# Help w/ this force problem please

1. Oct 19, 2004

### viendong

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
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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 $$tan(\theta) critical$$

Last edited: Oct 19, 2004
2. Oct 19, 2004

### BLaH!

The force will become infinitely large whenever the denominator of your expression for the force goes to zero. Thus, the singularity occurs at

$$\mu \sin \theta - \cos \theta = 0$$

3. Oct 19, 2004

### viendong

I'm not quite understand how $$\mu \sin \theta - \cos \theta = 0$$
relate w/ $$tan(\theta critical)$$ ?

4. Oct 20, 2004

### viendong

hi, i think i got it . it's 1/mu after all :)