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Does the normal force of a block always equal the weight of the block? For example, a block on an incline?
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Absolutely not. In the case of an inclined plane we can impose the following coordinate system and diagram|dw:1420319540624:dw|The equations of motion will be (\(\theta\) is the angle of inclination of the floor)\[\sum F_x =mg \sin \theta - f = ma\]\[\sum F_y = N-mg \cos \theta=0\] Then it is clear that the normal force is not the weight but is \(N=mg \cos \theta\). Moreover, using the usual model for the friction force \(f=\mu N\) were \(\mu\) is the coefficient of friction, the acceleration will be not g but \[a=g( sen \theta - \mu \cos \theta)\]
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