Force-applied when Δp=0

  • #1
I am having trouble working this out

If a 1kg object under constant acceleration, g, starts out at rest and over the course of .5 seconds traverses 1m winding up at rest again, can you deduce exactly force F that acted on it? Or only the impulse?

Δp = 0
Δke = 0
Δx = 1
Δt = .5
g = 9.8

F - the unknown force
t1 - time F was applied
x1 - distance F was applied

gt = Ft1
gx = Fx1

That gives me x/t = x1/t1
and x1 = .5((F-g)/m)t12

I am having trouble working it out, but it seems at first glance that since x1= .5((F-g)/m)t12 is not linear there would be only 1 set of appropriate F, x1 and t1 that get me exactly 1m in .5s and back at rest. Any ideas?

Answers and Replies

  • #2
Nope, from newton's second law we know that
[tex] F = \frac{\Delta p}{\Delta t} [/tex]
So since there was no change in momentum the NET force was zero, now since you know at least two forces acted on this object (gravity and we'll assume just one other) and you know the work that was done by gravity you know the net work done by the other force, but you can't find the magnitude of the force without more information.
  • #3
there would be only 1 set of appropriate F, x1 and t1 that get me exactly 1m in .5s and back at rest. Any ideas?

Why you think there's only one set?I think there're infinite sets where the average force varies from 9.81N to infinite.
  • #4
That is what I would have originally thought and still seems more likely, but

I know A) the impulse of gravity must equal to impulse of the applied force, I also know B) The work done by gravity must match the work done by the applied force

To be clear, the NET impulse is 0, however, during the interval [t=0,t=t[SUB]1[/SUB]] there is a change of momentum, lets call it Δp1 that is equal and opposite to the change in momentum over the interval [t=t[SUB]1[/SUB],t=.5] for a total Δp=0

We are assuming two constant forces

The acceleration caused by the net force(applied force - gravity) for a time of t1 yields a specific displacement, x1, the acceleration of gravity slowing it to 0 then gives another specific displacement that must add to x

So it seems I should be able to relate those three facts to get one Force, t1 and x1
  • #5
Alright, so, I was doing some work on it and it seems that if I choose some arbitrary force between 9.81 and ∞, F = 49N And I plug in into both

gt = Ft1 = 9.8(.5) = 49t1
t1 = .1s

gx = Fx1 = 9.8(1) = 49x1
x1 = .2

But, for a 39.2(49-9.8)N Force, giving and acceleration of 39.2m/s2
x1 ≠ 1/2at12 It is close, but off
.2 ≠ .5(39.2).12
.2 ≠ .196

another, F = 10N

gt = Ft1 = 4.9 = 10t1, t1 = .49s
gx = Fx1 = 9.8 = 10x1, x1 = .98m

.98 ≠ 1/2(.2)(.492)
.98 ≠ .02401
  • #6
I was over at the math sub-forum ( ) and while it is yet to be verified for me there could maybe be only one Force-time-distance possibility for two constant forces acting in opposite directions over a specified overall time and distance with a net impulse of 0. I think it is only if we do not care about the overall distance traveled that an infinite set of pairs of force and time satisfy the conditions. The same is then true of requiring the work done be the same, an infinite set of forces and distances only applies if we do not care about the overall time. It sort of makes sense...but is a little weird.

I got the equation
F = g2t2/(gt2-2x)

So for the data listed above
g = 9.8, t = .5, x = 1

I get F = ((9.8)2(.5)2)/((9.8)(.5)2-2(1))
F = ((96.04)(.25))/((9.8)(.25)-2(1))
F = 24.01/(2.45-2)
F = 24.01/.45 = 53.35556

Which put into gt = Ft1 gives
9.8(.5) = 53.35556t1
t1 = .09184

and gx = Fx1
9.8(1) = 53.35556x1
x1 = .18367

Which satisfies x/t = x1/t1 to within 99.99% due to rounding

and x1 = .5(F-g)t1^2 to 99.98%, as compared to the 98% and 2.45% of the arbitrarily picked Force used above.

But does that mean that if I constantly accelerate something against a constant force for a given time, it's average velocity at that time is the average velocity when it comes to rest.
  • #7
Umm, bump, could anyone please take a look at this and tell me if and where I am messing up?
  • #8
Doc Al
Not quite clear what you're trying to do here. You want to find a constant force that, when added to gravity, moves something a distance with zero speed at the start and finish? Let me know if my understanding is correct.
  • #9
Yes, what brought it up was a question about weight-lifting, I understand that in weight-lifting it is most likely not a constant force. A situation that actually fits this description would be if I had an object on a table, tied a weight that hung off the table and let it drop, gravity applies a constant force until the weight hits the ground, if the object were to continue sliding and stop(from the constant force of friction), if I knew the force of friction, could I find the force of gravity? i.e. if the weight were of unknown mass, and dropped from an unknown height, and all I had was Force of friction, total time elapsed and total displacement

*and mass of the object on table
  • #10
Doc Al
A situation that actually fits this description would be if I had an object on a table, tied a weight that hung off the table and let it drop
If the hanging weight is enough to overcome friction and get the object moving, why would that change? The object would simply accelerate. (And when you 'let it drop', realize that you are changing the force.)

There's no way to meet your criteria with a constant force. The net impulse is zero, since the speed doesn't change. And the only way to get a zero impulse with a constant force would be if the force were zero.
  • #11
I apologize, that description was fairly jumbled as well.

I am talking about a kinetic force of friction on the object that is smaller than the force of gravity on the weight. I am considering the static friction to be negligible. So that what is happening is the object is accelerating under the constant force of gravity minus the constant force of friction, and then the weight stops at the ground, and the object decelerates under the constant force of friction. Thus the net impulse is 0, and the net work is 0, but there is a point x1, t1 that the object is at a non-zero velocity(It's maximum velocity since that is also when the weight stop pulling it), but then decelerates at a constant rate to a zero velocity.
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  • #12
In thinking about it, I am not sure if that could work....scrap that, lets go back to what I was talking about in the first place, lets say I apply a constant upward force greater than gravity for a time t1, it gains a velocity of (F-g)t1/m as per the impulse and change in momentum, the force stop being applied, and gravity slows the object down until it stops some time later, giving an over all time. Since the net impulse is 0 the force applied * t1 must equal force of gravity * total time.

But my question is concerning if I only know the total time, total distance, mass, and acceleration of gravity. There must be a displacement since the force applied was greater than gravity, it just has Δp because it both accelerated and decelerated.
  • #13
I must make a correction to the equation mentioned in #6, I stupidly calculated it as just the accelerations by assuming the mass to be 1, but when I checked it out, it wasn't working, so I redid it, and the correct version is:

F = g2t2/(gt2-2xm)

So if there were two rockets were on either side of a ship pointed in opposite directions that were designed to give the same impulse, but at different forces, and F were the larger force. It would end up at rest because of the equal impulse, but would have a displacement of x.

For example,
A 20kg box with a rocket on either side, facing opposite directions
Lets say rocket A is gives 180N of thrust to the left for 10s(assume weight doesn't change from using fuel)
rocket B gives 900N to the right for 2s.

For the first 2 seconds there is a net force of 720N to the right. That gives the box an acceleration of 36m/s2, it will travel a distance of x1 = .5(36)22 = 72m, it will have a kinetic energy of 720N*72m = 51840J, it will take the 180N force a distance of 288m to bring that KE to 0.

That is a total distance of 360m in 10s, so knowing only this, the mass, and the force of the 180N rocket, you could use the equation above to get the force of the other rocket.

F = (1802)(102)/(180(102)-2(20)(360))
F = (32400)(100)/(180(100)-40(360))
F = 3240000/(18000 - 14400)
F = 3240000/(3600)
F = 900

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