Friction question, maximum incline

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SUMMARY

The discussion focuses on determining the maximum incline angle (alpha) for a 50 kg object resting on a rough plane, supported by a string with a breaking strain of 200 N and a coefficient of friction of 0.2. The equation derived from resolving forces parallel to the incline is 50g sin(alpha) - 10g cos(alpha) = 200. The solution involves expressing the left-hand side as a single trigonometric function to find the critical angle where the string remains intact.

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Homework Statement



An object of mass 50kg rests on a rough plane incline at an angle alpha to the horizontal. It is supported in this position by a light string parallel to the plane. The string has a breaking strain of 200N and the coefficient of friction between the object and the plane is 0.2. Find the largest value of alpha for the string to remain intact.

The Attempt at a Solution



Resolving parallel to the plane:

50gsin(alp) = 200 + (0.2x50gcos(alp))
= 200 + 10gcos(alp)

So it seems that I have to find the maximum value of alpha for which the above equation is true, but I have no idea how to do that.

Any help would be much appreciated. Thanks :)
 
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why not take sin(α)=x and cos(α)=√(1-x2)
 
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You can rewrite as $$50g \sin \alpha - 10 g \cos \alpha = 200$$ and express the LHS as one trig function and a corresponding phase.
 

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