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!

Air Resistance & Weight

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    This isn't exactly homework...but I'm not sure where to post it. I was reading a textbook, and got myself in a mental predicament. I understand the reasons the book gave, but I want to *really* understand.

    So my question:
    "Why does air resistance decrease as the surface area to weight ratio decreases (i.e. same surface area, more weight)?"

    (Specifically, I was thinking about why it takes less time for a coffee filter WITH paper clips in it to fall than it does WITHOUT any paper clips in it.)

    2. Relevant equations

    I haven't taken any physics really yet; I'm working on Electricity and Magnetism now (for fun), but that doesn't really help here...so I don't know any relevant equations.

    3. The attempt at a solution

    OK, so here was how I got myself in an "I understand this but I don't really understand this" state:
    1) I know that a falling object is going to accelerate until it reaches its terminal velocity (that is, when the upward force of air resistance equals the downward pull of gravity)
    2) I also know that, air resistance aside, gravity affects all objects equally (Galileo). BUT, when air resistance is an important factor, then different weighted objects (assuming constant weight) will fall at different speeds.
    3) So the downward pull of gravity remains constant. This means that, somehow, the air resistance must decrease as the surface area-to-weight ratio decreases.
    4) ...And here I got stuck.
  2. jcsd
  3. Mar 14, 2014 #2


    User Avatar

    Staff: Mentor

    Hi scioly! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    By decreasing the area:weight ratio, I believe they mean that you keep the same weight but
    decrease the surface area! :smile: Try looking at it that way and see how you go. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]
    Last edited by a moderator: May 6, 2017
  4. Mar 14, 2014 #3
    Well if you decrease the surface area, then yeah, the number of air molecules that hits the object will be less, and thus the air resistance will be less. But that doesn't explain a scenario like this one:

    Experiment: Take a coffee filter and drop it. Then stick a paper clip in the coffee filter and drop it. Then stick 2 paper clips in the coffee filter and drop it. The surface area of the object remains constant, but the weight increases, and the time it takes to drop decreases. Why?

    Or this one: Why do smaller water drops fall slower than bigger ones?

    I get why there'd be more air resistance as surface area decreases, but why is there less air resistance as mass increases? And why does it only "affect" the object at low weights?

    Thanks!! :)
  5. Mar 14, 2014 #4
    Oh, and I know I stated before that I didn't really have any physics knowledge...I'm in 9th grade, and so have not taken a formal physics class, but that being said, I've studied thermodynamics, electricity, and magnetism. I've also read through (most of) a Saxon physics book, although I was focusing on the electricity and magnetism parts. So if you want to use formulae and stuff to help me understand this, feel free to. I haven't taken calculus yet...so any equations involving that I can't do, but anything below that I should be able to understand.
  6. Mar 14, 2014 #5


    User Avatar

    Staff: Mentor

    The important formula here is

    F = m.a

    where F is the nett force acting downwards on the body.

    Perhaps you can use that to explain your observations?
  7. Mar 14, 2014 #6
    So as the mass increases, the force downwards increases as well?
  8. Mar 14, 2014 #7


    User Avatar

    Staff: Mentor

    You would have to explain your thinking more precisely before I could give you any marks for that answer.

    Another equation that crops up here is W = m.g

    Can you explain this one?
  9. Mar 14, 2014 #8
    Wait...no...I was wrong. The force downwards is constant (gravity) So we have an increase in mass, but downwards force (gravity) remains constant. So it's:

    F/m = a

    meaning that as the mass increases, with a constant downwards force of gravity (9.8 m/s free fall) the acceleration decreases. But somehow as the mass increases, the object's terminal velocity is also supposed to increase...
    Last edited: Mar 14, 2014
  10. Mar 14, 2014 #9
    And terminal velocity is when the force upwards (air resistance) equals the force downwards (gravity) So I think it's established that the downwards force remains constant (Galileo's experiments). It has to be the upward force of air resistance then.
  11. Mar 14, 2014 #10
    So F = ma and this force isn't "constant" like gravity? So then as the mass is increased, the upwards force of the air resistance will increase? No, that can't be right...I'm getting something wrong.

    As F is increased, ma will increase as well in order for F to be equal to ma. Therefore, if m increases, the product of mass and acceleration will also increase, and therefore the force will increase. But I'm not sure if the acceleration of the object remains constant as the mass is increased. I would think that the opposite would be true: As the mass increases, the acceleration decreases.
  12. Mar 14, 2014 #11
    OK, so here's another way I was thinking about it:

    Assume air to be a fluid, and the coffee filter with or without paperclips to be a denser object moving through the fluid. As the density of the object increases (i. e. more paperclips in filter) the object moves faster. However, this effect is only seen with low weight, high surface area substances (kind of like low density...in my analogy) because with higher "density" substances, the effect of the relatively low resistance exerted by the air on the object is practically negligible, and all we're dealing with is the constant force of gravity.

    So then my question becomes: "Why do denser objects move faster through air than less-dense ones?"

    Or is this just a completely wrong way of looking at it?
  13. Mar 14, 2014 #12
    I'm taking AP Chem this year, so I'm interested in what's happening at the molecular level too (and I should be able to understand, if you want to use those sorts of examples)
  14. Mar 14, 2014 #13


    User Avatar

    Staff: Mentor

    We can identify a number of different forces in this situation, each arising from different origins.

    The object moves according to the nett force, this being the sum of all the forces.
  15. Mar 14, 2014 #14
    OK, so W = m.g

    weight = mass X gravity

    Force of gravity remains constant (9.8 m/s not thinking about air resistance), so as the mass of an object (amount of "matter" in the object) increases, the weight of the object will increase as well.
  16. Mar 14, 2014 #15
    Net force on falling object downwards = (Force downwards) - (Force upwards)


    Net force = gravity's force (not dependent on mass) - air resistance's force (dependent on mass)

    I still don't get it :(
  17. Mar 14, 2014 #16


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't seem to have taken the hint that others have given about the nett force acting on the filter and investigated what the forces are. One of the forces is drag due to air resistance. I suggest you google/Wikipedia drag force and in particular look up it's relation to velocity.

    Perhaps think about the forces involved when the filter falls at it's terminal velocity.
  18. Mar 14, 2014 #17


    User Avatar

    Staff: Mentor


    Wrong. Try this again.
  19. Mar 14, 2014 #18
    Net force = Force of gravity - Force of air resistance


    gravity: not dependent on mass/velocity of object
    air resistance: dependent on terminal velocity of object (because as terminal velocity increases, the drag forces decrease (i. e. air resistance decreases)) But why do the drag forces decrease as the velocity increases, and why does a greater mass mean a greater terminal velocity? (I mean on a molecular level...)
  20. Mar 14, 2014 #19
    Wait...is the force of gravity dependent on mass? I remember this equation:

    F = mg

    This would seem to say that ALTHOUGH gravity is constant, the total gravitational force on an object is dependent on its mass. I don't get why, and I can't reconcile this in my mind with Galileo's experiments, but...if this is true then I think I understand why the terminal velocity would increase...the air resistance (upward force) would stay the same, but the force downwards would increase; hence, objects that weigh more would fall faster. But that's not what my book said...grr.
  21. Mar 14, 2014 #20
    Hi CWatters,

    I tried googling drag force, but it didn't help me much. It had a bunch of stuff about the shape of the object and the flow around it. I found this:

    F = 0.5 C ρA V2

    A = Reference area in m2.
    C = Drag coefficient, unitless.
    F = Drag force, N.
    V = Velocity, m/s.
    ρ = Density of fluid, kg/m3.

    But I don't understand it yet.

    Also, as for thinking about the forces involved...as I said before, I haven't taken a physics course yet, but I'm interested in the subject...so here goes:

    Downwards forces:
    1) Gravity: F=mg. As mass increases, gravitational force increases. On Earth, g is 9.8 m/s (I think)
    2) ?? I can't think of any others

    Upwards forces:
    1) Air resistance (sheer number of molecules hitting against surface area): dependent upon surface area
    2) Drag force: dependent on mass?? I don't get it yet...

    Thanks for your help; I know it's probably quite frustrating working with me.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Air Resistance & Weight
  1. Air Resistance project (Replies: 3)

  2. Air resistance (Replies: 1)

  3. Air Resistance (Replies: 17)