Expert: Expert - 3/1/2006

Question
Hello. I wonder if you could help me with a problem I’m having  grasping the concept of the normal force.

If I put a 5 newton object on a table,  as I understand it the object will, due to gravity, press DOWN on the table with a 5 newton force, and because of the 3rd law the table will press UP on the object (i.e., in the exact opposite direction to the force imposed on the table)  with a 5 newton force. So far so good.

If you put the object on an inclined plane, it seems to me that gravity still causes the object to impose a directly downward force on the plane, which should react by imposing an upward force on the object. But all the texts and websites I’ve looked at show a “normal force� imposed by the plane on the object which is not upward but, rather, perpendicular to the plane, and thus at an angle (other than 180 degrees!) to the downward force imposed on the plane. I have not been able to figure out why this does not contravene Newton's 3rd law.

Thanks.
David  

Answer
If you don't understand how one vector can be broken into two parts, then the following (indeed, any explanation I can give) will make no sense.

This image explains it best

http://en.wikipedia.org/wiki/Image:Freebodydiagram1.png

In my explanation (o) means the "theta" symbol.

You correctly note that the weight force W is operating on the block, as is the normal force N. But also note what happens when we break W into two components.

Wsin(o) is that portion of W pushing down on the inclined plane. It is that portion of the force that the normal force "reacts" to, exactly cancelling Wsin(o), and thus preventing the block from going through the inclined plane.

Wcos(o) is the other portion of W, that is pushing the block parallel to the inclined plane. If this force on the block is not opposed by any other force (friction, for example), the block will accelerate down the inclined plane.