abhikesbhat said:
Yea I drew a free-body diagram on paper. I don't understand by what you meant when you said
"The first is that you work parametrically, and only substitute for the question values at the very end."
Thank you for your help.
By that I mean that you work things out without substituting for the numerical values given in the question.
So your final answer would be:
\mu_k=\frac{F_{boy}\cos{\beta}-mg\sin{\alpha}}{mg\cos{\alpha}-F_{boy}\sin{\beta}}
You'll soon meet questions which ask you about different situations (Things like, what if the boy exerted twice as much force and things like that). They would be painfully frustrating to solve your way, since you'd end up having to solve everything from the start.
But if you solve parametrically, you get a full view of what's at play, and you have an easy way to solve any following questions.
A very powerful tool you then get is a way to analyze your answer for correctness. You can look at extreme cases and compare what you get to what you think should happen.
For instance, in the above question, \mu_k is defined as a positive pure number. The first test is dimensional analysis. What you get on the right side has to be the same as what you get on the left, what you get, has to be a pure number.
But the really interesting case is when you consider what happens when the value turns negative. For the value to turn negative, \beta, which I have defined as the angle of the force relative to the incline, would have to be negative. That means that the boy pushes the sled into the incline harder and hard until the normal is so great, that you would need an infinite coefficient of friction to keep the sled at equilibrium. :)
I may have rambled on a bit, I hope I got my point across. :x
