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Q7.8:

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 $36.9^\circ$ and $53.1^\circ$ respectively. The bar is 2 m long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

Solution :

Step 1: Establishing the Coordinate System and Force Components

Let the left end of the bar be at the origin $(0,0)$ on the $x$-axis. The bar has a length $L = 2\,\text{m}$. The weight $W$ acts downwards at the center of gravity located at distance $d$ from the origin. The tensions in the left and right strings are $T_1$ and $T_2$ respectively, making angles $\theta_1 = 36.9^\circ$ and $\theta_2 = 53.1^\circ$ with the vertical.

W 0 d 2m

Step 2: Resolving Forces into Components

For the system to be at rest, the sum of horizontal forces ($\sum F_x = 0$) and vertical forces ($\sum F_y = 0$) must be zero [Per Newton's First Law of Equilibrium].

  • Horizontal: $T_1 \sin(36.9^\circ) = T_2 \sin(53.1^\circ)$
  • Vertical: $T_1 \cos(36.9^\circ) + T_2 \cos(53.1^\circ) = W$

Using trigonometric values ($\sin 36.9^\circ \approx 0.6, \cos 36.9^\circ \approx 0.8$; $\sin 53.1^\circ \approx 0.8, \cos 53.1^\circ \approx 0.6$):

$0.6 T_1 = 0.8 T_2 \implies T_1 = \frac{4}{3} T_2$

Substituting into the vertical equation:

$(\frac{4}{3} T_2)(0.8) + T_2(0.6) = W$

$T_2 (\frac{3.2}{3} + 0.6) = W \implies T_2 (\frac{3.2 + 1.8}{3}) = W \implies T_2 = \frac{3}{5} W = 0.6W$

$T_1 = \frac{4}{3} (0.6W) = 0.8W$

Step 3: Calculating Center of Gravity via Torque Balance

Taking the torque ($\tau$) about the left end (origin) to be zero for rotational equilibrium [Per the Second Condition of Equilibrium, $\sum \tau = 0$]:

$\sum \tau_{origin} = (W \cdot d) - (T_2 \cdot L \cdot \cos(53.1^\circ)) = 0$

Note: The vertical component of $T_2$ creates the torque about the left end at the full length $L=2\,\text{m}$.

$W \cdot d = (0.6W) \cdot (2) \cdot (\cos 53.1^\circ)$

$W \cdot d = (0.6W) \cdot (2) \cdot (0.6)$

$W \cdot d = 1.2W \cdot 0.6$

$d = 0.72\,\text{m}$

Final Solution: The distance $d$ of the centre of gravity from the left end is $0.72\,\text{m}$.


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