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viendong
Oct19-04, 06:52 PM
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.
picture (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

BLaH!
Oct19-04, 11:42 PM
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

viendong
Oct19-04, 11:49 PM
I'm not quite understand how \mu \sin \theta - \cos \theta = 0
relate w/ tan(\theta critical) ?

viendong
Oct20-04, 10:01 AM
hi, i think i got it . it's 1/mu after all :)