Pushing a Lawnmower: Find tanθ_critical

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The discussion focuses on determining the critical angle, θ_critical, at which the force required to push a lawnmower becomes infinitely large due to friction. The force exerted by the handle, F_h, is derived from the equations of motion, leading to the expression F_h = (-μw)/(-cosθ + sinθμ). It is established that F_h approaches infinity when the denominator equals zero, specifically when cosθ_critical = μsinθ_critical. Consequently, the relationship tanθ_critical = 1/μ is identified as the solution. This critical angle indicates the limit beyond which pushing the lawnmower becomes physically unfeasible.
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Homework Statement


Consider a lawnmower of weight w which can slide across a horizontal surface with a coefficient of friction \mu. In this problem the lawnmower is pushed using a massless handle, which makes an angle theta with the horizontal. Assume that F_{h}, 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://img408.imageshack.us/img408/2788/mfscf8ayr4.jpg
http://g.imageshack.us/img408/mfscf8ayr4.jpg/1/

A:
Find the magnitude, F_{h} of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

B:


The solution for F_{h} has a singularity (that is, becomes infinitely large) at a certain angle \theta_{critical}. For any angle \theta > \theta_{critical}, the expression for F_{h} will be negative. However, a negative applied force F_{h} 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}


2. The attempt at a solution

okay, so i got part A fine:


fnet=0
x:
-cos\theta*F_{h}+F_{f}
y:
n-sin\theta*F_{h}-w

multiply the y equation by \mu to eliminate n
-cos\theta*F_{h}+\mun=\mun-sin\thetaF_{h}\mu

rearrange to get fh...
F_{h}= (-\muw)/(-cos\theta+sin\theta\mu)

part b i just not sure where to start..the force becomes infinite when the denominator becomes 0... but i don't know how to express that in terms of tan\theta_{critical}.

thanks for any help
lol i see now someone had the exact same question.. no answer though..here
 
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Welcome to PF!

phthiriasis said:
F_{h}= (-\muw)/(-cos\theta+sin\theta\mu

part b i just not sure where to start..the force becomes infinite when the denominator becomes 0... but i don't know how to express that in terms of tan\theta_{critical}.

Hi phthiriasis! Welcome to PF! :smile:

(have a theta: θ and a mu: µ :smile:)

Yes, you're right …

the force becomes infinite when the denominator becomes 0 …

ie cosθcritical = µsinθcritical

so tanθcritical = … ? :smile:
 
1/µ
seems so simple now... thanks tim
 
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