Finding minimum value of force P given static friction

[Zipzap]

Homework Statement

A crate of weight Fg is pushed by a force P on a horizontal
floor. (a) If the coefficient of static friction is us and P
is directed at angle theta below the horizontal, show that the
minimum value of P that will move the crate is given by

P = ( us*Fg*sec(theta)) / (1 - us*tan(theta))


(b) Find the minimum value of P that can produce motion
when us " 0.400

Homework Equations

The Attempt at a Solution

All right, I managed to do (a) on my own, but I'm dead stuck on (b). I've only gotten as far as inputting us = 0.400 in, and all I know is that I need to find an angle such that I get a minimum for P. I'm unsure as to how to solve this problem without inputting a bunch of angles and using trial-and-error to get my answer. Can anyone help me?

 

[Pagan Harpoon]

Differentiate your expression for P with respect to Theta and set it to 0, find the minimum in the usual way. I think that will work.

 

[Zipzap]

I tried that, and then I got some insane result that I have no idea how to solve:

http://www.wolframalpha.com/input/?i=derivative+of+(0.4secx)+/+(1-0.4tanx)

That's what I get, and I'm once again stuck from here...

 

[Pagan Harpoon]

You will need to solve that equation. It's not as bad as it looks. Straight away you can simplify it by multiplying top and bottom of the second fraction by 1-0.4tan(x), then the denominators for both are the same, so you can throw them away and you have:

$$\frac{0.16}{cos^{3}(x)}+\frac{0.4sin(x)}{cos^{2}(x)}-\frac{0.16sin^{2}(x)}{cos^{3}(x)}=0$$

You can solve this to get the root you want in a simpler form than it is given on the WolframAlpha page.

Keep in mind that 1-sin^2(x)=cos^2(x), this property is useful in the process of solving the equation.

 

[rwisz]

To find the minimum value of P for a coefficient of static friction of 0.400, we can use the equation from part (a) and set it equal to P. This will give us an equation in terms of only theta, which we can then solve for to find the minimum angle.

P = ( us*Fg*sec(theta)) / (1 - us*tan(theta))

Substituting us = 0.400, we get:

P = (0.400*Fg*sec(theta)) / (1 - 0.400*tan(theta))

To find the minimum value of P, we need to find the minimum value of theta that will give us a positive P. This means we need to find the maximum value of the denominator, since the numerator will always be positive.

Setting the derivative of the denominator equal to zero and solving for theta, we get:

d/dtheta (1 - 0.400*tan(theta)) = 0

-0.400*sec^2(theta) = 0

sec^2(theta) = 0

theta = pi/2

So, the minimum value of P for a coefficient of static friction of 0.400 occurs when theta = pi/2, or when the force P is perpendicular to the floor. Plugging this value into the equation, we get:

P = (0.400*Fg*sec(pi/2)) / (1 - 0.400*tan(pi/2))

P = 0.400*Fg*sec(pi/2)

P = 0.400*Fg*infinity

Since sec(pi/2) is infinite, we can see that the minimum value of P is also infinite. This means that the crate will not move unless the force P is greater than infinity, which is impossible. Therefore, there is no minimum value of P for a coefficient of static friction of 0.400 and the crate will not move.