Physics Homework Question on terminal velocity and weight

AI Thread Summary
The discussion revolves around understanding terminal velocity and the forces acting on a skydiver. Participants clarify that at terminal velocity, the weight of the skydiver equals the air resistance, resulting in zero acceleration. The specific case discussed indicates that if the terminal velocity is 5 m/s, the corresponding weight is 600 N. It is emphasized that the weight must match the drag force for the skydiver to maintain constant speed. The conversation concludes with insights on how changes in weight affect terminal velocity and the graphical representation of drag forces.
OscarF
Messages
21
Reaction score
1
Homework Statement
From the graph, show the force of air resistance on our sky diver's parachute changes with her speed of fall
Relevant Equations
w=mg
Below I've attached the question - I don't know why this question is so difficult, perhaps I missed a lesson or such, however I've Benn working at it for ages and got nowhere...
 

Attachments

  • Screenshot 2020-10-30 at 10.36.41.png
    Screenshot 2020-10-30 at 10.36.41.png
    62.4 KB · Views: 302
  • Screenshot 2020-10-30 at 10.36.48.png
    Screenshot 2020-10-30 at 10.36.48.png
    72.5 KB · Views: 353
Physics news on Phys.org
OscarF said:
Homework Statement:: From the graph, show the force of air resistance on our sky diver's parachute changes with her speed of fall
Relevant Equations:: w=mg

Below I've attached the question - I don't know why this question is so difficult, perhaps I missed a lesson or such, however I've Benn working at it for ages and got nowhere...
Is it point a) you are missing?
 
greg_rack said:
Is it point a) you are missing?
No point a) was the only one I could do...
 
OscarF said:
No point a) was the only one I could do...
What could you state for a body traveling at terminal velocity, in terms of forces and acceleration acting on it?
 
What do you get if you apply Newton's 2nd law at the state where we have terminal velocity, i.e velocity constant and total acceleration zero.
 
  • Like
Likes greg_rack
greg_rack said:
What could you state for a body traveling at terminal velocity, in terms of forces and acceleration acting on it?
They are balanced...
 
Delta2 said:
What do you get if you apply Newton's 2nd law at the state where we have terminal velocity, i.e velocity constant and total acceleration zero.
I mean you would get that f=mg, but then how can we find out m?
 
OscarF said:
They are balanced...
What do you mean? Be more precise.
If an object is traveling at terminal velocity, its acceleration is 0.
According to Newton's 2nd law, the resultant force acting on it must be even equal to zero.
In this particular case, which are the forces you must take into account?

OscarF said:
I mean you would get that f=mg, but then how can we find out m?
You don't need the mass. Think in terms of forces and forget ##F=mg## for now.
 
greg_rack said:
What do you mean? Be more precise.
If an object is traveling at terminal velocity, its acceleration is 0.
According to Newton's 2nd law, the resultant force acting on it must be even equal to zero.
In this particular case, which are the forces you must take into account?You don't need the mass. Think in terms of forces and forget ##F=mg## for now.
We must take into account weight and air resistance (drag)
 
  • #10
OscarF said:
We must take into account weight and air resistance (drag)
That's right. So, in order to have that parachutist traveling at terminal velocity, you must have the weight equal in magnitude to the drag. Right?
 
  • #11
I think another thing I may be missing is I'm not sure where on the graph terminal velocity is reached...
 
  • #12
greg_rack said:
That's right. So, in order to have that parachutist traveling at terminal velocity, you must have the weight equal in magnitude to the drag. Right?
yes... I still don't understand though how the weight is found out from that being that we are only given speed. Although, I guess that's the next step - I'm just stumped.
 
  • #13
OscarF said:
I think another thing I may be missing is I'm not sure where on the graph terminal velocity is reached...
The graph just tells you the intensity of the resistive force acting, depending on the speed at which the body is traveling.
Once you understand the value the resistive force you must have in order to satisfy Newton's 2nd law, you can find out the terminal velocity just by looking at the graph
 
  • #14
greg_rack said:
The graph just tells you the intensity of the resistive force acting, depending on the speed at which the body is traveling.
Once you understand the value the resistive force you must have in order to satisfy Newton's 2nd law, you can find out the terminal velocity just by looking at the graph
Would that value be 600N?
 
  • #15
OscarF said:
Would that value be 600N?
Which case are you referring to?
 
  • #16
greg_rack said:
Which case are you referring to?
When the parachutist is traveling at 5/ms
 
  • #17
OscarF said:
When the parachutist is traveling at 5/ms
Yes, when the parachutist is traveling at 5m/s, the resistive force acting is ABOUT 600N(you can't tell exactly since you have no precise reference points/intersections in the graph).
 
  • #18
greg_rack said:
Yes, when the parachutist is traveling at 5m/s, the resistive force acting is ABOUT 600N(you can't tell exactly since you have no precise reference points/intersections in the graph).
so then if the weight were 600N, that point would be terminal velocity. I still don't understand why 600N MUST be the parachutist's weight though.
 
  • #19
OscarF said:
so then if the weight were 600N, that point would be terminal velocity. I still don't understand why 600N MUST be the parachutist's weight though.
That's exact.
If her weight equals the resistive force of 600N and is thus equal to 600N, the acceleration acting is ##0m/s^2## and the speed remains constant: we have reached terminal velocity
 
  • Like
Likes Delta2
  • #20
greg_rack said:
That's exact.
If her weight equals the resistive force of 600N and is thus equal to 600N, the acceleration acting is ##0m/s^2## and the speed remains constant: we have reached terminal velocity
but why is her weight 600N, if we had done this at a different point we would just be concluding that her weight is 800N for example.
 
  • #21
OscarF said:
but why is her weight 600N, if we had done this at a different point we would just be concluding that her weight is 800N for example.
The given of the problem is that his particular skydiver's terminal velocity is 5 m/s. From this information we derive the skydivers weight as 600 N.

If the given of the problem were some other terminal velocity for some other skydiver we could derive some other result for that skydiver's weight.
 
  • Like
Likes greg_rack
  • #22
OscarF said:
but why is her weight 600N, if we had done this at a different point we would just be concluding that her weight is 800N for example.
The statement is telling you that her terminal velocity is 5m/s, so her weight MUST be 600N because of how this problem is formulated.
If the weight was 800N, her terminal speed would have indeed been higher.
 
  • #23
greg_rack said:
The statement is telling you that her terminal velocity is 5m/s, so her weight MUST be 600N because of how this problem is formulated.
If the weight was 800N, her terminal speed would have indeed been higher.
oh ok, makes sense. Just quickly for question c, would the terminal velocities be 400N and 1000N
 
  • #24
OscarF said:
oh ok, makes sense. Just quickly for question c, would the terminal velocities be 400N and 1000N
For point c) you are being given the weights of two other parachutists.
Since, when traveling at terminal velocity their weight must be equal to the drag force, just find on the graph the corresponding speed values to those of the resistive forces for each parachutist.
 
  • #25
greg_rack said:
For point c) you are being given the weights of two other parachutists.
Since, when traveling at terminal velocity their weights must be equal to the drags intensities, just find on the graph the corresponding speed values to those of the resistive forces for each parachutist.
great, makes sense! And finally, part d, is it basically just the same graph but with the drag values doubled?
 
  • #26
OscarF said:
great, makes sense! And finally, part d, is it basically just the same graph but with the drag values doubled?
Exactly, you will have that graph vertically dilated:
Double the area ##\rightarrow## double the drag
 
Back
Top