Unbalanced forces acting on a ball at maximum altitude

[Martyn Arthur]

TL;DR Summary

I am studying Questions In Science at the Open University. Is a ball projected vertically upwards stationer at maximum altitude, does the concept of unbalanced gravity contradict Newton? (I’m not bothered about the mark I got, just the question).

When a ball projected vertically upright reaches its maximum height is it instantaneously stationery (calculus provides for instantaneous events) before it starts its downwards journey?

is gravity acting on the object at that instant?

Newton provides that an object at rest must be subject to balanced forces.

Is gravity is acting does it thus follow that an equal opposite force.

The OU perspective from my tutor, in its marking notes on this says the following (I’m not bothered about the mark I got, just the question).

'
Hi Martyn,
Thanks for your mail. As a non-physicist this answer always makes me ponder as well! Here's the full version from my marking notes:

"At the top of the bounce (maximum height) the only force acting on the ping pong ball is its weight (gravity). At the top of its trajectory the ping pong ball is still accelerating under gravity at 9.8 m s−2. An unbalanced force is acting on the ping pong ball and changing its motion - there is a change in the velocity of the ping pong ball."

The gist of it as I understand it is that you can't ever remove the effects of gravity so even at the top of the bounce the velocity is changing from positive through zero to negative but always in the process of changing so there is an acceleration and hence a force active. The table top isn't moving so no change in velocity.
Hope this makes sense!
Cheers,
'

 

[Dale]

Newton provides that an object at rest must be subject to balanced forces.

This is not correct. Can you explain your reasoning for this?

 

[Martyn Arthur]

Thank you, I stand well and truly corrected, apologies 'Newton 1 A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.'
Is it then the case that the ball in question, at the instantaneous moment when it is stationery it is in a stat of rest, and has no forces acting on it. If so what happens to force of gravity?

 

[anorlunda]

A ball at the highest point of a throw is not in "uniform motion" it is being accelerated. Newton's 2nd law says f=ma. It seems that you are visualizing f=mv, zero forces zero velocity. That is not correct.

The force of gravity on Earth is accelerating the ball towards the center of the Earth. If the ball starts with positive velocity (up) and accelerates toward negative velocity (down) it must pass through zero velocity (the highest point of the throw).

Uniform motion means constant velocity.

 

[PeroK]

Thank you, I stand well and truly corrected, apologies 'Newton 1 A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.'
Is it then the case that the ball in question, at the instantaneous moment when it is stationery it is in a stat of rest, and has no forces acting on it. If so what happens to force of gravity?

If there were no gravity at the maximum height, then the ball would remain (at rest) there: it would hover stationary at the maximum height. That the ball does not remain at rest shows that Newton's first law does not apply and the ball is subject to the downward force of gravity.

An object being instantaneously at rest is not the same as remaining at rest (over time).

 

[Dale]

Newton 1 A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

This says that a body which is acted on by no force and which is already at rest will remain at rest. It does not in any way imply that a body at rest is not acted on by any force.

Is it then the case that the ball in question, at the instantaneous moment when it is stationery it is in a stat of rest, and has no forces acting on it.

No. It is not the case. When it is at rest it still has forces acting on it. These forces cause it to not remain at rest.

If the force of gravity were to disappear for an object at rest then you would throw a ball up, and when it reaches the peak it would stay there. That is not what happens. It falls back down because gravity continues to act on it.

 

[Martyn Arthur]

I take the point that the ball would hover thank you. I am not being awkward, however what then is the diffference between being "at rest over time", and "at rest instantaneously", and hovering instantaneously? Does at rest instantaneouslyhave a prescribed time limit? surely that's what calculus and instantaneous is about, but here we don't approach zero, we obtain it?

 

[PeroK]

I take the point that the ball would hover thank you. I am not being awkward, however what then is the diffference between being "at rest over time", and "at rest instantaneously", and hovering instantaneously? Does at rest instantaneouslyhave a prescribed time limit? surely that's what calculus and instantaneous is about, but here we don't approach zero, we obtain it?

Instantaneously at rest means at rest for zero time. The ball in this case will move during any finite time interval.

In one sense, you can never measure the accelerating ball to be at rest "instantaneously". It is implied by the mathematics (calculus) that there is a single time (not time interval) where the ball is at rest.

In that respect, being at rest is no different from all other points on the ball's trajectory, where it has some velocity $v$, say, for an instant. You could equally claim that "for an instant it has constant velocity $v$ and is, therefore, not accelerating".

Acceleration means a change in velocity over time. By definition of an instant, it cannot change its velocity over an instant.

 

[Martyn Arthur]

Thats clear than you!
Martyn

 

[Dale]

what then is the diffference between being "at rest over time", and "at rest instantaneously",

”At rest” means the velocity is zero: $v=0$. ”At rest over time” means the derivative of velocity is zero: $v=0$ and $\dot v=0$. “At rest instantaneously” means the derivative of velocity is non-zero: $v=0$ and $\dot v \ne 0$

 

[PeroK]

It might be easier to look at an example velocity against time graph:

The velocity is zero instantaneously at times $t = 0, t = 30, t = 55$ minutes. But, at all these times the object is accelerating. The point at $t = 30$ minutes is similar to the point of maximum height for a ball. The straight line indicates that the object has acceleration in the negative direction from $t = 15$ mins until $t = 40$ mins. There is a change of direction at $t = 30$ minutes.

The object has no acceleration from $t =10$ minutes until $t =15$ minutes. The velocity during this time is constant. Therefore, there is no net force on the object for $5$ minutes. Note that the velocity is not constant at any other time.

Note also that the points where $v = 0$ are not fundamentally different from points where $v \ne 0$ in terms of whether or not the object is accelerating.

The slope of the graph indicates acceleration.

 

[Martyn Arthur]

Thank you for your time, it is appreciated. Bear in mind please that I am just a student, but witha burning desire to understand! Now I have become confused again.The point of differentiation is to find the rate of change of a quantity, physically, here velocity. Since zero never changes, it is clearly a constant. So its derivative must be zero, no velocity.

In relation to the graph (leaving aside the 10 -15 minutes gap which is an aside) it relates to constant motion with a change of direction, like the orbit of the Earth around he sun, but here a short curve around a point. If you drew a stright line vertically down at time = 20 (min) velocity (m/min) zero you would set that, in any other context, the movement of a car say, as velocity zero.

This is not a curve, the ball rises vercally, stops, and then moves again, an entirely differEnt thing. If I understand correctly all explanations dictate the same thing, there is a point in time when the ball is not in motion, as velocity dictates a need for motion there is no velocity.
So we have a ball, with velocity zero (differentiated or not) at a point in time, with gravity present, but not applying a force that has effect, at that point in time.

Looking at it from a space time concept (I am just a student seeking to learn and not making assertions) wthere appears to be an instant when mass is not affecting space curvature?
Martyn

 

[PeroK]

In relation to the graph (leaving aside the 10 -15 minutes gap which is an aside) it relates to constant motion with a change of direction

The graph is velocity against time. It's not position against time. The position against time graph would be curved. An exercise for you would be to sketch the displacement/position against time graph.

The graph shows period of constant acceleration. Not constant velocity. In particular the middle section is essentially the same as a ball thrown up and falling back down under gravity.

 

[Dale]

not applying a force that has effect, at that point in time.

The force does have an effect at each point in time. The effect of the force is that $\dot v \ne 0$ at each point in time.

Since zero never changes, it is clearly a constant. So its derivative must be zero, no velocity.

I am not exactly sure what you are trying to say here, but it sounds wrong. It sounds like you are trying to say that $v=0$ implies $\dot v=0$, which is not correct.

 

[Martyn Arthur]

Thank you all very much for your time, it really is appreciated and I am not trying to 'say' anything, just understand!

Mathematics and science are what I am here (at age 70) to learn, but I need also to understand things in the context of the percived world.

The bottom line here, if I may please say, is that the ball is not following a curve of continuous moton (accelearation), it moves (accelerates) to a point, stops moving (accelerating) , and then reverses (starts accelerating) its direction.

When it stops moving (accelerating) it is stationary, devoid of velocity, it is not accelerating and changing direction.

Newton says that an object in that situation is not subject to an unbalanced force.
The contention here is that the unbalanced force of gravity continues to apply, but it has no application and does not cause any form of acceleration.

As I have said before, I am not being awkard, I just want to understand!
If there is a pragmatic answer, not enshrined in mathematical technicalities, then I would relly like please to be able to understand it.
Martyn

 

[berkeman]

It sounds like you still are a bit confused about this. Consider the following simple graphs showing the position, velocity and acceleration versus time for a ball thrown straight up. You can see that the acceleration is constant (down), the vertical velocity is decreasing at a constant rate, and the position goes up and then down. Does the graph help?

https://slidetodoc.com/presentation_image/3f212aeff67173ecc638a0fb80dc8078/image-9.jpg

 

[Martyn Arthur]

People, thank you for all your time and replies, it all is really very much appreciated. I have learned a lot about the mathematical issues here in a very short time!
From many years as an accountant I have seen how figures etc can be arranged to show many different facets.

We can't disregard the maths but the question is a pragmatic one.
The ball is stationary, when it is stationary, the force of gravity (the potential of mass distorting space time) exists.

The ball maintains position, not accelerating and expeiencing velocity at that time, unmoved by any unbalanced forces?

Martyn

 

[berkeman]

The ball is stationary, when it is stationary, the force of gravity (the potential of mass distorting space time) exists.

The ball maintains position, not accelerating and expeiencing velocity at that time, unmoved by any unbalanced forces?

Even when the velocity of the ball crosses the horizontal time axis (where its velocity is instantaneously zero), it is accelerating downward. That is the force of gravity on the ball causing that constant downward acceleration. Please keep those 3 graphs in your mind when thinking about this problem. They can help you avoid confusion, IMO.

 

[berkeman]

It's probably already been mentioned in this thread, but as you can see from the graphs, the velocity is the change in position versus time, and the acceleration is the change in velocity versus time.

 

[PeroK]

People, thank you for all your time and replies, it all is really very much appreciated. I have learned a lot about the mathematical issues here in a very short time!
From many years as an accountant I have seen how figures etc can be arranged to show many different facets.

We can't disregard the maths but the question is a pragmatic one.
The ball is stationary, when it is stationary, the force of gravity (the potential of mass distorting space time) exists.

The ball maintains position, not accelerating and expeiencing velocity at that time, unmoved by any unbalanced forces?

Martyn

If you want to learn physics you must focus on the basics as set out in post #16 above. This confusion of ideas you've posted here will get you nowhere.

 

[Dale]

When it stops moving (accelerating) it is stationary, devoid of velocity, it is not accelerating and changing direction.

This is wrong. Devoid of velocity ($v=0$) in no way whatsoever implies that it is not accelerating ($\dot v=0$)

 

[Martyn Arthur]

I'm sorry if I am being obtuse, boring but I am not confused about the question.
The question is a pragmatic one and does not require graphs etc to addres it.

I seek a prgamatic answer,aside from the mathematical dictate of theory.
The explantions so far fail completely to addressthe fundamentals of physics that I am learning in my first year.

My OU tutor, a Dr but not a physicist has expressed the same questions that I raise.
The ball is stationery at the end of its vertical movement, by the physics I have learned so far it has no velocity, and no acceleration.

To use a word expresed previously it hovers.
If the ball is experiencing any form of force it is not respondint to, or acting on it, it hovers.

After this year on my science learning curve I want really to understand the maths of calculus et al., but I do not want to have to sacrifice the pragmatic

 

[Dale]

The ball is stationery at the end of its vertical movement, by the physics I have learned so far it has no velocity, and no acceleration.

As has been told to you many times already, this is simply false. It has no velocity but it does have acceleration. You have no justification for your claim that it has no acceleration, and it is wrong as has already been pointed out many times.

I seek a prgamatic answer,aside from the mathematical dictate of theory.
The explantions so far fail completely to addressthe fundamentals of physics that I am learning in my first year.

Sorry, you don't get to reject math here. That is not your prerogative to tell physicists that physics should be done without math.

Thread closed.