Force vs. time graph (Impulse)

In summary, the conversation discusses a 4-lb collar on a frictionless vertical rod being acted upon by a force P that varies in magnitude. The goal is to determine the collar's velocity at two different times. The equation used to solve the problem is mv_1 + Area under F vs. t graph = mv_2, taking into account the weight of the collar. The correct answer for part (a) is 29.0 ft/s, and the reason for this is explained through the graph and the change of momentum. The conversation also includes a discussion on the weight of the collar and its impact on the solution. The full solution can be found
  • #1
JJBladester
Gold Member
286
2

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

impulse%20momentum%20vertical%20rod.jpg


Homework Equations



[tex]mv_1+\sum Imp_{1\rightarrow 2}=mv_2[/tex]

The Attempt at a Solution



[itex]mv_1+[/itex] Area under F vs. t graph [itex]=mv_2[/itex]

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

I did not get the book's answer of 29.0 ft/s for part (a). Why?
 
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  • #2
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
PhanthomJay said:
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).

[itex]mv_1+[/itex] Area under F vs. t graph [itex]+Wt = mv_2[/itex]

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

Closer, I believe... But not the book's answer.
 
  • #4
But this is still incorrect...the applied force acts up and the weight acts down...
 
  • #5
PhanthomJay said:
But this is still incorrect...the applied force acts up and the weight acts down...

[itex]mv_1+[/itex] Area under F vs. t graph [itex]-Wt = mv_2[/itex]

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

Closer...
 
  • #6
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
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
Quinzio said:
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 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
And that's what I get also!
 
  • #10
The collar will not start to lift/move until F becomes equal to 4lb.Think graph and shifting your x-axis upwards.
 
  • #11
JJBladester said:
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.

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
Quinzio said:
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.
 
  • #13
JJBladester said:
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.
 
  • #14
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
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.

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

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
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.
 
  • #17
Dadface said:
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):

impulse%20momentum%20collar%20graph.png
 

FAQ: Force vs. time graph (Impulse)

What is a force vs. time graph?

A force vs. time graph, also known as an impulse graph, is a graphical representation of the relationship between the force applied on an object over a period of time. It is used to analyze the change in momentum of an object due to an external force.

How is the impulse calculated from a force vs. time graph?

The impulse is calculated by finding the area under the force vs. time graph. This can be done by dividing the graph into smaller shapes, such as rectangles or triangles, and calculating the area of each shape. The sum of all these areas will give the impulse.

What does the slope of a force vs. time graph represent?

The slope of a force vs. time graph represents the magnitude of the force acting on an object. A steeper slope indicates a larger force, while a flatter slope indicates a smaller force. The units of the slope are force divided by time, which is equivalent to acceleration.

Why is the force vs. time graph important in physics?

The force vs. time graph is important in physics because it allows us to understand the relationship between force and momentum. It also helps us analyze the motion of objects and predict their behavior under the influence of external forces.

How does the shape of a force vs. time graph change for different types of collisions?

The shape of a force vs. time graph can vary depending on the type of collision. For an elastic collision, the graph will have a symmetrical curve, indicating equal and opposite forces acting on the objects. In an inelastic collision, the graph will have a flat section, indicating a constant force acting on the objects. For a completely inelastic collision, the graph will have a sharp peak, indicating a large force acting on the objects for a short period of time.

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