# Changing force, linearly changing.

1. Jun 13, 2010

### Beamsbox

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.
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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?
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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!

2. Jun 13, 2010

### pgardn

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?

3. Jun 13, 2010

### Beamsbox

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...

4. Jun 13, 2010

### pgardn

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?

5. Jun 13, 2010

### Beamsbox

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 eachother, 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...

6. Jun 13, 2010

### pgardn

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?

7. Jun 13, 2010

### Beamsbox

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]

Last edited by a moderator: Apr 25, 2017
8. Jun 13, 2010

### pgardn

Last edited by a moderator: May 4, 2017
9. Jun 13, 2010

### Beamsbox

Yes, that very much makes sense.

I'm thinking perhaps a derivative problem?

10. Jun 13, 2010

### pgardn

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. Jun 13, 2010

### Beamsbox

Last edited by a moderator: Apr 25, 2017
12. Jun 13, 2010

### pgardn

Last edited by a moderator: May 4, 2017
13. Jun 13, 2010

### Beamsbox

Yes, but it's been a while. About 6 years, actually.

14. Jun 13, 2010

### pgardn

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

Last edited: Jun 13, 2010
15. Jun 14, 2010

### Beamsbox

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. Jun 14, 2010

### pgardn

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 dont 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 dont 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. Jun 14, 2010

### Beamsbox

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!