- #1

- 186

- 0

First, what is a force-time graph?

...How do i read from a force-time graph?

What can i derive from a force-time graph?

any advice?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter gunblaze
- Start date

- #1

- 186

- 0

First, what is a force-time graph?

...How do i read from a force-time graph?

What can i derive from a force-time graph?

any advice?

- #2

- 13,081

- 645

So what u can derive is simply the characteristics of the movement.

Daniel.

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 964

You can, of course, "read" directly off the graph the force on an object at each time. Since "force equals mass times acceleration", As dextercioby says, if mass is a constant,you can then derive the acceleration of the object. If you are very clever and can calculate the "area under the graph" between two times, that will give you the change in speed between the two times.

- #4

- 35,977

- 4,676

Recall that

[tex]F=\frac{dp}{dt}[/tex]

Since we don't know the scenario of your F-t graph, we don't know if the graph was obtained under constant m situation (which is most often the case). So let's first do this without assuming anything. I can then write, using the above equation

[tex]\int F dt = \int dp[/tex]

Now, if you have done calculus you will know that for any function y(x),

[tex]\int y(x) dx[/tex]

is simply the area underneath the y(x) curve.

Thus,

[tex]\int F dt[/tex]

is simply the area under the F(t) curve, and this is equal to the momentum change. This is what is typically known as an Impulse in the limit of very small dt.

Now, what if you do have a constant mass in this situation? Then we know that F=ma, or

[tex]F = m \frac{dv}{dt}[/tex]

Again, doing the same thing, we get

[tex]\int F dt = m \int dv[/tex]

You literally end up with the same thing, i.e. the right hand side is nothing more than

[tex]m(v_f - v_i)[/tex]

where the two v's correspond to the limits of integration. This is nothing more than momentum change for an object with constant mass, a similar conclusion we have drawn from above in the general case. So in this case, you have an additional set of information, that you can also deduce the change in velocity of the object, in addition to knowing the change in momentum.

On the other hand, if you don't know, or not expected to know calculus, then this posting is utterly irrelevant to you. :)

Zz.

- #5

- 13,081

- 645

On the other hand,let's assume for simplicity the constant mass case.Then,by a mere translation || to the Oy/vertical axis,u can get the graph acceleration vs.time.Since time passes anyway,the function a(t) is continuous.If this function is positive,the body of mass "m" is accelerating.If it's zero,then it moves at constant velocity,if it's negative,it's decelerating.If it's constant,bu nonzero,it's either uniformly accelerating,or uniformly decelerating.

This graph can't tell u,however if the body is at rest or moving with constant velocity.Why?Ask Galilei & Newton about the principle of inertia...

Daniel.

Share: