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

[Neek 007]

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

F_m-F_fr= ma + mg

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


-F_fr= ma + mg - F_m

-F_fr = mg

= -2.45 N
the friction force greater than the weight? Doesnt sound right

 

[Doc Al]

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

F_m-F_fr= ma + mg

Fix this.

 

[Neek 007]

So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?

 

[Doc Al]

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?)

 

[Neek 007]

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

F_fr - F_m - w = ma_y

F_fr = ma_y + F_m + 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.

 

[Doc Al]

Fm is the meteor's weight, yes.

So gravity plays no part in this?

What do you think gravity is?

 

[Neek 007]

Okay, I'm settling with this.

F_fr - F_m = ma_y

F_fr = ma_y + F_m

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

F_fr = 4.75 N

 

[Doc Al]

Okay, I'm settling with this.

F_fr - F_m = ma_y

F_fr = ma_y + F_m

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

F_fr = 4.75 N

Almost. What's the sign of the acceleration?

 

[Neek 007]

negative

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

F_fr = -4.75 N

Thanks a bunch!

 

[Doc Al]

negative

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

F_fr = -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!

 

[Neek 007]

Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help

 

[Doc Al]

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.

 

[Dacaron79]

I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N

 

[Doc Al]

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.