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Skydiver falling

  • #1
599
6

Homework Statement



I just had an exam in physics and there was this one easy question that I just couldn't wrap my head around.

A skydiver (85kg) is accelerating at 3m/s^2. What is the force of the air resistance on him?

Homework Equations


f=ma and other mechanics


The Attempt at a Solution


I solved for the F of the skydiver which would equal 255 N. I then used Newton's third law to conclude that the air resistance must be 255 N up. But then I started thinking... If he's accelerating, there must be a Fnet... How, then, can he accelerate if the friction from the air equals the force of gravity?

What should I have done?
 

Answers and Replies

  • #2
537
1
So you basically say that the air resistance is equal to and opposite to that of the force due to gravity.

If the air resistance is *equal and opposite* to the force due to gravity, would he be accelerating?
 
  • #3
599
6
That's what I mean. That's what threw me off. I have no idea how to solve for the air friction... maybe something with 10m/s^2 and 3m/s^2?
 
  • #4
537
1
If they're equal and opposite, then

[itex]F_{g} = -F_{res}[/itex]

therefore

[itex]F_{net} = F_{g} + F_{res} = 0[/itex]

can you accelerate if there is zero net force?
 
  • #5
rcgldr
Homework Helper
8,628
501
maybe something with 10m/s^2 and 3m/s^2?
Yes.
 
  • #6
PeterO
Homework Helper
2,425
46

Homework Statement



I just had an exam in physics and there was this one easy question that I just couldn't wrap my head around.

A skydiver (85kg) is accelerating at 3m/s^2. What is the force of the air resistance on him?

Homework Equations


f=ma and other mechanics


The Attempt at a Solution


I solved for the F of the skydiver which would equal 255 N. I then used Newton's third law to conclude that the air resistance must be 255 N up. But then I started thinking... If he's accelerating, there must be a Fnet... How, then, can he accelerate if the friction from the air equals the force of gravity?

What should I have done?
Newtons third law: For every force acting, there is an equal and opposite force acting

That 255N force is the net force - the vector sum of the acting forces.

You don't apply Newtons Third Law to the net force.

You should have been looking at the acting forces; which gave a net force of 255N
 
  • #7
599
6
Hmm... So... Would this be correct reasoning

85kg 10m/s^2 =800N
85kg 3m/s^2 = 255N

800-255= 545

The air friction is 545N

?
 
  • #8
PeterO
Homework Helper
2,425
46
Hmm... So... Would this be correct reasoning

85kg 10m/s^2 =800N
85kg 3m/s^2 = 255N

800-255= 545

The air friction is 545N

?
Numerically not bad - except that 85 * 10 is 850 not 800; meaning a final answer of 595N

More appropriately.

Acting force down = weight force which is mg. If you assume a value of g = 10, that means a weight force of 850N.
Often we use the valu of 9.8 for g, which mans a weight force of 833N.

The net Force is 255N down [as you already calculated].

The other acting force is air resistance up.

Weight of 850N down, in combination with an air resistance force of 595N up is how we get a net force of 255N down, resulting in an acceleration 3 m/s^2.
 
  • #9
599
6
Woops, you're right :D.
Well, at least I figured it out. Thank you.
 

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