Force vs. time graph (Impulse)

[JJBladester]

Homework Statement

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.

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

Homework Equations

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

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?

 

[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.

 

[JJBladester]

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.

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.

 

[PhanthomJay]

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

 

[JJBladester]

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

$mv_1+$ 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...

 

[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 ?

 

[Dadface]

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.

 

[JJBladester]

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 ?

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

 

[PhanthomJay]

And that's what I get also!

 

[Dadface]

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

 

[Quinzio]

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

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 ?

 

[JJBladester]

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 ?

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

 

[Quinzio]

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

The collar behaves no differently than you.

 

[Dadface]

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.

 

[JJBladester]

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.

 

[Dadface]

you have the right answer for part a but I am not sure how you got there.In the first 0.8 second the collar does not move.In the next 1.2 seconds it does move being subjected to a resultant force which varies from 0 to 6 in 1.2 seconds.The force then remains constant at 6 for a further 1 sec.Look at the graph you presented in post 1 and resketch it by drawing a line parallel to the x-axis and meeting the y-axis at 4.The change of momentum is then equal to the area under the graph going down to the horizontal line you have just drawn.

 

[JJBladester]

you have the right answer for part a but I am not sure how you got there.In the first 0.8 second the collar does not move.In the next 1.2 seconds it does move being subjected to a resultant force which varies from 0 to 6 in 1.2 seconds.The force then remains constant at 6 for a further 1 sec.Look at the graph you presented in post 1 and resketch it by drawing a line parallel to the x-axis and meeting the y-axis at 4.The change of momentum is then equal to the area under the graph going down to the horizontal line you have just drawn.

Now that's a good explanation. Whew, long tired day... Think I have the flu... But to all who responded for guidance, THANK YOU. Full solution posted here (with the book's answers being obtained):