Question

Three wires of mass m and length $l$ each are joined in the form of 'C' as shown in figure. Find the angle made by side AB with the vertical in equilibrium, if it is suspended freely from A.

• A. 37°
• B. 53°
• C. 45°
• D. 30°

Answer 1

Answer: (B)

53°

Answer 2

Here's how to solve this problem:

1. Define Coordinate System and Wire Segments:

   Let point A be the origin (0,0). Based on the figure, the wires are:

   • Wire BA: Horizontal, from B(-l,0) to A(0,0).
   • Wire BC: Vertical, from B(-l,0) to C(-l,-l).
   • Wire CD: Horizontal, from C(-l,-l) to D(0,-l).

   Each wire has mass $m$ and length $l$.

2. Calculate Center of Mass (CM) for each wire:

   • CM1 (for BA): Midpoint of B(-l,0) and A(0,0) is $(-l/2, 0)$.
   • CM2 (for BC): Midpoint of B(-l,0) and C(-l,-l) is $(-l, -l/2)$.
   • CM3 (for CD): Midpoint of C(-l,-l) and D(0,-l) is $(-l/2, -l)$.

3. Calculate Overall Center of Mass ($X_{CM}$, $Y_{CM}$):

   Total mass $M = 3m$.

   $X_{CM} = \frac{m(-l/2) + m(-l) + m(-l/2)}{3m} = \frac{-l/2 - l - l/2}{3} = \frac{-2l}{3}$

   $Y_{CM} = \frac{m(0) + m(-l/2) + m(-l)}{3m} = \frac{-l/2 - l}{3} = \frac{-3l/2}{3} = -l/2$

   The overall CM is at $(-2l/3, -l/2)$.

4. Determine Equilibrium Angle:

   In equilibrium, the center of mass G must lie vertically below the point of suspension A. This means the line segment AG must be vertical.

   Let $\theta$ be the angle made by side AB with the vertical. Initially, side AB is horizontal (along the negative x-axis).

   The vector $\vec{AG}$ from A(0,0) to G($-2l/3, -l/2$) has components $X_{CM} = -2l/3$ and $Y_{CM} = -l/2$.

   Let $\phi$ be the angle that the line AG makes with the negative y-axis (the vertical direction downwards).

   From the coordinates, $\tan \phi = \frac{|X_{CM}|}{|Y_{CM}|} = \frac{|-2l/3|}{|-l/2|} = \frac{2l/3}{l/2} = \frac{2}{3} \times \frac{2}{1} = 4/3$.

   So, $\phi = \arctan(4/3) \approx 53^\circ$.

Therefore, the angle made by side AB with the vertical in equilibrium is approximately 53°.