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Confusing slope question

by SecretSnow
Tags: confusing, slope
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SecretSnow
#1
Dec13-12, 05:05 AM
P: 66
Hi guys, I'm very confused by a slope question...I've read that for a person with mass m on an inclined frictionless slope, it's normal force perpendicular to the surface is always equal to the y-component of the weight. Why? I'm always thinking that since the person slides down, he can also be accelerating downwards which means there's a net resultant y-component force? Or is that impossible because at any point the net y force should be zero? Why then? Is that an assumption that the person never accelerate downwards? Thanks!!
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#2
Dec13-12, 05:40 AM
P: 123
I think there is confusion about the coordinates system. Since something perpendicular to the surface, is on the y-axis direction, then the coordinates system x,y is the one that x-axis is parallel to the slope and y-axis perpendicular to it. I think that you considered that the coordinates system is the one where the y-axis is in the direction of gravity field, but this is not the case. Anyway, it doesn't matter what coordinates system you use in order to analyze the problem; actually there is no need to set a coordinates system. Just draw all the forces acting on the body, considering that the gravity force component that is perpendicular to the slope, is equal to the normal force. This should happen in order to remain always on the slope because if the perpendicular forces didn't cancel each other, then there would be perpendicular displacement.
SecretSnow
#3
Dec13-12, 07:00 AM
P: 66
Actually, is going down the slope (a displacement in y-distance) relevant to the question? Meaning if there's no y-velocity or force how can it travel in the y axis direction?

haruspex
#4
Dec13-12, 05:03 PM
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Confusing slope question

Quote Quote by SecretSnow View Post
Actually, is going down the slope (a displacement in y-distance) relevant to the question? Meaning if there's no y-velocity or force how can it travel in the y axis direction?
You're defining the y-axis as parallel to the slope? If so, the normal force is along the x-axis, and will be the x-component of g. If not, please clarify your axes.
SecretSnow
#5
Dec14-12, 08:09 AM
P: 66
I took my axes with x parallel to the slope surface. Even if I took the x to be entirely horizontal and a vertical y axis, the ball will still be descending in y right? Thanks!!
haruspex
#6
Dec14-12, 02:28 PM
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Quote Quote by SecretSnow View Post
I took my axes with x parallel to the slope surface.
Then I'll interpret your OP in that light:
since the person slides down, he can also be accelerating downwards which means there's a net resultant y-component force?
Your y direction is normal to the surface. Since the person neither leaps off the surface nor sinks into it, there's no movement in the y direction, so no acceleration in that direction either.
SecretSnow
#7
Dec15-12, 07:44 AM
P: 66
Quote Quote by haruspex View Post
Then I'll interpret your OP in that light:

Your y direction is normal to the surface. Since the person neither leaps off the surface nor sinks into it, there's no movement in the y direction, so no acceleration in that direction either.
Oh I got it already!!! There is no y gain or decrease cuz the x axis is parallel to the slope!! Thanks a ton bro!!!


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