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Block on inclined plane

  • Thread starter Parallel
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  • #1
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hello,i'm having some trouble with this problem.

a block with mass m,rests on a frictionless plane,inclined at angle alpha relative to the horizon.

what is the acceleration 'a' of the plane,if we dont want the block to slide down the plane?

here's what i've tried:
the x axis will be parallel to the plane,so

x axis:
mgsin(/alpha) = macos(alpha)
==>a= g tan(alpha)

Is this o.k?

because it looks weird,if I dont want the block to slide,why am I "saying" in the equation it does?
or maybe because it's relative to an inertial frame?

thanks
 
Last edited:

Answers and Replies

  • #2
radou
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It looks ok. In order for the block to rest, another force must act on in in the direction of the incline to balance the component of the weight along the incline. This very force must come from the acceleration of the block.

You can think about your result, too. If you put alpha = 0, then you get a = 0, which makes sense, doesn't it?
 
  • #3
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o.k

so now i'm thinking what if alpha=90 deg.

tangent doesnt like pi/2 :)
 
  • #4
radou
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Yes, because it's undefined at pi/2. You could take a limit when alpha -> Pi/2, and you'd get infinity. Which would mean you'd need an infinite acceleration. :biggrin:
 
  • #5
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are you saying that,if I put a block on a vertical plane,there's no way I could keep this block at rest,by accelerating the plane?

thanks for your help
 
  • #6
radou
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Parallel said:
are you saying that,if I put a block on a vertical plane,there's no way I could keep this block at rest,by accelerating the plane?

thanks for your help
Nope, at least not in this world.
 
  • #7
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do you have an intuitive explanation for this phenomena?
 
  • #8
radou
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The only explanation you need is that gravity acts downwards all the time, and to balance it, you need a vertical force which acts in the opposite direction, which can't be accomplished if the force has no vertical component, i.e. if it only acts in the horizontal direction.
 
  • #9
42
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I got it.

thanks alot
 

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