Block on a frictionless, accelerating, inclined plane

  • Thread starter chronos98
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  • #26
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I think this is a poorly worded question. I don't see where it says the acceleration acts (at least not clearly).
 
  • #27
rl.bhat
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A 2.0 kg block rests on a frictionless wedge that has a 60 deg incline and an acceleration LaTeX Code: \\vec{a} to the right

In the problem it is clearly stated that the wedge is accelerating to the right. So there must be a force acting on the wedge.
 
  • #28
Well, I tried the problem my self, and as my instructor had mentioned in class, I took the direction of motion (i.e. direction of acceleration) to be the positive x-axis.

The weight of the 2.0 kg block is PERPENDICULAR to the direction of motion, but points downwards, so 'mg' is in the -ve y-direction.

If you follow the co-ordinate system mentioned here, you will have to resolve just the normal force Fn into its components.

Fn (along +ve-x axis) = Fn (cos 60)
Fn (along +ve-y axis) = Fn (sin 60)

Now, trying to sum up the forces along the +ve x-co-ordinate system,



{EQUATION 1} Total F(x) = Fn (cos 60) = Force applied F (ma)




Also, summing up the forces along the +ve y-co-ordinate system,




{EQUATION 2} Total F(y) = Fn (sin 60) - mg = 0






This F(y) is EQUAL to ZERO, since there is no acceleration in the y-direction.

On dividing equation 1 by equation 2, you get:


1/ [tan (60)] = (a/g)




The final answer comes out to be a = 5.66 m/s^2


I think it makes more sense if you draw out the FBD ... I could be wrong, but that is the only way it made some sense to me. :)
 

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