How to calculate angle given coefficent of friction

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    Angle Friction
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To calculate the angle at which two objects begin to slide on a platform, the relationship between static friction and normal force is essential. The coefficient of static friction (0.4) can be used in the equation Fs ≥ μFn, leading to the condition mgsinθ ≥ μmgcosθ for motion. This simplifies to tanθ = μ, allowing the angle θ to be isolated as θ = arctan(μ). The mass of the objects cancels out, indicating that the angle is independent of their weights. Understanding this relationship helps determine the critical angle for sliding without confusion over mass effects.
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Homework Statement


two objects are suspended on a platform one weighs 10 kg and the other weighs 20 kg, the coefficent of static friction is 0.4


Homework Equations



how to find angle each of the objects start to slide

The Attempt at a Solution



OK so Us=Fs/Fn
0.4= Fs/Fn
Fg=9.8(10) for the 10 kg object
using the vertical and horizontal components of the fg vector i get
Fn= ma cos(theta)
Fs= ma sin(theta)
U=fs/fn
0.4=sin(theta)
cos(theta) mass and acceleration cancel out
this is where i get stuck because i have theta over theta they should cancel out and i will be left with a broken equation :( any help would be appreciated.

Also if mass cancels out how am i supposed to get different angles for each of the boxes which weigh different amounts?
 
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You should have for motion that Fs≥μFn

so you have mgsinθ ≥ μmgcosθ. (the minimum angle of theta would be when they are equal)

Which gives you a simple relation with θ and μ.
 
rock.freak667 said:
You should have for motion that Fs≥μFn

so you have mgsinθ ≥ μmgcosθ. (the minimum angle of theta would be when they are equal)

Which gives you a simple relation with θ and μ.

i don't understand how you would isolate theta without it canceling out or being 1
 
Eats Dirt said:
i don't understand how you would isolate theta without it canceling out or being 1

Remember the identity tanθ = sinθ /cosθ ?
 
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