tracker890 Source h
No Effort - Member warned that some effort must be shown
Dear Everyone：
?

Thank you for your time and consideration.

Staff Emeritus
Gold Member
You're trying to prove that ##\sqrt{3} w/2## is the smallest value P can take? What have you done so far to try to solve this? We can help identify what the next step is easier if we see what your work is so far.

mfb
tracker890 Source h
Dear Office_Shredder
Thank you for reminding, and this problem has been solved.

Problem：
Determine the Pmin for the sys equilibrium.

sol/

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Master1022
Dear Everyone：
I know this has been solved now, but another method to do this (other than differentiation) is to notice that the denominator can be expressed as a double-angle formula. The coefficients $\sqrt 3 /2$ and $1/2$ seem to stand out to me. For example:
$$\frac{1}{2} cos\theta + \frac{\sqrt 3}{2} sin \theta = sin(\theta + \frac{\pi}{6})$$
We can now see that we want the minimum of $P$ which occurs when the denominator is a maximum. Max(sin) = 1 and thus we get the answer as required (and we can see that $\theta = \pi / 3$ for the sin() to equal 1)