Writing a force equation!

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  • #1
MysticDude
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Homework Statement



Write the equation used to calculate net force on a block of mass m that is accelerating down a ramp.


Homework Equations





The Attempt at a Solution


Okay I think that we are going to have to use the equation F - f = ma. Since the question says to find the net force, does this mean the I'm going to have to add all of the vector forces and get the magnitude? Here is my free body diagram:
[PLAIN]http://img7.imageshack.us/img7/4929/nov24physicsnum4.jpg [Broken]
 
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Answers and Replies

  • #2
berkeman
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Homework Statement



Write the equation used to calculate net force on a block of mass m that is accelerating down a ramp.


Homework Equations





The Attempt at a Solution


Okay I think that we are going to have to use the equation F - f = ma. Since the question says to find the net force, does this mean the I'm going to have to add all of the vector forces and get the magnitude? Here is my free body diagram:
[PLAIN]http://img7.imageshack.us/img7/4929/nov24physicsnum4.jpg[/QUOTE] [Broken]

That looks pretty close. The sum of the forces will equal ma for the block. The forces are gravity down, ramp pushing back at an angle, and friction retarding the acceleration. I prefer to keep the FBD in the original orientation, though, but that's up to your preference. (Well, if a human is grading a test paper and trying to understand your work, they would probably prefer the traditional orientation too...)
 
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  • #3
MysticDude
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That looks pretty close. The sum of the forces will equal ma for the block. The forces are gravity down, ramp pushing back at an angle, and friction retarding the acceleration. I prefer to keep the FBD in the original orientation, though, but that's up to your preference. (Well, if a human is grading a test paper and trying to understand your work, they would probably prefer the traditional orientation too...)
Well, do you have any hints?

I remember learning that Fnet is all of the forces acting on a body combined. That is why I thought that I was going to have to add up all of the vector forces. Isn't that right?
 
  • #4
Delphi51
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Is something pulling the block down the ramp (other than gravity)?
If not, you should omit the applied force vector.
I would use the unrotated diagram, and I would show the mg vector split into a component along the ramp plus a component into the ramp. The component into the ramp is exactly canceled by the normal force (since there is no acceleration in that direction).

The net force is the sum of the other forces and the F in F = ma.
 
  • #5
MysticDude
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Is something pulling the block down the ramp (other than gravity)?
If not, you should omit the applied force vector.
I would use the unrotated diagram, and I would show the mg vector split into a component along the ramp plus a component into the ramp. The component into the ramp is exactly canceled by the normal force (since there is no acceleration in that direction).

The net force is the sum of the other forces and the F in F = ma.
I only have the information which the question stated, so I guess you are right and I should take out the applied forces vector. You kind of confused me with the mg vector component part.

In your last part, should I use F - f = ma? If I do, then my equation would be ma - μkmgcos(θ) = ma. Which means μkmgcos(θ)= 0. I'm not sure what to do for this question as I'm confused big time.
 
  • #6
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Yeah, get rid of the applied force, and your mg X component will take care of that role for you.

Your equation shouldn't include F as one of its forces, so that's what's messing it up. :)
In the X direction, is Fnet the same as the F in F=ma?
 
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  • #7
Delphi51
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should I use F - f = ma
Not as you have interpreted it, anyway. You need
component of mg down the ramp - friction force = ma
 
  • #8
MysticDude
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Not as you have interpreted it, anyway. You need
component of mg down the ramp - friction force = ma
So what you are saying is (mg)cos(θ) - μkmg = ma which is mg(cos(θ) - μk) = ma. OR is it mgcos(θ)(1 - μk) = ma?
 
  • #9
MysticDude
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So, is this correct?
 
  • #10
Delphi51
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Maybe (mg)cos(θ) - μkmg*sin(θ) = ma
Has to be sin(θ) in one place, cos in the other. No diagram, can't tell what you are calling θ.
 
  • #11
MysticDude
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Maybe (mg)cos(θ) - μkmg*sin(θ) = ma
Has to be sin(θ) in one place, cos in the other. No diagram, can't tell what you are calling θ.
Okay sorry about that. If we are using Normal diagram, in other words, the way things are without me changing the axes, would it then be (mg)sin(θ) - μkmg*cos(θ) = ma?
 
  • #12
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i would recommend drawing a final FBD breaking mg into components making the mg cos theta and mg sin theta
 
  • #13
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I too would draw a final FBD so that you can take inti account your additional x and y forces representing sin theta and cos theta and from there you can set your forces equal to one another
 

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