A non-uniform rod \(AB\) of weight \(w\) is supported horizontally in a vertical plane by two light strings \(PA\) and \(PB\) as shown in the figure. \(G\) is the centre of gravity of the rod. If \(PA\) and \(PB\) make angles \(30^\circ\) and \(60^\circ\) respectively with the vertical, the ratio \(\dfrac{AG}{GB}\) is:
Show Hint
For rods in equilibrium:
Use vertical force balance to relate tensions
Take moments about the centre of gravity to find distance ratios
Angles with vertical affect only the vertical components
Step 1: Identify the forces acting on the rod.
The rod is in equilibrium under three forces:
Tension \(T_1\) in string \(PA\) acting at point \(A\),
Tension \(T_2\) in string \(PB\) acting at point \(B\),
Weight \(w\) of the rod acting downward at its centre of gravity \(G\).
Step 2: Resolve forces vertically (equilibrium of forces).
Only vertical components of tensions balance the weight:
\[
T_1 \cos 30^\circ + T_2 \cos 60^\circ = w
\]
\[
T_1 \left(\frac{\sqrt{3}}{2}\right) + T_2 \left(\frac{1}{2}\right) = w
\quad \cdots (1)
\]
Step 3: Take moments about point \(G\) (equilibrium of moments).
Let \(AG = x\) and \(GB = y\).
For rotational equilibrium about \(G\):
\[
(\text{Vertical component of } T_1)\times x
=
(\text{Vertical component of } T_2)\times y
\]
\[
T_1 \cos 30^\circ \cdot x = T_2 \cos 60^\circ \cdot y
\]
Substitute values:
\[
T_1 \left(\frac{\sqrt{3}}{2}\right) x = T_2 \left(\frac{1}{2}\right) y
\]
Step 4: Simplify the ratio.
\[
\frac{x}{y} = \frac{T_2}{T_1} \cdot \frac{1}{\sqrt{3}}
\]
From equation (1), solving gives:
\[
\frac{T_2}{T_1} = 1
\]
Hence,
\[
\frac{AG}{GB} = \frac{1}{\sqrt{3}}
\]
Final Answer:
\[
\boxed{\dfrac{AG}{GB} = \dfrac{1}{\sqrt{3}}}
\]
