Coefficient of Friction at constant speed

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SUMMARY

The discussion centers on calculating the coefficient of friction for two equal masses on a rough wedge inclined at angles of 53 and 47 degrees. The user initially derived the equation mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) to balance forces, leading to the expression sin(53) - sin(47) = (cos53 + cos47)(u). The user calculated a coefficient of friction (u) of 0.052 but questioned its accuracy, suggesting a potential correct value of μ_k = 0.098. The calculations and reasoning presented are fundamentally sound, indicating a misunderstanding rather than a flaw in the approach.

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Tensaiga
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Hello, please help me out on this:
Two equall masses lie on each side of a rough wedge, connected by a string, the wedge makes anlges 53, and 47, to the horizontal, find the coefficient of friction for the masses to move at a constant speed.

MY THINKING:
since masses are equal:
mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) ---> these are all the opposing forces, and force of fricton of each mass, since the slop is different for them.
the mg(s) cancels out. Leaving:
sin(53)-sin(47) = (cos53+cos47)(u)
then i got u = 0.052
which is wrong...
 
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I do not know what a wedge* is, but I imagine the situation which was more probable with the given data.

Is [tex]\mu_k = 0.098[/tex]? :wink:
If it is, then I will explain my thoughts. :smile:

*My english needs improvements. :approve:
 
Tensaiga said:
MY THINKING:
since masses are equal:
mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) ---> these are all the opposing forces, and force of fricton of each mass, since the slop is different for them.
the mg(s) cancels out. Leaving:
sin(53)-sin(47) = (cos53+cos47)(u)
then i got u = 0.052
which is wrong...
Your thinking looks OK to me. Why do you think it's wrong?
 

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