Question 7

Author Message   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Monday, April 14, 2003 - 10:44 pm:    A smooth hemispherical bowl of radius a is fixed with its rim uppermost and horizontal. A smooth uniform rod of length 2l (l > a) rests with one end inside the bowl and leaning on the rim. Find the length of the rod that overhangs the bowl.   Ming Username: 971004 Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 12:40 am:    Let d be the length of the rod inside the bowl, R be the reaction force of the tip of the rod inside the bowl. Take moment about the rim. R．dsinθ=W(d-2L/2)cosθ => R=(W/dsinθ)(d-L)cosθ Resolving forces along the rod. Rcosθ=Wsinθ => R=Wtanθ ∴Wtanθ=(W/dsinθ)(d-L)cosθ dtan2θ=d-L L=d(1-tan2) =d(2-sec2θ) =d[2-(2a/d)2] =d[(2d2-4a2)/d2] =2(d2-2a2)/d 2d2-Ld-4a2=0   Ming Username: 971004 Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 12:49 am:    By Quadratic Formula: => d=[L-開方(L2-4(2)(-4a)2)]/4 =(L-開方(L2+32a2)/4 as d is positive(thus, no"+" sign) ∴ The length of the rod that overhangs the bowl =2L-d =2L-(L-開方(L2+32a2)/4 =[7L-開方(L2+32a2)]/4 打死人啦！！！ 打死人啦！！！ 打死人啦！！！   Ming Username: 971004 Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 01:26 am:    The Force Diagram is the same as the diagram in Unit2B/PPR/P.11. I dont draw it here la.   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 10:54 am:    Good definition and description. But there is typing mistake in your first post. Anyway, there is a conceptual mistake in your second post. As the product of root is negative, this means one root is +ve and the other is -ve. You take -ve sign, that means you take the -ve root! So your answer is incorrect.   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 11:02 am:    咁你得閒可以試下幫我打solution…哈哈！   Ming Username: 971004 Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 02:04 pm:    By Quadratic Formula: => d={L+-開方[L2-4(2)(-4a)2]}/4 =[L+-開方(L2+32a2)]/4 ∴ The length of the rod that overhangs the bowl =2L-d =2L-[L+-開方(L2+32a2)]/4 =[7L-開方(L2+32a2)]/4 (For the "+"root) OR =[7L+開方(L2+32a2)]/4 (For the "-" root) >2L (rejected) ∴The length of the rod that overhangs the bowl is [7L-開方(L2+32a2)]/4 Like this??   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Tuesday, April 15, 2003 - 03:23 pm:    In fact, you should reject the -ve sign instead of the +ve sign, as the -ve sign will produce the -ve root!