Drawing Free Body Diagrams

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  • #1
MysticDude
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Homework Statement


Draw free body diagrams:
A rectangular block being accelerated on a horizontal, frictionless surface.
A skydiver who had just jumped from a plane and hasn't reached terminal velocity.
A cube being accelerated up a 30° incline with μk > 0.
A pendulum bob that has just been released at an angle of 25° from the vertical.


Homework Equations





The Attempt at a Solution


I'm pretty sure about the first three, but not so sure about the last one.
[PLAIN]http://img510.imageshack.us/img510/1808/physicsec.png [Broken]

Sorry but I forgot about the theta, I'll put up the new picture soon.
 
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Answers and Replies

  • #2
Doc Al
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In your last diagram, what does 'normal force' refer to? And what do you mean by 'force of bob'?
 
  • #3
MysticDude
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Now that I think more about it, I think that "force of bob" should just be T for tension. I guess that that counteracts mg so there really is no normal in the free body diagram?
 
  • #4
Doc Al
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Now that I think more about it, I think that "force of bob" should just be T for tension.
Good.
I guess that that counteracts mg so there really is no normal in the free body diagram?
A normal force is a force between surfaces in contact. There's no surface touching the pendulum bob.
 
  • #5
MysticDude
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Good.

A normal force is a force between surfaces in contact. There's no surface touching the pendulum bob.
So it's just like:

|\
..\
....\...Tension at 25°
......\
.......O
.......|
.......|...mg
.......|
.......V

Hopefully you can understand this, if not, I'll upload a picture.
 
  • #6
Doc Al
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So it's just like:

|\
..\
....\...Tension at 25°
......\
.......O
.......|
.......|...mg
.......|
.......V

Hopefully you can understand this, if not, I'll upload a picture.
Exactly!
 
  • #7
MysticDude
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Exactly!
Thanks for your help!
By the way, were the other ones correct?
 
  • #8
Doc Al
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By the way, were the other ones correct?
They look OK. The only one I would question is the 'jumping out of the plane' one. It depends what you mean by 'just jumped'. Are you ignoring the sideways force due to the rushing air when you jump out? (Let's say forget that detail.) So assume you drop with zero initial speed with respect to the air. What's the air resistance at that first instant before picking up speed?

But these are probably nit-picking details.
 
  • #9
MysticDude
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They look OK. The only one I would question is the 'jumping out of the plane' one. It depends what you mean by 'just jumped'. Are you ignoring the sideways force due to the rushing air when you jump out? (Let's say forget that detail.) So assume you drop with zero initial speed with respect to the air. What's the air resistance at that first instant before picking up speed?

But these are probably nit-picking details.
Well, my teacher didn't go into too much detail about these problems. My teacher does put attention to detail sometimes though. So if the skydiver "just jumped" then there would be a the force of the air rushing against him. In other words, should it be something like this:
O--------> air force against skydiver
|
|
|
V gravitational force

, or is their another force retarding the air force against the skydiver?
 
  • #10
Doc Al
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In other words, should it be something like this:
O--------> air force against skydiver
|
|
|
V gravitational force
That's what I would say. The air resistance is always in the same direction as the relative motion of the air.

But if you wanted, say, the FBD for a brick (or a skydiver) that was dropped from high up with no initial speed, before it reached terminal velocity, then your diagram would be fine. Until it reaches terminal speed, the gravitational force will exceed the air resistance.
 
  • #11
MysticDude
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That's what I would say. The air resistance is always in the same direction as the relative motion of the air.

But if you wanted, say, the FBD for a brick (or a skydiver) that was dropped from high up with no initial speed, before it reached terminal velocity, then your diagram would be fine. Until it reaches terminal speed, the gravitational force will exceed the air resistance.
Okay thanks, you've been a big help!
I think I'm going to go with out idea since it seems more reasonable.
Once again, thanks a lot!
 

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