# Pushing a lawnmower

by phthiriasis
Tags: lawnmower
 P: 2 1. The problem statement, all variables and given/known data 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. 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$$\theta$$*$$F_{h}$$+$$\mu$$n=$$\mu$$n-sin$$\theta$$$$F_{h}$$$$\mu$$ rearrange to get fh... $$F_{h}$$= (-$$\mu$$w)/(-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 dont 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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P: 26,157
 Quote by phthiriasis $$F_{h}$$= (-$$\mu$$w)/(-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 dont know how to express that in terms of tan$$\theta_{critical}$$.
Hi phthiriasis! Welcome to PF!

(have a theta: θ and a mu: µ )

Yes, you're right …

the force becomes infinite when the denominator becomes 0 …

ie cosθcritical = µsinθcritical

so tanθcritical = … ?
 P: 2 1/µ seems so simple now... thanks tim

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