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

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SUMMARY

Force decomposition is essential for understanding the physics of friction, particularly in analyzing the equilibrium of a block subjected to gravitational force (W). The normal force (Fn) is calculated as Fn = W cos(θ), while the force of friction (Fr) is expressed as Fr = W sin(θ). Decomposing the weight vector does not introduce additional forces; rather, it provides a clearer representation of the forces acting on the block. The discussion emphasizes that forces must be treated as vectors, where direction is crucial for accurate calculations.

PREREQUISITES
  • Understanding of vector decomposition in physics
  • Familiarity with Newton's laws of motion
  • Knowledge of trigonometric functions (sine and cosine)
  • Concept of equilibrium in mechanics
NEXT STEPS
  • Study vector addition and subtraction in physics
  • Learn about the applications of Newton's laws in real-world scenarios
  • Explore the role of friction in different materials and surfaces
  • Investigate advanced topics in mechanics, such as dynamics and kinematics
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Students of physics, educators teaching mechanics, and engineers involved in designing systems where friction plays a critical role will benefit from this discussion.

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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