Changing force, linearly changing.

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The discussion revolves around calculating the velocity of an object subjected to a linearly changing force over time. Given initial conditions of velocity and mass, the user attempts to find the final velocity after two seconds but struggles with the correct application of formulas. Key concepts introduced include impulse, momentum, and the relationship between force, acceleration, and velocity. The conversation highlights the importance of understanding the area under the force vs. time graph to determine changes in momentum and velocity. Ultimately, the user successfully resolves the problem by calculating acceleration and finding the area under the graph, leading to the final velocity.
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I have a problem that's a bit more complicated than this, but the basic idea is this: a force is changing over time, it changes linearly. And I need to find the velocity. after multiple changes over time.
------------------------
given:
velocity, v0 = 3 m/s
mass, m = 3 kg.

at t = 0, F = 2N
at t = 2, F = 6N
(This change is linear throughout the given two seconds.)

what is the v at T = 2?
------------------------
I tried to average the change in force over the given time, using the formula:
Favg = delta F/t
Favg = 4/2 = 2N

Then, I took the formula: F = ma (mass x acceleration)
a = F/m
a = 2N/3kg
a = 2/3 m/s2

Then I took the formula: v = at (acceleration x time)
v = (2/3 m/s2)(2s)

But this doesn't give me the correct answer... hmmm...

Basically the problem doesn this multiple times, but I need to figure out how to do it once first!
 
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Beamsbox said:
I have a problem that's a bit more complicated than this, but the basic idea is this: a force is changing over time, it changes linearly. And I need to find the velocity. after multiple changes over time.
------------------------
given:
velocity, v0 = 3 m/s
mass, m = 3 kg.

at t = 0, F = 2N
at t = 2, F = 6N
(This change is linear throughout the given two seconds.)

what is the v at T = 2?
------------------------
I tried to average the change in force over the given time, using the formula:
Favg = delta F/t
Favg = 4/2 = 2N

Then, I took the formula: F = ma (mass x acceleration)
a = F/m
a = 2N/3kg
a = 2/3 m/s2

Then I took the formula: v = at (acceleration x time)
v = (2/3 m/s2)(2s)

But this doesn't give me the correct answer... hmmm...

Basically the problem doesn this multiple times, but I need to figure out how to do it once first!

What is Force multiplied by the time the force is applied called?

And then what does it (Force for a period of time) change in an object?
 
Force times time is impulse, even though we haven't gotten there yet in the book. I remember it from long ago.

What does the impulse change in an object?
It changes the momentum...

Not sure what you're getting at, where you're leading me. The force itself is changing over time...
 
Beamsbox said:
Force times time is impulse, even though we haven't gotten there yet in the book. I remember it from long ago.

What does the impulse change in an object?
It changes the momentum...

I still don't get it. This question is denoted as difficult, in a chapter that's thus far quite easy.

So in the chpt you are currently in it has nothing about impulse and change in momentum?

Does it have anything about F v. time graphs?
 
No, the chapter is "Force and Motion I", "Force and Motion II" comes next, lol. I haven't seen a force vs time graphat all, all it's talked about is vector, adding components, multiple bodies interacting on each other, people in elevators, etc. The only movement associated with force thus far is acceleration, whose formula was just given.

This is the last problem in the chapter, so I think they're just gearing us up for things to come, but this one seems like a great leap from the previous questions...
 
Beamsbox said:
No, the chapter is "Force and Motion I", "Force and Motion II" comes next, lol. I haven't seen a force vs time graphat all, all it's talked about is vector, adding components, multiple bodies interacting on each other, people in elevators, etc. The only movement associated with force thus far is acceleration, whose formula was just given.

This is the last problem in the chapter, so I think they're just gearing us up for things to come, but this one seems like a great leap from the previous questions...

well there is a longer way to do this...

Does it make sense that if the net Force on the mass is changing then the acceleration of the mass must be changing?
 
Simple little graph, but basically the slpe of the line changes many times in the program. So I assume it needs to be broken into groups, and then the velocities just added together...

Not sure how to find the velocity, though...[URL]http://s51.photobucket.com/albums/f362/BeamsBox/?action=view&current=MathGraph.jpg[/URL]
 
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Beamsbox said:
Simple little graph, but basically the slpe of the line changes many times in the program. So I assume it needs to be broken into groups, and then the velocities just added together...

Not sure how to find the velocity, though...[PLAIN]http://s51.photobucket.com/albums/f362/BeamsBox/?action=view&current=MathGraph.jpg[/QUOTE]

Can you put the graph up again? It does not show up.
 
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pgardn said:
well there is a longer way to do this...

Does it make sense that if the net Force on the mass is changing then the acceleration of the mass must be changing?

Yes, that very much makes sense.

I'm thinking perhaps a derivative problem?
 
  • #10
Beamsbox said:
Yes, that very much makes sense.

I'm thinking perhaps a derivative problem?

Actually its the opposite. If you have an equation that describes the acceleration of an object thru time, you can find the change in velocity over that time. But its the antiderivative or the integral... have you done these?
 
  • #11
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  • #12
Beamsbox said:
I'll try... sorry, I'm on a 28k dialup connection in Alaska...

[PLAIN]http://s51.photobucket.com/albums/f362/BeamsBox/?action=view&current=MathGraph.jpg[/QUOTE]

No problem. The graph still does not show up. So the force starts out at 2N at t = 0s and climbs to 6N at t =2s and the graph is perfectly linear (up diagonally)? Ok it is this, the picture showed up.

So can you write an equation for the acceleration with respect to time for this graph? It should be in a simple y = mx + b form...And have you done integration in the chapter?
 
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  • #13
pgardn said:
Actually its the opposite. If you have an equation that describes the acceleration of an object thru time, you can find the change in velocity over that time. But its the antiderivative or the integral... have you done these?

Yes, but it's been a while. About 6 years, actually.
 
  • #14
Beamsbox said:
Yes, but it's been a while. About 6 years, actually.

So integration is not in the chapter? Or a chapter before it?

Have you had anything about areas under curves?
 
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  • #15
pgardn said:
So integration is not in the chapter? Or a chapter before it?

Have you had anything about areas under curves?

We have had some info about area under curves. I have had calculus multiple times, but of course I need refreshers here and there.

We have talked about the relationships of the position, velocity, and acceleration and their relations regarding the curves and the areas underneath. We have found derivatives as well, but we have yet to cover anti-derivatives, or to do the process backwards... I wonder why they'd throw that in here with no explanation...
 
  • #16
Beamsbox said:
We have had some info about area under curves. I have had calculus multiple times, but of course I need refreshers here and there.

We have talked about the relationships of the position, velocity, and acceleration and their relations regarding the curves and the areas underneath. We have found derivatives as well, but we have yet to cover anti-derivatives, or to do the process backwards... I wonder why they'd throw that in here with no explanation...

Bottom line is that the area under the "curve" in the F v. time graph you posted is the change in momentum. Its that easy. You really don't have to integrate to find the area as its clean geometry. But you would need integrate if the force was not linear through time. And you have the initial momentum and you have the mass so just solve for Vf...

You could also make an acceleration versus time graph from your Force v. time graph since you know m. The area under this "curve", again its linear so its nice neat geometry, is the change in velocity. You know what the intitial velocity is, and the area is the change in velocity, so you can find the velocity at 2s...

Same thing. You don't have to make a new graph with impulse/change in momentum. But if you have not covered momentum, the paragraph above would be the way to go.
 
  • #17
Well, that seems to have done it for me!

I just calculated A, from force and mass, and plotted that with respect to time. Then found the area under the graph. I tend to make these seem so difficult.

Thanks once more for your superior help! Always appreciated!
 

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