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I'm confused about the formula for the force of friction on inclined surface.

It says that $f$ (force of friction) = $\mu N$ (where $\mu$ is the coefficient of the friction).

Is: $$f = mg\cos(a)\mu~?$$

So basically, is Normal force on inclined surface equal to the y component of the gravity force?

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Yes but no.

The normal force on an inclined surface is the component of gravity perpendicular to the surface. That is not the same thing as the $y$ component of the gravity force, unless your coordinate system has $x$ along the surface and $y$ perpendicular to it (which you did not specify).

The equation you wrote is correct, if $a$ is the angle of the incline to the horizontal.

enter image description here

Consider adding a diagram to your question if this does not clear it up for you.

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  • $\begingroup$ Hey, so if i chose the coordinate system along the surface and there is a friction between the box and surface, would the normal force still be the y component of the mg? $\endgroup$ – student123 Oct 21 '15 at 18:21
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We are taking components of the weight of the block. One is perpendicular to the inclined surface and another is parallel. At equilibrium, the normal reaction is equal to the cosine component and the sine component = the frictional force. Now if you take X and Y axes in the perpendicular and parallel to the inclined surface(see the picture) then Normal force=the X component..enter image description here

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The µ part only comes in as the maximum. If you find that the frictional force necessary to keep the system at equilibrium is more than µFn, then the frictional force will be µs Fn (kinetic frictional coefficient), and so your system will be at equilibrium.

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