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