Maximum angle of inclined plane before falling off the plane

  • Thread starter Sinnaro
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  • #1
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I've been thinking about this for a model I'm devising. Assuming that you have an object of mass m, height h, coefficient of friction u, how large can you make the angle between the ground and the inclined plane. Otherwise, at what angle does the torque from the center of mass of the object overcome the force that is keeping the object on the inclined plane?
 

Answers and Replies

  • #2
Filip Larsen
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Welcome to PF!

If the incline is completely flat and the coefficient of friction is for static friction, then you can find the largest angle for which the mass wont slide by noting that the maximum force along the slide is usually modeled simply as the normal force times the coefficient, and you can equate that with the force from gravity down the incline, that is,

[tex] \mu F_n = \mu mg \cos(\alpha) = F_g = mg \sin(\alpha) \ \Rightarrow \ \tan(\alpha) = \mu [/tex]

where [itex]\alpha[/itex] is the angle and [itex]\mu[/itex] the coefficient of static friction.
 
  • #3
tiny-tim
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Welcome to PF!

Hi Sinnaro! Welcome to PF! :wink:

I'm not sure whether you're talking about sliding (which as Filip Larsen :smile: says depends on whether the tangent exceeds the coefficient of static friction), or toppling.

If it's toppling, then all that matters is whether a vertical line through the centre of mass goes outside the base. :wink:
 
  • #4
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To clarify: it is assumed that the mass is sliding down the plane. I'm looking for the angle (as the angle approaches 90 degrees) for which the mass will no longer be in contact with the plane (falls off).

Picture of what I'm talking about. Assume that the mass looks similar to my drawing (tall vertical height with wide base):

http://i.imgur.com/kIwSt.png
 

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