Question 2

Author Message   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Sunday, April 13, 2003 - 12:31 am:    A uniform ladder is placed with its foot on horizontal ground and its upper end against a vertical wall. The angle of friction at both points of contact is 30°. Find the greatest possible inclination of the ladder to the vertical. Please post your solution.   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Sunday, April 13, 2003 - 01:08 am:      971142 Username: 971142 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 07:33 pm:    Resolving horizontally, R1 = F2 Resolving vertically, F1 + R2 = W F1 = R1 tan30¢X F2 = R2 tan30¢X (¡Ô3) F1 = R1 (¡Ô3) F2 = R2 ¡ï (¡Ô3) F1 = F2 let the inclination of the ladder to the vertical be £c, F2 / F1 = tan£c Then, £c= 60¢X   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 09:19 pm:    Why F2 / F1 = tan q ? This step seems to be incorrect.   971142 Username: 971142 Registered: 04-2003 Posted on Thursday, April 17, 2003 - 11:18 am:    incorrect ar? so i dunno how to do wor   971142 Username: 971142 Registered: 04-2003 Posted on Thursday, April 17, 2003 - 01:10 pm:    Let l be the length of the ladder, By conservation of moment, (R1+F2)cos£c(l /2) = (R2-F1)sin£c(l /2) Tan£c = (R1+F2)/ (R2-F1) After solving, Tan£c = ¡Ô3 £c= 60¢X   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Thursday, April 17, 2003 - 04:51 pm:    Okay. But you should state where you take the moment, e.g. by taking moment at the centre of the ladder. The other thing is ... you jump too many steps!   971142 Username: 971142 Registered: 04-2003 Posted on Thursday, April 17, 2003 - 05:54 pm:    i didn't jump steps ar u need to combine all the things i posted ga