Physics - statics?

Question

A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown in the figure:(http://img183.imageshack.us/my.php?image=physics29nb2.jpg). One cord makes the angle θ = 21.5° with the vertical; the other makes the angle φ = 49.1° with the vertical. If the length L of the bar is 5.60 m, compute the distance x from the left end of the bar to its center of mass.

Answer 1

You just need to resolve the forces and moments step by step.

First, some definitions. Tension in the first rope = T1, tension in the second rope = T2, weight of bar = W.

Vertical equilibrium:

T1 cos(21.5) + T2 cos(49.1) = W

Horizontal equilibrium:

T1 sin(21.5) = T2 sin(49.1)

So T1 = T2 * (sin(49.1) / sin(21.5) ) ~= 2.06T2

Substitue into vertical equilibrium:

(2.06T2).cos(21.5) + T2cos(49.1) = W

(1.9188 + 0.65474)T2 = W

T2 = 0.38857 W

Now take moments about left-hand end:

W*x = T2 cos(49.1) * L = (0.38857W)*0.65474 * L

So x = 0.2544L = 1.42 metres