• Support PF! Buy your school textbooks, materials and every day products Here!

Force and acceleration leading to weight of a parachutist

  • Thread starter Mukilab
  • Start date
  • #1
73
0

Homework Statement



http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_PC_2008_QP.pdf

Question 8

Homework Equations



f=ma

The Attempt at a Solution



Sure since f=ma
A) would have force of 0, since there is no acceleration
B) force would be 800N
C) force of 0N
D) force of 0N
E) Force of 1000N (10 m/s^2 due to gravity)

so surely E would be the answer? I just don't understand why A is the answer (it is in the mark scheme that A is the answer).

Plus, how on earth do you add in the 'extra-large parachute' thing?

Any help would be great thanks
 

Answers and Replies

  • #2
537
1
well, remember that the question is just asking for the one with the largest upward force, not the one with the largest net force.

And remember which direction the force of gravity points.
 
  • #3
73
0
well, remember that the question is just asking for the one with the largest upward force, not the one with the largest net force.

And remember which direction the force of gravity points.
Well drag = kAv^2

But for some of the parachutes we don't know the area of 'extra-large' so what can we do?

(if we ignore extra-large etc, A is right, thank you very much for your help)
 
  • #4
537
1
well if you look at kAv2, we can see that the velocity has the largest impact on the force

let's say the normal parachute has area A. Let's also ignore k for the moment since I assume it's the same for each parachute.

So for each of them (other than E) we have 36A, A, 16A and then 4 times the area of the large parachute.

The extra large parachute would have to be nine times as large as the normal parachute to get the same amount of drag force. Since nine times the area is a bit excessive, I'd go with answer A.
 
  • #5
73
0
well if you look at kAv2, we can see that the velocity has the largest impact on the force

let's say the normal parachute has area A. Let's also ignore k for the moment since I assume it's the same for each parachute.

So for each of them (other than E) we have 36A, A, 16A and then 4 times the area of the large parachute.

The extra large parachute would have to be nine times as large as the normal parachute to get the same amount of drag force. Since nine times the area is a bit excessive, I'd go with answer A.
Thank you very much, problem resolved :)
 

Related Threads for: Force and acceleration leading to weight of a parachutist

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
5
Views
4K
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
Top