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marlon
#3
Nov2-04, 05:19 PM
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P: 4,006
Quote Quote by tricky_tick
can you please explain how you arrived at that conclusion?
You have four forces. Suppose the x-acis is along the ramp and the force F for pulling the object is aligned along this x-axis. Y-axis perpendicular to x-axis.

gravity
[tex]-mg\sin(\theta)*e_{x} - mg\cos(\theta)*e_{y}[/tex]

normal force
[tex] N * e_{y} [/tex]

friction
[tex]-{\mu}N*e_{x} [/tex]

pull-force
[tex]F*e_{x} [/tex]

The sum of y-componets yield :

[tex] N = mg\cos(\theta) [/tex]

The sum of x-componets yield :

[tex] -{\mu}N -mg\sin(\theta) + F = 0 [/tex]

So we have that [tex] F = {\mu}mg\cos(\theta) + mg\sin(\theta) [/tex]

or [tex] F = mg(\mu\cos(\theta) + \sin(\theta)) [/tex]

let's take the derivative with respect to the angle theta and set this equal to 0. Thus we get :

[tex] 0 = mg(-{\mu}\sin(\theta) + \cos(\theta)) [/tex]
or
[tex] 0 =-{\mu}\sin(\theta) + \cos(\theta) [/tex]

or
[tex] \mu = \frac {\cos(\theta)}{\sin(\theta)} = \cot(\theta)[/tex]
Thus [tex]\tan(\theta) = \frac {1}{\mu}[/tex]

marlon.