Finding minimum value of force P given static friction

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The discussion focuses on determining the minimum force P required to move a crate on a horizontal surface, given the coefficient of static friction us and an angle theta. The formula for P is derived as P = (us*Fg*sec(theta)) / (1 - us*tan(theta)). While the initial part of the problem was solved, the user struggled with finding the minimum value of P when us is 0.400. A suggested approach involves differentiating the expression for P with respect to theta and setting it to zero to find the minimum. The conversation emphasizes simplifying the resulting equation to facilitate solving for the angle that minimizes P.
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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?
 
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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.
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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