-Ball collision is elastic, then the ball bounces back with the same speed as before. Therefore, the final momentum of the ball is also $mv$ (pointing up). The total change in momentum during this collision is then $\Delta p = mv-(-mv) = 2mv$.
Now, suppose that the collision occurs in time $t$. If the ball is perfectly rigid then the collision happens in an instant and $t=0$. More realistic collisions always occur over a small time period $t$ is small and non-zero.
Now, since the momentum of the ball is changing by $2mv$ in time $t$, there must be a force that acts on it. This is the force that the Earth exerts on the ball during the collision and is $$ F = \frac{2mv}{t}~. $$
Now, by Newton's 3rd law, since the Earth exerts a force on the ball, the ball must exert an equal and opposite force on the Earth. Thus, the force on the Earth is
$$ | F_{\text{earth}} |= \frac{2mv}{t}~. $$
Note that the force on the Earth has nothing to do with the acceleration of the ball due to gravity, $g$. If you drop a ball from a height $h$, then when it reaches the ground its speed is $v = \sqrt{2gh}$ so that $$ | F_{\text{earth}} |= \frac{2m}{t} \sqrt{2gh}~. $$
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Force does equal mass times acceleration. However $9.8~\text{m/s^2}$ is the acceleration of the ball imposed by gravity. The acceleration that the ball experiences upon impact with the ground is instead proportional to its current velocity.
Upon impact, $a=-\frac{v}{t}$, where v is current velocity and t is the time impact lasts.
If the ball were travelling $100~\text{m/s}$ and impact lasts 2 seconds, the acceleration upon impact would be $-50~\text{m/s^2}$.
The time impact lasts is related to the properties of all the materials in question. Striking a solid concrete surface, a much shorter impact, striking a pool of water, longer. And the related deceleration is not necessarily distributed evenly over time, it can be much higher in the beginning of impact and lesser near the end, etc, but that's beyond the scope of the question.
Finally, even if the ball has stopped accelerating just before it hits the ground (reaches terminal velocity), sudden impact with the ground will have the ball rapidly decelerate, which is the definition of Kinetic Force.
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3$\begingroup$ This is wrong / a misunderstanding: "If the ball were travelling 100 m/s, then the acceleration upon impact would be -100 m/s/s". The force can be arbitrary and depends of the amount of time taken for deceleration. $\endgroup$ Commented Jan 21, 2015 at 2:39
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$\begingroup$ Corrected to remove all inaccuracies, I hope. $\endgroup$– iPherianCommented Jan 22, 2015 at 9:50