# Force decomposition

1. Aug 11, 2010

### ErikD

I can decompose W to get Fn and -Fr.

Fn = W cos($$\theta$$)
-Fr = W sin($$\theta$$)

I know I'm allowed to decompose forces like that but I'm a bit confused as to why. Cause |Fn| + |Fr| > |W| (for the angle in this picture). So by decomposing W am I not introducing force that isn't there?

Last edited by a moderator: May 4, 2017
2. Aug 11, 2010

### Staff: Mentor

I don't quite understand your diagram. What's Fr? Is the block in equilibrium?

You can certainly write any vector in terms of its components.

I don't understand the significance of adding the magnitudes of these components. The sum of the magnitudes of the components will be greater than the magnitude of the vector itself. So?
No. You're just describing the same force in a different way. Nothing's been added or removed.

3. Aug 11, 2010

### ErikD

Sorry I should have been a bit more clear. Yes the block is in equilibrium. Fr is the force of friction. Fr = W sin($$\theta$$) so the block isn't moving.

Isn't the magnitude of a force the amount of newtons the force is strong? So don't the magnitudes of the components have more newtons than the force itself?

4. Aug 11, 2010

### Staff: Mentor

OK.

Sure.

You mean "Is the sum of the magnitudes of the components greater than the magnitude of the force itself?" Sure! So what?

Note that the components are in different directions--they are perpendicular to each other--so adding the magnitudes has no special meaning. Only adding them as vectors has any meaning.

The moral is that you must treat force as a vector, not a scalar. Given two vectors (in this case, the components of the weight) you must add them as vectors to find the total. You can't just add the magnitudes. Direction matters! (Take two 10 N forces. Depending on their directions, the total of those forces can be anything from 0 N to 20 N. Adding the magnitudes only makes sense if they point in the same direction.)

5. Aug 11, 2010

### ErikD

Thanks, that clears up my confusion.