Question 6

Author Message   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Monday, April 14, 2003 - 10:37 pm:    A uniform rod AB of length 2a and weight W is inclined at 30° to the horizontal with its lower end A on rough horizontal ground, the angle of friction being 30°. The rod rests in contact with a smooth peg C (AC < AB). Calculate the height of the peg above the ground and the reaction at the peg if the rod is in limiting equilibrium.   961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:33 am:    At limiting equilibrium, F = R1tanλ = R1tan30° = (√3／3) R1 Resolve forces vertically, W = R1 + R2cos30° = R1 + (√3/2)R2 … (1) Resolve forces horizontally, F = R2sin30° = 1/2 R2 height of the ladder = (2a)sin30° = a   961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:34 am:    Let h be the height of the peg By similar triangle, 2a/a = AB/h AB = 2h Take moment about center of gravity, (a)R1cos30&#8728;= (a)Fsin30&#8728;+ R2(2h – a) (√3/2)R1(a) = (√3/6)R1(a) + R2(2h – a) (√3/2)R1(a) = (√3/6)R1(a) + (2√3/3)R1(2h – a) (√3/2)(a) = (√3/6)a + (2√3/3)(2h – a) 3√3(a) = √3(a) + (4√3)(2h – a ) 2h = (1/4√3)(6√3)(a) h = (3/4)a From (1), W = (3/2√3 )R2 + (√3/2)R2 2√3 = 3R2 + 3R2 R2 = (√3/3)W   961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:39 am:      961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:44 am:      961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:49 am:    Take moment about center of gravity, (a)R1cos30° = (a)Fsin30° + R2(2h – a) (√3/2)R1(a) = (√3/6)R1(a) + R2(2h – a) (√3/2)R1(a) = (√3/6)R1(a) + (2√3/3)R1(2h – a) (√3/2)(a) = (√3/6)a + (2√3/3)(2h – a) 3√3(a) = √3(a) + (4√3)(2h – a ) 2h = (1/4√3)(6√3)(a) h = (3/4)a From (1), W = (3/2√3 )R2 + (√3/2)R2 2√3 = 3R2 + 3R2 R2 = (√3/3)W   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:24 pm:    It should be AC=2h and AB=2a. Besides, you would better take moment at the point where the most number of forces intersect. For example, you can take moment at A to find R2 directly. Anyway, a good job.   961183 Username: 961183 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 02:27 pm:    我明白了﹗多謝你的提點﹗