# More optimization problems

• Mar 22nd 2008, 01:13 PM
pakman134
More optimization problems
[IMG]http://i115.photobucket.com/albums/n297/pakman134/optimizationsled.jpg[/IMG]
• Mar 22nd 2008, 04:42 PM
bobak
This really belongs in the "advance applied math section"

For question 1. you should know that the maximum value that a friction force is $\displaystyle \mu R$ where R is the normal reaction force between the object and the surface it in in contact with and $\displaystyle \mu$ is the coefficient of friction.

(if you don't follow what I say here ask me to post a diagram)

the normal reaction force is given by $\displaystyle R = mg - F \cos \theta$. therefore the maximum friction force is $\displaystyle \mu mg - \mu F \cos \theta$. If there is to be motion the forward must be greater than the maximum friction force (such that there is a net resultant force forwards).

so we get $\displaystyle F \sin \theta > \mu mg - \mu F \cos \theta$

which rearranges to $\displaystyle \tan \theta = \frac{1}{\mu}$

this can rearrange to give $\displaystyle F > \frac{\mu mg }{\sin \theta +\mu \cos \theta }$

to find the max/min or F you differentiate using the chain rule. to get $\displaystyle F' = \frac{-\mu mg ( \cos \theta - \mu \sin \theta) }{(\sin \theta +\mu \cos \theta )^2}$

we require $\displaystyle \cos \theta - \mu \sin \theta = 0$

which rearranges to $\displaystyle \tan \theta = \frac{1}{\mu}$

using some simple trig you get $\displaystyle \sin \theta = \frac{1}{\sqrt{\mu^2 + 1}}$ and $\displaystyle \cos \theta = \frac{\mu}{\sqrt{\mu^2 + 1}}$

put this into the formula (with your value of $\displaystyle /mu$) I get the minimum value of $\displaystyle F = 0.148 mg$.

you should be able to notice that F will be a maximum when $\displaystyle \theta = \frac{\pi}{2}$ in which case $\displaystyle F = \mu mg$

Edit: F is actually maximum in the case when $\displaystyle \theta = 0$ in this case you would just be pulling the sledge vertically
• Mar 22nd 2008, 05:02 PM
pakman134
hm...
I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm.....
• Mar 22nd 2008, 05:10 PM
bobak
Quote:

Originally Posted by pakman134
I somewhat follow what you're saying and when I follow your process I come up with the same answers as you did, but with your answers they're still coming up as wrong. Maybe it is a simple math error but I can't seem to see anything really wrong with your process. Hm.....

Well the maximum value of F should be $\displaystyle mg$ as it occurs when $\displaystyle \theta = 0$ I have check my working again, I am fairly confidient the minimum value of F is $\displaystyle 0.148mg$ but maybe this software your using requires it answer to greater accuracy, have you tired $\displaystyle 0.1483mg$ ?
• Mar 22nd 2008, 05:21 PM
pakman134
Quote:

Originally Posted by bobak
Well the maximum value of F should be $\displaystyle mg$ as it occurs when $\displaystyle \theta = 0$ I have check my working again, I am fairly confidient the minimum value of F is $\displaystyle 0.148mg$ but maybe this software your using requires it answer to greater accuracy, have you tired $\displaystyle 0.1483mg$ ?

ugh i hate these stupid online homework things... You were right about just having to take it out another decimal place. thank you very much!!