Coefficient of friction and static friction

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The discussion centers on the relationship between static and dynamic friction when an object slides down a slope at a constant speed. It clarifies that static friction applies when two bodies are stationary relative to each other, while dynamic friction applies during relative motion. The calculations presented suggest that the coefficients of static and dynamic friction can be equal under specific conditions, particularly when the slope angle balances the forces involved. However, in general, static friction tends to be greater than dynamic friction, and the two coefficients are typically different. The conversation concludes that while there are exceptions, the coefficients are not universally the same across all scenarios.
p.tryon
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If an object slides down a slope at a constant speed is the coefficient of static friction the same as the coefficient of dynamic friction? If yes, is this true in every situation?
 
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p.tryon said:
If an object slides down a slope at a constant speed is the coefficient of static friction the same as the coefficient of dynamic friction? If yes, is this true in every situation?
I'm afriad that I don't understand the question. Static friction occurs when two bodies are stationary relative to each other, whereas kinetic friction occurs when the two bodies are in relative motion.

When two bodies moving relative to each other are in contact, there is said to be no static friction.
 
Imagine a box of weight W sliding at a constant speed down a slope at an angle of 45 degrees. Imagine the same box at rest on the same slope. The two types of friction are different (static and kinetic) and the two coefficients of friction should be different (or should they?)

According to the following calculations they are the same

From here on:

Coefficient of friction C
Frictional force F
Normal Force Fn

F = C.Fn
Therefore
C = F / Fn

1. At rest:

F = WSin45 and Fn = Wcos45
Therefore

C = WSin45 / Wcos45

2. Sliding at a constant speed:

Since F is equal to the opposite component of gravity (WSin45) - Newton's first law! the equation and therefore answer is exactly the same for the coefficient of dynamic friction:

C = WSIn45 / Wcos45
 
p.tryon said:
Imagine a box of weight W sliding at a constant speed down a slope at an angle of 45 degrees. Imagine the same box at rest on the same slope. The two types of friction are different (static and kinetic) and the two coefficients of friction should be different (or should they?)
In general, they would be different.

According to the following calculations they are the same

From here on:

Coefficient of friction C
Frictional force F
Normal Force Fn

F = C.Fn
Therefore
C = F / Fn
The relationship F = μN is only true for kinetic friction.

For static friction, μN represents the maximum value, so F ≤ μN. Big difference!
 
p.tryon said:
If an object slides down a slope at a constant speed is the coefficient of static friction the same as the coefficient of dynamic friction?
This would only imply that the slope angle and coefficient of dynamic friction were "balanced" such that they resulted in equal and opposing forces from gravity and dynamic friction (therefore no acceleration), and that dynamic friction was independent of speed within the speed range experienced, excluding a speed of zero. The constant speed (not including zero speed) case occurs when tan(slope angle) = coefficient of dynamic friction.

If the speed were zero, then the friction would be due to a normally higher coeffcient of static (than dynamic) friction, and the slope could be increased without the box moving. If the box was pushed, then the box would accelerate down the steeper slope instead of sliding at constant speed.

There are pairs of surfaces where static and dynamic friction are about the same, such as teflon on teflon.
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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