How Is the Retarding Force on a Meteor Calculated Using Newton's Laws?

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Homework Help Overview

The discussion revolves around calculating the retarding force acting on a meteor falling through Earth's atmosphere, considering both gravitational and frictional forces. The subject area includes dynamics and Newton's laws of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the meteor, including gravitational force and retarding force due to atmospheric drag. There are attempts to express the net force and clarify the roles of different forces, with some questioning the setup and assumptions regarding the direction of forces.

Discussion Status

The discussion has seen various attempts to clarify the equations governing the forces. Some participants have provided guidance on the correct formulation of the forces, while others are still exploring different interpretations and calculations. There is an ongoing examination of the signs and magnitudes of the forces involved.

Contextual Notes

Participants are grappling with the implications of including both gravitational and retarding forces in their calculations, as well as the correct application of Newton's laws in the context of the problem. There is also a consideration of the direction of acceleration and forces, which has led to some confusion.

Neek 007
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Homework Statement


A meteor of mass .25 kg is falling vertically through Earth's atmosphere with an acceleration of 9.2 m/s. In addition to gravity, a vertical retarding force (due to the frictional drag of the atmosphere) acts on the meteor. What is the magnitude of this retarding force?


Homework Equations


F = ma

but I think, because of "in addition to gravity"

F = ma + mg


The Attempt at a Solution



Fm-Ffr= ma + mg

Fm = (.25)(9.2) = 2.3 N


-Ffr= ma + mg - Fm

-Ffr = mg

= -2.45 N
the friction force greater than the weight? Doesnt sound right
 
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Neek 007 said:
F = ma

but I think, because of "in addition to gravity"

F = ma + mg
No, you were right the first time:
ƩF = ma

The Attempt at a Solution



Fm-Ffr= ma + mg
Fix this.
 
So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?
 
Neek 007 said:
So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?
Why do you think it equals 0? (I assume Fm is the object's weight, right?)
 
Fm is the meteor's weight, yes.

So gravity plays no part in this?

Now I am thinking this
(im switching to Y+ up, i forgot to consider that i was making Y+ down)
ƩF = ma

Ffr - Fm - w = may

Ffr = may + Fm + W

I think this is correct because there is 1 force pulling the meteor back, the Force of friction, and 2 forces acting down, the weight of the meteor(with Earth's gravity) and the meteor moving with the 9.2 m/s2 acceleration.

Or am I just adding the weight of the meteor twice?
I think I am overthinking this.
 
Neek 007 said:
Fm is the meteor's weight, yes.

So gravity plays no part in this?
What do you think gravity is?
 
Okay, I'm settling with this.

Ffr - Fm = may

Ffr = may + Fm

Ffr = (.25kg)(9.2m/s2) + (.25kg)(9.8m/s2)

Ffr = 4.75 N
 
Neek 007 said:
Okay, I'm settling with this.

Ffr - Fm = may

Ffr = may + Fm

Ffr = (.25kg)(9.2m/s2) + (.25kg)(9.8m/s2)

Ffr = 4.75 N
Almost. What's the sign of the acceleration?
 
negative

Ffr = (.25)(-9.2) + (.25)(-9.8)

Ffr = -4.75 N

Thanks a bunch!
 
  • #10
Neek 007 said:
negative

Ffr = (.25)(-9.2) + (.25)(-9.8)

Ffr = -4.75 N

Thanks a bunch!
The acceleration of the object is -9.2; the weight is just mg, not -mg. (You already included the direction of the weight--downward--in your first equation.)

One more time!
 
  • #11
Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help
 
  • #12
Neek 007 said:
Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help
Good! And you're welcome.

It's always useful to do a 'sanity check' of your answers. In this case, you know the acceleration is just a little bit less than free fall acceleration. That should tell you that the resistance is small compared to the weight.
 
  • #13
I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N
 
  • #14
Dacaron79 said:
I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N
No. Only two forces act: gravity and the retarding force. You can find the net force directly using ∑F = ma.
 

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