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A particle mass $m$ moves down a frictionless inclined plane. Does a normal force cancels a normal component of $m$ g?

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Thanks. I am ok with drawing a diagram and replacing a force by its components, but I am not sure about intensity of a normal force. I have read in the article you attached that normal component of weight balances a normal force. Does it mean that $F_{normal}$ and normal component has equal intensity, but opposite directions? (I am not a physics student, so I don't know much about these stuffs). –  user23709 Jun 28 '13 at 16:15
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marked as duplicate by Manishearth Jul 12 '13 at 5:47

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Think about the free body representation. If you have a mass on the inclined plane at some angle $\theta$, the weight (mg) is in the -y direction. The normal force points NORMAL to the surface, which means it will have both an x and a y component. Through some simple trig, we can see that $F_{normal}=mg cos(\theta)$. This is not completely mg, because the other component ($mg sin(\theta)$) is the force responsible for the mass sliding down the inclined plane.

So I am not completely sure what you mean by a normal force cancels a normal component of m g but there are different components to each.

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I have inclined plane at angle $\theta$ to the surface. I took a coordinate system such that inclined plane lies on x-axes. Then I wrote $m$ g as $m$ g $=mg\sin \theta$ i $-mg\cos \theta$ j. If I take such coordinate system, $F_{normal}$ would have a direction of $y$-axes, right? What I meant by "cancels" is that I have written that $m$ g $+$ N $=mg\sin \theta$ i, which made me conclude that $F_{normal} =mg \cos \theta$ j, i.e. that it is equal to normal component of $m$ g, but opposite direction. –  user23709 Jun 28 '13 at 15:42
    
That is all correct –  yankeefan11 Jun 28 '13 at 16:46
    
Does it ALWAYS hold that $F_{normal}$ has equal intensity as normal component has, but it has the opposite direction? –  user23709 Jun 28 '13 at 16:52
    
Well, the $F_{Normal}$ does not have the same intensity as the weight. If that we the case, the block would not slide down. I assume you mean by has equal intensity as normal component has that it equals the weight? This is not true. F normal is the normal component so yes(?) they are equal –  yankeefan11 Jun 28 '13 at 17:09
    
No, I don't mean that it is equal to weight but to normal component of weight. If they are equal (which I can see from the problem I wrote above, and you said it was all correct), there is still another component of the weight that force the block to move. –  user23709 Jun 28 '13 at 17:14
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