• Support PF! Buy your school textbooks, materials and every day products Here!

Applications of Newton's Laws

  • Thread starter Neek 007
  • Start date
  • #1
41
0

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
 

Answers and Replies

  • #2
Doc Al
Mentor
44,912
1,170
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.
 
  • #3
41
0
So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?
 
  • #4
Doc Al
Mentor
44,912
1,170
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?)
 
  • #5
41
0
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
Doc Al
Mentor
44,912
1,170
Fm is the meteor's weight, yes.

So gravity plays no part in this?
What do you think gravity is?
 
  • #7
41
0
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
Doc Al
Mentor
44,912
1,170
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?
 
  • #9
41
0
negative

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

Ffr = -4.75 N

Thanks a bunch!
 
  • #10
Doc Al
Mentor
44,912
1,170
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
41
0
Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help
 
  • #12
Doc Al
Mentor
44,912
1,170
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
1
0
I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N
 
  • #14
Doc Al
Mentor
44,912
1,170
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.
 

Related Threads on Applications of Newton's Laws

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
21
Views
1K
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
3K
Top