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Inclined ramp problem

  • Thread starter mcstink
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  • #1
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Homework Statement



A block of mass 2.1 kg is placed on a frictionless inclined ramp of mass 20.0 kg, with the angle θ= 31o as shown below.

http://s4.lite.msu.edu/res/msu/physicslib/msuphysicslib/09_Force_and_Motion/graphics/prob91_plane2.gif

Calculate the size of the net horizontal force that must be applied to the ramp to prevent the block from accelerating with respect to the ramp?

Homework Equations



mgsin(theta) = F of the small block

The Attempt at a Solution



i've used the data to find the acceleration of the small block (5.0474). but i'm not sure where to go from there. i've tried multiplying by the mass of both block and that didn't work. anyone want to help?
 
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Answers and Replies

  • #2
Doc Al
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i've used the data to find the acceleration of the small block (5.0474).
You've calculated the acceleration (parallel to the incline) that the small block would have if it slid down a fixed frictionless ramp--but that's not relevant here. (After all, you are trying to prevent the small block from sliding down.)

Hint: Start by analyzing the forces acting on the small block.
 
  • #3
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Start with drawing a free body diagram and identifing all the forces.

Look at:

[tex]\sum F_{x}=\\w_{x}\\-f_{s}=\\0[/tex]
[tex]\sum F_{y}=\\n_{y}-\\w_{y}=\\0[/tex]
and [tex]f_{s}=\\u_{s}\\n[/tex]
 
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  • #4
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okay i'm not quite sure i know all of those abbreviations.

but i think the only forces acting on the small box is the force as it slides down the plane (from gravity) and the force from the ramp from it being pulled. the normal force cancels out with mgcos(theta) and there is no friction. how would i set this up?
 
  • #5
force from gravity is weight and it acts all the time not only as it slides down...
use Newton's II Law,
and, yes there are only 2 forces
 
  • #6
Doc Al
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but i think the only forces acting on the small box is the force as it slides down the plane (from gravity) and the force from the ramp from it being pulled.
Right. There are two forces acting on the small box: gravity and the normal force from the ramp.
the normal force cancels out with mgcos(theta) and there is no friction.
Not sure what you mean here.

Hint: The vertical component of the normal force must balance the weight of the small box, since it has no vertical acceleration.

Your first goal is to find the acceleration that the small box must have for it not to slide down the ramp.
 
  • #7
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Well,
Since: [tex]\sum F_{x}=\\w_{x}\\-f_{s}=\\0[/tex] Newtons First Law because we want no net force/accleration
[tex]f_{s}=\\w_{x}[/tex]
[tex]f_{s}=\\mgsin(\theta)[/tex]
 
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  • #8
Doc Al
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44,892
1,143
Well,
Since: [tex]\sum F_{x}=\\w_{x}\\-f_{s}=\\0[/tex] Newtons First Law because we want no net force/accleration
[tex]f_{s}=\\w_{x}[/tex]
[tex]f_{s}=\\mgsin(\theta)[/tex]
Careful about using the ramp as a reference, which I think is what you are doing here. Since the ramp is accelerating it is a non-inertial frame.
 
  • #9
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hi i'm having trouble with this question s well, though i have different numbers. but just in terms of setting it up, i was able to calculate the vertical forces acting on the small block, but what is the acceleration supposed to be?

I have

Fy=F_s - mgsin(theta)=ma

what is a supposed to be? 0? I think it should be 0, because you want zero acceleration, but then what to do with the ramp forces..?
 
  • #10
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or should the accelerations of the block and ramp be the same?
 

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