Friction, object pulled at an angle, given only (mue) and max tension hmmm

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Having difficulties with this, I assume that something's supposed to cancel out at some point, but I must be missing something important. If someone could point em in the right direction... Thanks for all your help, once again!

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Any ideas?
 

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Very interesting problem. Your initial thought patterns are the same as mine.

I would suggest you try using calculus. I'm not certain this will work though, I'm quite tired.

EDIT: Okay, just worket it through. You've said that Tcos(theta)=muFn.
Rearrange to get Fn.
Substitute this into Fn + Tsin(theta) = mg.

Differentiate mg with respect to theta and let it equal to 0 at maximum.
This should get you the required angle.

What is the correct answer?


BTW, there are probably easier methods, but this is the first one that came to mind.
 
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Is this a calculus based physics course? Looking at this the only way i can think of how to do it is with calculus. I'm going to assume it is.
The question is asking at which angle is the mass maximized. There fore you need to find [tex]\frac{dm}{d\theta}[/tex] and set it equal to zero. then you can solve for the angle.
 
The answer in the book is 19 degrees. The course sells itself as having a calculus component, but it's few and far in between in application. I haven't had calc for a handful of years, so I have to thank you all for your help!

I've showed my workings, but still have no verification that I did it correctly. I got 19.3 and assume the book just rounded down. Please let me know if you think this is correct, or if there's something else.
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Cheers!
 
Yes, you got it.
There is another way not using calculus. From the last equation (before you take the derivative of it): [tex]A=\frac{0.35mg}{T}=0.35sin\theta + cos\theta[/tex].
Apply the Bunyakovsky inequality: [tex](ab+cd)^2\leq (a^2+c^2)(b^2+d^2)[/tex], plus that [tex]sin^2\theta + cos^2\theta = 1[/tex], you will have [tex]A\leq \sqrt{1^2+0.35^2}[/tex]. The both sides equal only when [tex]\frac{a}{c}=\frac{b}{d}[/tex] or [tex]tan\theta = \frac{sin\theta }{cos\theta }=0.35[/tex].
 
Thanks for the feedback.

I've never heard of Bunyakovsky... I'll have to look into that.
Thanks once more!