Q.A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. 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.

Setup: Let $T_1$ (left string) make angle $\theta_1=36.9°$ with the vertical and $T_2$ (right string) make angle $\theta_2=53.1°$ with the vertical — the familiar 3-4-5 triangle, so $\sin36.9°=0.6,\ \cos36.9°=0.8,\ \sin53.1°=0.8,\ \cos53.1°=0.6$.

Step 1 — Horizontal force balance. The bar is at rest, so the horizontal components of the two tensions must cancel:

Step 2 — Vertical force balance. The vertical components together support the weight $W$:

Substituting $T_1=\tfrac43T_2$:

$\tfrac{4}{3}T_2(0.8) + T_2(0.6) = W \;\Rightarrow\; \tfrac{3.2}{3}T_2 + 0.6T_2 = W \;\Rightarrow\; \tfrac{5}{3}T_2 = W \;\Rightarrow\; T_2 = \tfrac{3}{5}W$ …

Setting up

Let $T_1$ (left string) make angle $\theta_1=36.9°$ with the vertical, and $T_2$ (right string) make angle $\theta_2=53.1°$ with the vertical. These are the familiar 3-4-5 triangle angles: $\sin36.9°\approx0.6$, $\cos36.9°\approx0.8$, $\sin53.1°\approx0.8$, $\cos53.1°\approx0.6$. The bar has length $L=2$ m, and its weight $W$ acts at the (unknown) centre of gravity, a distance $d$ from the left end.

Force balance

Horizontal ($\sum F_x=0$): the two tensions' horizontal components must cancel:

Vertical ($\sum F_y=0$): the two tensions' vertical components support the weight:

Substituting $T_1=\tfrac43T_2$:

Torque balance about the left end …

Concept: Torque Balance on a Suspended Bar (Static Equilibrium)

A rigid bar of weight $W$ hangs at rest from two strings — both the net force and net torque on the bar must be zero.

Step 1: Set up the geometry

Let $T_1$ (making $36.9°$ with the vertical) act at the left end, and $T_2$ (making $53.1°$ with the vertical) act at the right end, $L = 2$ m away. Using the 3-4-5 triangle: $\sin36.9°=0.6,\ \cos36.9°=0.8,\ \sin53.1°=0.8,\ \cos53.1°=0.6$.

Step 2: Force balance

Horizontal: $T_1\sin36.9° = T_2\sin53.1° \Rightarrow 0.6T_1 = 0.8T_2 \Rightarrow T_1=\frac43T_2$

Vertical: $T_1\cos36.9°+T_2\cos53.1°=W \Rightarrow 0.8T_1+0.6T_2=W$

Step 3: Solve for $T_2$

Substituting: $0.8\left(\frac43T_2\right)+0.6T_2 = \frac53T_2 = W \Rightarrow T_2=\frac35W$ …