回複：Thank you so much !! If I did not learn Calculus,how could I

Let me try.
Remember we solved the pulling force

F = mg*mk/(mk*sinq + cosq)

We need to find q than makes F minimum. Or to find the q value that makes (mk*sinq + cosq) maximum.

mk is the friction coefficient which should be a number between 0 and 1. We can define a new variable b as

b = arctan(mk)

and therefore mk=tan(b)

(mk*sinq + cosq) gets maximum means

tan(b)*sin(q)+cos(q) gets maximum. Multiply by cos(b) on both sides, we have

sin(b)*sin(q) + cos(q)*cos(b) reaches maximum.

Remember the formulus

sin(b)*sin(q) + cos(q)*cos(b)=cos(b-q)

when b-q=0 the expression reach max value 1.

therefore q=b and

tan(q) = tan(b)=mk.

Yahoo!

所有跟帖： 

• Thanks! I get your idea. -pingshi- ♂ (0 bytes) () 10/02/2009 postreply 23:54:18