How do I simplify this inequality

AI Thread Summary
To simplify the inequality Mg * abs(sin(θ) - cos(θ)) <= μMg * (cos(θ) + sin(θ)), start by dividing both sides by cos(θ). This leads to the expression sin(θ)/cos(θ) - 1 <= μ(1 + tan(θ)). Isolate tan(θ) to derive the final result, which shows that if tan(θ) >= 1, then tan(θ) <= (1 + μ) / (1 - μ). This process clarifies the relationship between the trigonometric functions and the coefficient of friction, μ. Understanding these steps is crucial for solving similar physics problems.
MeMoses
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Homework Statement


So I'm following along with my physics book and I get to the point where
Mg * abs(sin(θ) - cos(θ)) <= μMg * (cos(θ) + sin(θ)
Next they say: If tan(θ) >= 1 then
sin(θ) - cos(θ) <= μ(cos(θ) + sin(θ)) => tan(θ) <= (1+μ) / (1-μ)


Homework Equations





The Attempt at a Solution


How do I go from sin(θ) - cos(θ) <= μ(cos(θ) + sin(θ)) to tan(θ) <= (1+μ) / (1-μ)?
Could someone walk me through this or at least get me started? Thanks
 
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MeMoses said:

Homework Statement


So I'm following along with my physics book and I get to the point where
Mg * abs(sin(θ) - cos(θ)) <= μMg * (cos(θ) + sin(θ)
Next they say: If tan(θ) >= 1 then
sin(θ) - cos(θ) <= μ(cos(θ) + sin(θ)) => tan(θ) <= (1+μ) / (1-μ)

Homework Equations



The Attempt at a Solution


How do I go from sin(θ) - cos(θ) <= μ(cos(θ) + sin(θ)) to tan(θ) <= (1+μ) / (1-μ)?
Could someone walk me through this or at least get me started? Thanks
You get to do most of the walking. (Hopefully, you know by now that that's how we do things here at PF.)

Take
\sin(\theta) - \cos(\theta) \le \mu(\cos(\theta) + \sin(\theta))​
and divide both sides by cos(θ) .

Simplify, and the only trig function remaining is tan .

Now, try to isolate tan(θ).
 
Thanks, that's all I needed
 
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