Friction at an angle verification

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SUMMARY

The discussion focuses on calculating the maximum coefficient of friction for boxes sliding down a 30° ramp that is 5.4 meters long, with a time constraint of 3.3 seconds. The user correctly determined the acceleration to be 0.99 m/s² using the equation 5.4 = 0.5a(3.3)². They applied the formula mgsin(θ) - μmgcos(θ) = ma, leading to a calculated coefficient of friction (μ) of 0.46. The method used is confirmed as correct by another participant in the discussion.

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warmfire540
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Hey, i just want to make sure I'm doing this right:

At the end of a factory production line, boxes start from rest and slide down a 30o ramp 5.4 m long. If the slide is to take no more than 3.3 s, what is the maximum coefficient of friction between the box and the slide?


I first found out the acceleration:
5.4=0.5a(3.3)^2
a=.99m/s^2
so i then did
mgsinTHETA-umgcosTHETA=ma (mass cancels out)
4.9-8.5u=0.99
-8.5u=-3.91
u=.46...the coefficient of friction, it seems to work..

however i still need help on my post "Tension and coefficient force revisited"
THANKS for alll ya'lls help
 
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warmfire540 said:
Hey, i just want to make sure I'm doing this right:

At the end of a factory production line, boxes start from rest and slide down a 30o ramp 5.4 m long. If the slide is to take no more than 3.3 s, what is the maximum coefficient of friction between the box and the slide?I first found out the acceleration:
5.4=0.5a(3.3)^2
a=.99m/s^2
so i then did
mgsinTHETA-umgcosTHETA=ma (mass cancels out)
4.9-8.5u=0.99
-8.5u=-3.91
u=.46...the coefficient of friction, it seems to work..

however i still need help on my post "Tension and coefficient force revisited"
THANKS for alll ya'lls help

I didn't check the numbers, but your method seems correct.
 

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