# Applications of Newton's Laws

1. Aug 18, 2012

### Neek 007

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
F = ma

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

F = ma + mg

3. 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

2. Aug 18, 2012

### Staff: Mentor

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

Fix this.

3. Aug 18, 2012

### Neek 007

So, Fm-Ffr= ma

-Ffr = ma - Fm

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

4. Aug 18, 2012

### Staff: Mentor

Why do you think it equals 0? (I assume Fm is the object's weight, right?)

5. Aug 18, 2012

### 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

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 earths 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.

6. Aug 18, 2012

### Staff: Mentor

What do you think gravity is?

7. Aug 18, 2012

### Neek 007

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

8. Aug 18, 2012

### Staff: Mentor

Almost. What's the sign of the acceleration?

9. Aug 18, 2012

### Neek 007

negative

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

Ffr = -4.75 N

Thanks a bunch!

10. Aug 18, 2012

### Staff: Mentor

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. Aug 18, 2012

### Neek 007

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

finally, 5th times the try.

Thank you very much for the help

12. Aug 18, 2012

### Staff: Mentor

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. Jan 5, 2014

### Dacaron79

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

14. Jan 5, 2014

### Staff: Mentor

No. Only two forces act: gravity and the retarding force. You can find the net force directly using ∑F = ma.