# Force vs. time graph (Impulse)

1. Oct 16, 2010

1. The problem statement, all variables and given/known data

A 4-lb collar which can slide on a frictionless vertical rod is acted upon by a force P which varies in magnitude as shown. Knowing that the collar is initially at rest, determine its velocity at (a) t = 2s (b) t = 3s.

(a) 29.0 ft/s
(b) 77.3 ft/s

2. Relevant equations

$$mv_1+\sum Imp_{1\rightarrow 2}=mv_2$$

3. The attempt at a solution

$mv_1+$ Area under F vs. t graph $=mv_2$

$$v_2=\frac{\left (\frac{1}{2} \right )(2)(10)}{\frac{4}{32.17}}=80.43ft/s$$

I did not get the book's answer of 29.0 ft/s for part (a). Why?

2. Oct 16, 2010

### PhanthomJay

I don't get either answer. Don't forget that the net force acting on the collar must include the weight of the collar acting down.

3. Oct 16, 2010

Right, forgot about the weight force and its associated impulse (Wt).

$mv_1+$ Area under F vs. t graph $+Wt = mv_2$

$$v_2=\frac{0+\left (\frac{1}{2} \right )(2)(10) + (4)(2)}{(4/32.17)}=144.8ft/s$$

Closer, I believe... But not the book's answer.

4. Oct 16, 2010

### PhanthomJay

But this is still incorrect...the applied force acts up and the weight acts down...

5. Oct 16, 2010

$mv_1+$ Area under F vs. t graph $-Wt = mv_2$

$$v_2=\frac{0+\left (\frac{1}{2} \right )(2)(10) - (4)(2)}{(4/32.17)}=16.09ft/s$$

Closer...

6. Oct 16, 2010

### Quinzio

mmm closer....

stop doing formulas and think.
If the collar weight was 100 lb, what would be the speed after 2s or 3s, or 10s, 20s or any seconds ?

7. Oct 16, 2010

I'm not familiar with the units used in this question and I had to read between the lines to use the 32.17 factor.Non SI units are a pain.Anyway my answers agree with the books answers.

8. Oct 16, 2010

The speed (velocity) at any time would be vi-gt, if we weren't including an external force P. There is an external force present, so there's more to it than that.

9. Oct 16, 2010

### PhanthomJay

And that's what I get also!

10. Oct 16, 2010

The collar will not start to lift/move until F becomes equal to 4lb.Think graph and shifting your x axis upwards.

11. Oct 16, 2010

### Quinzio

Now you are sitting on a chair I suppose. If a force of 4 lb was applied to you upward, would you be moving at vi-gt ???

12. Oct 16, 2010

I weigh 170lb, so it would take 170lb of force to counter my weight, then any additional force would start me upward.

13. Oct 16, 2010

### Quinzio

The collar behaves no differently than you.

14. Oct 16, 2010

For your graph think of the relevant resultant force on the collar.This doesn't vary from 0 to 10 in 2 seconds and then levels off,it varies from 0 to six(10-4) in 1.2 seconds and then levels off.

15. Oct 16, 2010

I think I'm starting to understand this. It takes the pulling force P 4lb to even get the collar started.

From the graph in the problem, we have y=5x from t=0 sec to t=2 secs. So, in order for the pulling force P to even move the collar, we need to know what the time of the equation is when pulling force = 4lb. It turns out to be 4/5, or 1.2.

$$\frac{\frac{1}{2}(1.2)(6)+(0.8)(10)-(4)(2)}{(4/32.17)}=29 ft/s$$

The first term in the numerator is P acting from t=0 to t=1.2 seconds to counteract the collar's weight. The 2nd term is what happens from 1.2 seconds to 2 seconds (after the weight of the collar is overcome), and the third term in the numerator is to subtract the weight*time of the collar. The denominator is the mass of the collar.

16. Oct 16, 2010