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Confusing slope question 
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#1
Dec1312, 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 ycomponent 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 ycomponent 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!!



#2
Dec1312, 05:40 AM

P: 123

I think there is confusion about the coordinates system. Since something perpendicular to the surface, is on the yaxis direction, then the coordinates system x,y is the one that xaxis is parallel to the slope and yaxis perpendicular to it. I think that you considered that the coordinates system is the one where the yaxis 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.



#3
Dec1312, 07:00 AM

P: 66

Actually, is going down the slope (a displacement in ydistance) relevant to the question? Meaning if there's no yvelocity or force how can it travel in the y axis direction?



#4
Dec1312, 05:03 PM

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Confusing slope question



#5
Dec1412, 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!!



#6
Dec1412, 02:28 PM

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#7
Dec1512, 07:44 AM

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