Help w/ this force problem please

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The discussion revolves around solving a physics problem involving a lawnmower being pushed at an angle with friction. The user successfully derived the force required to slide the lawnmower at constant speed but encountered difficulties with the critical angle where the force becomes infinitely large. The critical angle is determined by the condition that the denominator of the force equation equals zero, leading to the equation μ sin(θ) - cos(θ) = 0. This relationship simplifies to finding tan(θ critical) as 1/μ. The user ultimately confirms their understanding of the critical angle's relationship to the coefficient of friction.
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hi, I've approach the first question, but i don't know how to start w/ this question. I already have my body diagram and everything. Could someone tell me how ?
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Consider a lawnmower of weight which can slide across a horizontal surface with a coefficient of friction . In this problem the lawnmower is pushed using a massless handle, which makes an angle with the horizontal. Assume that , the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward.
http://session.masteringphysics.com/probhtml/MFS.cf.8_a.jpg
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The first question...
Find the magnitude Fh , of the force required to slide the lawnmower over the ground at constant speed by pushing the handle. Express the required force in terms of given quantities.
>>> For this I got
Fh=(-(\mu*w))/(-cos(\theta)+\mu*sin(\theta))
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This question I've problem with
The solution for Fh has a singularity (that is, becomes infinitely large) at a certain angle theta critical . For any angle \theta>\theta critical , the expression for Fh will be negative. However, a negative applied force Fh would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all.

Find an expression for tan(\theta) critical
 
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The force will become infinitely large whenever the denominator of your expression for the force goes to zero. Thus, the singularity occurs at

\mu \sin \theta - \cos \theta = 0
 
I'm not quite understand how \mu \sin \theta - \cos \theta = 0
relate w/ tan(\theta critical) ?
 
hi, i think i got it . it's 1/mu after all :)
 
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