1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force and acceleration leading to weight of a parachutist

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

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

    Question 8

    2. Relevant equations

    f=ma

    3. 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
     
  2. jcsd
  3. Mar 6, 2012 #2
    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.
     
  4. Mar 6, 2012 #3
    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)
     
  5. Mar 6, 2012 #4
    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.
     
  6. Mar 6, 2012 #5
    Thank you very much, problem resolved :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force and acceleration leading to weight of a parachutist
Loading...