What is the Relationship Between Mu and Theta in the Inclined Plane Problem?

[tricky_tick]

empty post

empty post

 

[marlon]

i chose d

marlon

 

[marlon]

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
-mg\sin(\theta)*e_{x} - mg\cos(\theta)*e_{y}

normal force
N * e_{y}

friction
-{\mu}N*e_{x}

pull-force
F*e_{x}

The sum of y-componets yield :

N = mg\cos(\theta)

The sum of x-componets yield :

-{\mu}N -mg\sin(\theta) + F = 0

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

or F = mg(\mu\cos(\theta) + \sin(\theta))

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

0 = mg(-{\mu}\sin(\theta) + \cos(\theta))
or
0 =-{\mu}\sin(\theta) + \cos(\theta)

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

marlon.

 

[tricky_tick]

Answer d yields a maximum value of Work. We are looking for a minimum value.

 

[marlon]

Answer d yields a maximum value of Work. We are looking for a minimum value.

please prove your statement...

marlon

 

[marlon]

please prove your statement...

marlon

Besides i disagree because the first derivative is zero in \mu = \cot(\theta)

Yet if \mu > \cot(\theta) then thefirst derivative is negative
Yet if \mu < \cot(\theta) then the first derivative is positive

So the mu corrsponds to a minimal value here. You are mixing extrema with maxima. An extremum is a general name for both local max and min values of a function
marlon