Pushing a Lawnmower: Find tanθ_critical

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SUMMARY

The discussion focuses on determining the critical angle, θ_critical, for pushing a lawnmower with a handle at an angle θ, where the force exerted, F_h, becomes infinitely large. The equation derived for F_h is F_h = (-μw)/(-cosθ + sinθμ). The critical angle occurs when the denominator equals zero, leading to the relationship cosθ_critical = μsinθ_critical. Consequently, the expression for tanθ_critical is established as tanθ_critical = 1/μ.

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Homework Statement


Consider a lawnmower of weight w which can slide across a horizontal surface with a coefficient of friction [tex]\mu[/tex]. In this problem the lawnmower is pushed using a massless handle, which makes an angle theta with the horizontal. Assume that [tex]F_{h}[/tex], the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the positive 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, [tex]F_{h}[/tex] of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

B:


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


2. The attempt at a solution

okay, so i got part A fine:


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

multiply the y equation by [tex]\mu[/tex] to eliminate n
-cos[tex]\theta[/tex]*[tex]F_{h}[/tex]+[tex]\mu[/tex]n=[tex]\mu[/tex]n-sin[tex]\theta[/tex][tex]F_{h}[/tex][tex]\mu[/tex]

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

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[tex]\theta_{critical}[/tex].

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:
[tex]F_{h}[/tex]= (-[tex]\mu[/tex]w)/(-cos[tex]\theta[/tex]+sin[tex]\theta[/tex][tex]\mu[/tex]

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[tex]\theta_{critical}[/tex].

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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