Normal force when slope accelerates horizontally

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
9 replies · 4K views
mystmyst
Messages
57
Reaction score
0
We have a slope, with mass M, at x degrees. Mass m is on the slope. The slope moves right at constant acceleration [tex]a_0[/tex]. 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 [tex]a_0[/tex] but I don't know how to.

Can someone please help?
Thanks!
 
Last edited by a moderator:
Physics news on Phys.org
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
 
Last edited:
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 [tex]a_0[/tex] 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 [tex]a_0[/tex] 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 [tex]a_0[/tex] 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...)