How does force decomposition help us understand the physics of friction?

AI Thread Summary
Force decomposition allows for a clearer understanding of friction by breaking down the weight (W) into its components: the normal force (Fn) and the frictional force (Fr). While the magnitudes of these components may seem to exceed the original force, this does not imply the introduction of additional forces; rather, it is a different representation of the same force. The block in question is in equilibrium, meaning that the forces acting on it balance out, with Fr equating to W sin(θ). It's crucial to treat forces as vectors, as their directions affect the resultant force, and simply adding magnitudes without considering direction can lead to confusion. Understanding this distinction clarifies how force decomposition aids in analyzing frictional forces in physics.
ErikD
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[PLAIN]http://upload.wikimedia.org/wikipedia/en/7/7a/Friction.png

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?
 
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ErikD said:
I can decompose W to get Fn and -Fr.

Fn = W cos(\theta)
-Fr = W sin(\theta)
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 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).
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?
So by decomposing W am I not introducing force that isn't there?
No. You're just describing the same force in a different way. Nothing's been added or removed.
 
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?
 
ErikD said:
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.
OK.

Isn't the magnitude of a force the amount of Newtons the force is strong?
Sure.

So don't the magnitudes of the components have more Newtons than the force itself?
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.)
 
Thanks, that clears up my confusion.
 

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