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Homework Help: Friction Due to Jumping w Parachute

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A 500 g rock falls out of a 75 m tree. It takes it 5s to parachute to the ground. How much friction is upon it?

    2. Relevant equations
    possibly [tex]\Delta[/tex]d=v1[tex]\Delta[/tex]t+.5a([tex]\Delta[/tex]t)2 and other kinematics formulas

    3. The attempt at a solution
    I tried to find the net force, using 9.8 m/s2 as a, then gravitational force. I subtracted the gravitational force from the net force, which did not work at all. Next I tried to find a using a kinematics formula, and use that number to find the net force and subtract the new gravitational force from it, but still did not get a good answer.
  2. jcsd
  3. Jun 18, 2009 #2
    Assuming friction with the air is a constant force, not dependent on the speed of the falling object:

    What is the net force acting on the rock, and how does it accelerate it?
  4. Jun 18, 2009 #3
    F=(.5 kg)(9.8 m/s2)
    F=4.9 N

    Fg=(.5 kg)(9.8 m/s2)(75 m)
    Fg=367.5 N

    Ffriction=4.9 N-367.5 N
    Ffriction=-362.6 N

    75 m=0 m/s(5 s)+.5a(5 s)2

    F=(.5 kg)(6 m/s2)
    F=3 N

    Fg=(.5 kg)(6 m/s2)(75 m)
    Fg=225 N

    Ffriction=3 N-225 N
    Ffriction=-222 N

    And my answer's supposed to be 1.9 N.
  5. Jun 18, 2009 #4
    Consider Newton's second law of motion:
    ΣF = ma
    The sum of all real force on the rock, is equal to its mass times its TOTAL acceleration.

    It looks like you're confused about what happens in free-fall.
    Were the rock to fall in free-fall (Without friction slowing it down), the sum of the forces on it would be:
    ΣF = mg = Fnet
    ΣF = ma
    a=g //

    That -367.5N figure. It is completely unrelated, and is 100 times larger than any forces you should be dealing with in this problem.
    Fg = mg
    The initial height ENERGY is mgh and is completely irrelevant to the question at hand, unless you had the speed upon impact rather than the time of impact.

    Try and find two different ways of expressing a, the acceleration of the rock towards the ground and making an equation. You have the correct formula up there.

    And yes, the final answer is 1.9 N if you approximate g≈9.8 m/s²
    Last edited: Jun 18, 2009
  6. Jun 18, 2009 #5


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    This would not be correct.

    m*g*h is the potential energy of the object.

    m*g is the force from gravity on the object.
  7. Jun 18, 2009 #6
    Do you mean the acceleration of the rock towards the ground? haha.
    Anyway, I'm confused by what you said when finding two different ways of expressing a.

    If I use the kinematics equation, I get a=6m/s2

    So then,
    =(.5 kg)(a)
    =(.5 kg)(6m/s2)
    =3 N

    F=(.5 kg)(9.8 m/s2)
    F=4.9 N

    Subtracting these, I get 1.9 N?

    If I got the math all right, I'm trying to understand the reasoning behind it, though I think I get it know and that I'm completely stupid for not seeing it in the first place. When I use 9.8 m/s2 as acceleration, this is during free fall, but when I find a, that is the acceleration due to the parachute. When I find the forces, I can find their difference, which will be the friction because of the parachute slowing down the fall?
  8. Jun 18, 2009 #7
    Yes, I meant the acceleration of the rock towards the ground. :) Silly typo on my part.

    Yes, the kinematics equation nets you a numerical value for a, that is correct.
    But a is not always g, as you've just seen.

    Gravity on earth enacts a force of mg on a body, where m is its mass, and g is the acceleration brought on by gravity.
    g, for the purpose of this discussion*, is constant.

    mg is a force acting on the rock, just like any other. The rock's acceleration is determined by the SUM of all the forces acting on it.

    The approach you posted is wrong. You mixed up a and g.

    ΣF = ma = (½ kg) * (6 m/s²) = 3N
    What forces comprise ΣF, and the value of which one do you know?

    Once you start studying about gravity in class, you'll see what I mean. The force of gravity enacted by the earth on objects near its surface depends on their distance from its core.
    Fg = m*g(r)
    g(r) = (G*Mearth)/r² in the direction of the center of the earth, where r is the distance from the center of the earth to the object in question. This acceleration is approximately constant if you're very close to the earth.
    Don't worry if you don't quite understand this bit, it's completely unrelated and you'll get to it eventually in class.
  9. Jun 18, 2009 #8


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    That is a feeling we've all had. It comes with understanding, really understanding something.

    I think you've found what they want you to find with the problem. That the difference in acceleration, is a result of friction. Of course it is an average force over the distance, as the force varies with speed - being dependent on air viscosity and cross sectional area and such. But for the purpose of illustration, I think they want you to understand net accelerations.
  10. Jun 18, 2009 #9
    thanks for all of your help everyone!
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