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Homework Statement
A uniform board of length L and weight W is suspended between two vertical walls by ropes of length L/2 each. When a weight w is placed on the left end of the board, it assumes the configuration shown in the figure.
http://img260.imageshack.us/img260/3992/rw1234ok6.jpg
Find the weight w in terms of the board weight W.
Homework Equations
If the left rope corresponds to T_1, and the right rope to T_2, left and down is negative, right and up is positive, then:
Sum(F_x) = -T_1 sin(35) + T_2 sin(60) = 0
Sum(F_y) = T_1 cos(35) + T_2 cos (60) - W - w = 0
Sum(T_z) = - W(L/2) cos(9.2) + T_2 L sin(20.8 [just calculated this from graphical drawing]) = 0 [pivot at w].
The Attempt at a Solution
eq1. T_2 sin(60) = T_1 sin(35)
eq1. T_2 = T_1 sin(35)/sin(60)
eq2. T_1 (cos(35) + sin(35)cos(60)/sin(60)) - W = w
eq3. T_2 L sin(20.8) = W (L/2) cos(9.2)
eq3. T_2 sin(20.8) = W (1/2) cos(9.2)
eq3. T_1 sin(35)sin(20.8)/sin(60) = W (1/2) cos(9.2)
eq3. T_1 = [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)]
eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W * .4274425413] / [.203680986] * [1.150306554] = w
eq2 w = 2.41 W
The correct answer is 1.42 W. What did I do wrong?
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