A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in. The angles made by the strings with the vertical are $36.9°$ and $53.1°$ respectively. The bar is $2m$ long. Calculate the distance d of the centre of gravity of the bar from its left end.
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The free body diagram of the bar is shown in the following figure:
Length of the bar, $l = 2m T_1$ and $T_2$_ are the tensions produced in the left and right strings respectively. At translational equilibrium, we have, $\text{T}_1\sin36.9^{\circ}=\text{T}_2\sin53.1$
$\frac{\text{T}_1}{\text{T}_2}=\frac{\sin53.1^{\circ}}{\sin36.9}$
$=\frac{0.800}{0.600}=\frac{4}{3}$
$\text{T}_1=\frac{4}{3}\text{T}_2$ For rotational equilibrium, on taking the torque about the centre of gravity, we have, $\text{T}_1\cos36.9\times\text{d}=\text{T}_2\cos53.1(2-\text{d})$
$\text{T}_1\times0.800\text{d}=\text{T}_20.600(2-\text{d})$
$\frac{4}{3}\times\text{T}_2\times0.800\text{d}=\text{T}_2\big[0.600\times2-0.600\text{d}\big]$
$1.067\text{d}+0.6\text{d}=1.2$
$\therefore\ \text{d}=\frac{1.2}{1.67}$
$=0.72\text{m}$ Hence, the C.G. (centre of gravity) of the given bar lies 0.72m from its left end.
Length of the bar, $l = 2m T_1$ and $T_2$_ are the tensions produced in the left and right strings respectively. At translational equilibrium, we have, $\text{T}_1\sin36.9^{\circ}=\text{T}_2\sin53.1$
$\frac{\text{T}_1}{\text{T}_2}=\frac{\sin53.1^{\circ}}{\sin36.9}$
$=\frac{0.800}{0.600}=\frac{4}{3}$
$\text{T}_1=\frac{4}{3}\text{T}_2$ For rotational equilibrium, on taking the torque about the centre of gravity, we have, $\text{T}_1\cos36.9\times\text{d}=\text{T}_2\cos53.1(2-\text{d})$
$\text{T}_1\times0.800\text{d}=\text{T}_20.600(2-\text{d})$
$\frac{4}{3}\times\text{T}_2\times0.800\text{d}=\text{T}_2\big[0.600\times2-0.600\text{d}\big]$
$1.067\text{d}+0.6\text{d}=1.2$
$\therefore\ \text{d}=\frac{1.2}{1.67}$
$=0.72\text{m}$ Hence, the C.G. (centre of gravity) of the given bar lies 0.72m from its left end.
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