I If force is done in an instant, how is it calculated?

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The discussion revolves around the calculation of force and acceleration when an object, like a 1kg block, is pushed and accelerates in an infinitesimal amount of time. Participants debate the implications of instantaneous changes in velocity, questioning whether this results in infinite acceleration and force. They clarify that while an object cannot realistically transition from 0 to a speed instantaneously, the concept of impulse is used to understand momentum changes. Additionally, the conversation touches on the work done when lifting an object, emphasizing that the force applied does not need to correlate with the speed of lifting, as work is defined by the integral of force over distance. Ultimately, the discussion highlights the complexities of motion and the mathematical principles that govern these physical phenomena.
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Hello. I may have worded my question incorrectly. Let's say there is a 1kg block on a table, and I push it. It starts accelerating in an infinitesimal amount of time. So how do we calculate the acceleration (and therefore the force) if the velocity goes from say 0 m/hr to 10 m/hr in an infinitely small period of time?

If an object goes from 0 m/hr to 1 m/hr in an instant, isn't that infinite acceleration?
 
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CuriousBanker said:
Hello. I may have worded my question incorrectly. Let's say there is a 1kg block on a table, and I push it. It starts accelerating in an infinitesimal amount of time. So how do we calculate the acceleration (and therefore the force) if the velocity goes from say 0 m/hr to 10 m/hr in an infinitely small period of time?

You calculate the "impulse", i.e. the rate of change of momentum of the object. This is the type of estimation we do when, say we calculate the force on a ball being hit by a bat, for example.

Zz.
 
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The problem still remains though because it still happens in an infinitesimal amount of time.
 
CuriousBanker said:
The problem still remains though because it still happens in an infinitesimal amount of time.

Define "infinitesimal". It can't be diminishingly small, because it means that as dt→0, the force→∞!

Zz.
 
CuriousBanker said:
If an object goes from 0 m/hr to 1 m/hr in an instant, isn't that infinite acceleration?

Yes it is and the corresponding force goes infinite too. But in reality objects doesn't go from 0 m/hr to 1 m/hr in an instant.
 
Wouldn't the measuring time affect it a lot then? If you measured it over 1 second it would be 1 N. But as you measure it over smaller and smaller time intervals, even though the full acceleration doesn't happen immediately, the closer you get towards the start the acceleration gets larger, right?

Second question- for work- let's say I lift a 100kg barbell off the ground for 2 meters. I know the work done is approximately 1960 joules, but why? I know that gravity exerts a force downward, but does it matter how fast we lifted the bar? If we lift the bar faster,it accelerates faster, wouldn't the force then be greater making the work done greater? Why is the force exactly equal to the opposite of gravity?
 
CuriousBanker said:
Wouldn't the measuring time affect it a lot then? If you measured it over 1 second it would be 1 N. But as you measure it over smaller and smaller time intervals, even though the full acceleration doesn't happen immediately, the closer you get towards the start the acceleration gets larger, right?

I have no idea what this means.

You should not start a "second question" before settling the first one.

Zz.
 
CuriousBanker said:
Hello. I may have worded my question incorrectly. Let's say there is a 1kg block on a table, and I push it. It starts accelerating in an infinitesimal amount of time. So how do we calculate the acceleration (and therefore the force) if the velocity goes from say 0 m/hr to 10 m/hr in an infinitely small period of time?
Acceleration is change in speed over elapsed time.
If an object goes from 0 m/hr to 1 m/hr in an instant, isn't that infinite acceleration?
Yes, but "instant" and "infinitessimal" are not the same thing. I suspect you may be misunderstanding something about your own scenario; how much force or time? Is this supposed to match any realistic scenario?
 
  • #10
ZapperZ said:
I have no idea what this means.

You should not start a "second question" before settling the first one.

Zz.

I start at 0 m/s. After 1 second you see I am at 5 m/s. This did not happen "in an instant", so it happened over time. After half of a second maybe I was only at 3 m/s. Maybe after 1/5 of a second I was only at 2 m/s. But the instant I went from 0 m/s to SOMETHING, that happened in an infinitesimal period of time. So depending on when you took your sample points (at infinitesimal period of time, or after 1 second) you would have me accelerating infinitely, or you would have me accelerating at 5 m/s/s. Where did I go wrong?
 
  • #11
Sorry about the deleted post...we may come back to it though because it almost seems like you are confusing speed and acceleration with each other...
 
  • #12
russ_watters said:
Sorry about the deleted post...we may come back to it though because it almost seems like you are confusing speed and acceleration with each other...

I'm not confusing speed and acceleration. In the deleted post you said force doesn't have to do with speed or time. But f = ma, and the A in that equation has to do with speed and time.
 
  • #13
CuriousBanker said:
I start at 0 m/s. After 1 second you see I am at 5 m/s. This did not happen "in an instant", so it happened over time. After half of a second maybe I was only at 3 m/s. Maybe after 1/5 of a second I was only at 2 m/s. But the instant I went from 0 m/s to SOMETHING, that happened in an infinitesimal period of time. So depending on when you took your sample points (at infinitesimal period of time, or after 1 second) you would have me accelerating infinitely, or you would have me accelerating at 5 m/s/s. Where did I go wrong?
Ahh, it's an zero/infinitesimal problem. Now matter how small you make the time interval, the equation holds. Don't confuse yourself by thinking one goes to zero, but the other doesn't!

[Edit] btw, you should probably do the math instead of guessing at the numbers.
 
  • #14
CuriousBanker said:
I'm not confusing speed and acceleration. In the deleted post you said force doesn't have to do with speed or time. But f = ma, and the A in that equation has to do with speed and time.
Change in speed and change in time. Not the same as speed and time...but again, I'm not really sure that's your main issue.
 
  • #15
russ_watters said:
Ahh, it's an zero/infinitesimal problem. Now matter how small you make the time interval, the equation holds. Don't confuse yourself by thinking one goes to zero, but the other doesn't!

Ah, I see what you are saying, that makes sense; they both go to zero. Thanks!

What about my question about work? let's say I lift a 100kg barbell off the ground for 2 meters. I know the work done is approximately 1960 joules, but why? I know that gravity exerts a force downward, but does it matter how fast we lifted the bar? If we lift the bar faster,it accelerates faster, wouldn't the force then be greater making the work done greater? Why is the force exactly equal to the opposite of gravity?
 
  • #16
CuriousBanker said:
I start at 0 m/s. After 1 second you see I am at 5 m/s. This did not happen "in an instant", so it happened over time. After half of a second maybe I was only at 3 m/s. Maybe after 1/5 of a second I was only at 2 m/s. But the instant I went from 0 m/s to SOMETHING, that happened in an infinitesimal period of time. So depending on when you took your sample points (at infinitesimal period of time, or after 1 second) you would have me accelerating infinitely, or you would have me accelerating at 5 m/s/s. Where did I go wrong?

In addition to what Russ has stated, try this:

s(t) = kt2, where k is a constant for a particle of some mass m.

What is the acceleration at t=0, or at ANY instant during its motion? You will be using calculus to do this which means that you'll implicitly will be evaluating the value at an infinitesimal period of time at a given instant. Do you think this give an infinite force?

Zz.
 
  • #17
ZapperZ said:
In addition to what Russ has stated, try this:

s(t) = kt2, where k is a constant for a particle of some mass m.

What is the acceleration at t=0, or at ANY instant during its motion? You will be using calculus to do this which means that you'll implicitly will be evaluating the value at an infinitesimal period of time at a given instant. Do you think this give an infinite force?

Zz.

Well it would be 2kx, right? But that's assuming uniform acceleration, no?
 
  • #18
ZapperZ said:
You calculate the "impulse", i.e. the rate of change of momentum of the object.
Likely you mean to write "the total change in momentum over the infinitesimal interval"

Often, we do not know or care how much force was involved. We do not know or care how short a time it took. We do not know or care how short a distance was acted over. Often all we care about is how much momentum (or energy) was transferred.

So, for instance, we fire a bullet into a ballistic pendulum and measure how far it deflects as a result, thus obtaining an indirect measurement of impulse.
 
  • #19
CuriousBanker said:
Well it would be 2kx, right? But that's assuming uniform acceleration, no?

Why would uniform or non-uniform acceleration matters? If you wish, do kt5, if you like.

Zz.
 
  • #20
ZapperZ said:
Why would uniform or non-uniform acceleration matters? If you wish, do kt5, if you like.

Zz.

Does motion tend to obey such neat equations? Serious question. I see what you are saying that it would be 5kt^4 at any moment. Ok, makes sense to me.

What about my work question?
 
  • #21
CuriousBanker said:
What about my question about work? let's say I lift a 100kg barbell off the ground for 2 meters. I know the work done is approximately 1960 joules, but why? I know that gravity exerts a force downward, but does it matter how fast we lifted the bar? If we lift the bar faster,it accelerates faster, wouldn't the force then be greater making the work done greater? Why is the force exactly equal to the opposite of gravity?

Check out this thread, "Work done lifting a box":
https://www.physicsforums.com/threads/work-done-lifting-a-box.750539/
 
  • #22
Ah, now I get it. Thank you all.
 
  • #23
CuriousBanker said:
Does motion tend to obey such neat equations? Serious question.
For any motion at any point in time you can do a Taylor series expansion and approximate the motion with an arbitrarily accurate polynomial.
 
  • #24
Yes, in this case, you have modeled an infinite acceleration, which means that for a finite mass, there must be an infinite force - which is impossible. Now what is close to this is the idea of a very high, but finite, acceleration for a very low, but non-zero, amount of time. As this force is typically unsteady, this model is analyzed as a collision problem, in which the object is accelerated because of a transfer of momentum from another mass; this change in momentum is termed an impulse, and is the integral of force over some time period.
 
  • #25
CuriousBanker said:
What about my question about work? let's say I lift a 100kg barbell off the ground for 2 meters. I know the work done is approximately 1960 joules, but why? I know that gravity exerts a force downward, but does it matter how fast we lifted the bar? If we lift the bar faster,it accelerates faster, wouldn't the force then be greater making the work done greater? Why is the force exactly equal to the opposite of gravity?
It's not. You can accelerate at different rates as you move the weight. You can even overshoot and then drop back down, or go sideways, or whatever. But work W = ∫F⋅ds so any F(t) you instantaneously apply is accompanied by a ds and when you add them all up you get W = mgh where m = 100kg, g = 9.81 m/s, h = 2m. (vectors in bold type).

Average force over the vertical distance h = mg.

 
  • #26
Sorry, I am a latecomer to this but I would like to go back to OP's initial question. I always thought that pushing a block can be modeled as pushing a (very stiff) collection of springs. There is no spooky action at a distance. When you push one end of a block, the layer of of atoms at that end pushes the next layer which pushes the next layer and so on for about 1024 times. This creates a wave within the solid which travels at the speed of sound or at the phonon group velocity if you want to be more precise. So, roughly speaking, the information that one end has been pushed travels at the speed of sound. The speed of sound in wood is about 4000 m/s. If you suddenly push a 40 cm block at one end, the other end will start accelerating with a delay Δt = 0.4 m/4000 s = 100 μs.
 
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  • #27
CuriousBanker said:
The problem still remains though because it still happens in an infinitesimal amount of time.
Calculus takes care of this because, when used properly, an 'infinitesimal' is not zero. Calculus always considers the limiting value of a function as the infinitesimal approaches zero. Calculus works for 'well behaved functions' that are continuous and differentiable. So we believe that motion (for example) is continuous and the sums all work however small you choose your time interval. This happens for any 'law' you choose to apply to it.
 
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