Calculate Normal Force in Mass M & m Slope System with no Friction

AI Thread Summary
To calculate the normal force between mass M on a slope and mass m resting on it, the effect of the slope's acceleration (a_0) must be considered due to the absence of friction. The normal force is not simply N = mgcosx; the fictitious force resulting from the slope's movement must be included in the free body diagram. This fictitious force acts in the opposite direction of the slope's acceleration and should be represented in both x and y components. Properly incorporating this force allows for the correct application of Newton's laws in a non-inertial frame. Understanding these dynamics is crucial for accurately determining the normal force in this scenario.
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We have a slope, with mass M, at x degrees. Mass m is on the slope. The slope moves right at constant acceleration a_0. There is no friction between M and m, and no friction between M and ground.

Determine the Normal Force between M and m.

[PLAIN]http://img638.imageshack.us/img638/620/imaginaryforces2.png

The Attempt at a Solution



I'm a bit confused with this because the slope is moving.
I would just think the answer would be N = mgcosx.
But since the slope is moving, I have to take into consideration a_0 but I don't know how to.

Can someone please help?
Thanks!
 
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Hint: Use Pseudo force

Edit: For using Pseudo(fictitious) force, you have acc. of incline

These forces come into action when you need to apply Newton's laws in non inertial frame(i.e. a is not 0 of frame)

In this case it will be -mao, where - sign indiate that its direction is opposite to ao
 
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hi mystmyst! :smile:
mystmyst said:
I'm a bit confused with this because the slope is moving.
I would just think the answer would be N = mgcosx.
But since the slope is moving, I have to take into consideration a_0 but I don't know how to.

well, you know from the question that a fictitious force is involved …

what is that fictitious force in this case?

draw it on the free body diagram, and proceed as normal …

fictitious forces go on the fbd just like ordinary forces … that's the point! :biggrin:
 
tiny-tim said:
hi mystmyst! :smile:


well, you know from the question that a fictitious force is involved …

what is that fictitious force in this case?

draw it on the free body diagram, and proceed as normal …

fictitious forces go on the fbd just like ordinary forces … that's the point! :biggrin:

The fictitious force is a_0 but I don't know how to draw it on a fbd. Am I supposed to split it into x and y coordinates (I'm using the slope as the x-axis...)
 
as the acceleration of incline is along +X ... so this fictitious force will be along -X

as i told you ...


cupid.callin said:
In this case it will be -mao, where - sign indiate that its direction is opposite to ao
 
write all other forces like gravity, normal rxn.. and also include this force in your FBD
 
mystmyst said:
The fictitious force is a_0 but I don't know how to draw it on a fbd. Am I supposed to split it into x and y coordinates (I'm using the slope as the x-axis...)

a0 is horizontal, so you draw it horizontal

if you're using coordinates (you could use trig instead), then yes you'll have to split it into components along the axes you've chosen :smile:
 
Taking X along incline will be easier!
 
cupid.callin said:
Taking X along incline will be easier!

cupid.callin, he is taking X along the incline :frown:
mystmyst said:
Am I supposed to split it into x and y coordinates (I'm using the slope as the x-axis...)
 
  • #10
OH! sorry :bugeye:
 
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