Question

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Answer 1

Given,
Weight of bar = w
Length of bar = 2m
Let the distance of the center of gravity of the bar from its left end is ' d '

Break the components and see the figure.

The system is in equilibrium,
So, T1cos53.1° = T2cos36.9° { horizontal force balanced each other } -------(1)

Similarly, vertical forces balanced each other.
T1sin53.1° + T2sin36.9° = w ---(2)

And torque about point A ,
T2sin36.9° × 2m = w× d
T2 = w×d /2sin36.9° ----(3)

From eqns (2) and (3) ,
T1sin53.1° = w - T2sin36.9°
= w - wd/2
T1 = {w/sin53.1°} ( 1 - d/2)
now put this in eqn (1)

{w/sin53.1°}(1 - d/2)cos53.1° = {wd/2sin36.9°}.cos36.9°

(1 - d/2)/tan53.1° = d/2tan36.9°

Use, tan53.1° and tan36.9° values .

(1 - d/2)/1.3319 = d/2×0.7508

After solving this we get ,
d = 72.1 cm

Answer 2

Check the attachment......