A non-uniform bar of weight W and length L is λsuspended by two strings of negligible weight as

step 1 In this question we have given a non -uniform road of weight W and it is suspended in horizontal

A non-uniform bar of weight W and length L is λsuspended by...

A non-uniform bar of weight $W$ and length $L$ is $\lambda$ suspended by two strings of negligible weight as shown in figure. The angles $\hat{x} \rightarrow d^{n}$ made by the strings with the vertical are $\theta_{1}$ and $\theta_{2}$ respectively. The distance $d$ of the centre of gravity of the bar from its left end is (a) $L\left(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{1}}\right)$ (b) $L\left(\frac{\tan \theta_{1}}{\tan \theta_{1}+\tan \theta_{2}}\right)$ (c) $L\left(\frac{\tan \theta_{2}}{\tan \theta_{1}+\tan \theta_{2}}\right)$ (d) $L\left(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{2}}\right)$

A non-uniform bar of weight $W$ and length $L$ is $\lambda$ suspended by two strings of negligible weight as shown in figure. The angles $\hat{x} \rightarrow d^{n}$ made by the strings with the vertical are $\theta_{1}$ and $\theta_{2}$ respectively. The distance $d$ of the centre of gravity of the bar from its left end is
(a) $L\left(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{1}}\right)$
(b) $L\left(\frac{\tan \theta_{1}}{\tan \theta_{1}+\tan \theta_{2}}\right)$
(c) $L\left(\frac{\tan \theta_{2}}{\tan \theta_{1}+\tan \theta_{2}}\right)$
(d) $L\left(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{2}}\right)$

Key Concepts

Torque Equilibrium

This concept refers to the condition that the net torque (or moment) acting on a body in equilibrium must be zero. In problems involving suspended bodies, such as a bar held by strings, taking moments about a chosen pivot point is essential to establish relationships between unknown distances (like the center of gravity location) and the forces (or tensions) that create these moments.

Force Equilibrium

In static equilibrium, not only must the net torque be zero, but the sum of all forces acting on a system must also be zero. This means that the upward forces exerted by the strings must balance the downward weight of the bar. Understanding force equilibrium helps in establishing relations between the forces and ensures that the body is not accelerating linearly.

Center of Mass of a Non-Uniform Object

The concept of the center of mass (or center of gravity in a uniform gravitational field) is critical, especially when the mass distribution of an object is not uniform. It determines the point at which the weight of the object can be considered to act. In non-uniform objects, finding the center of mass is essential for applying torque and force equilibrium conditions correctly.

Trigonometric Resolution of Forces

When forces act at an angle, they are typically resolved into perpendicular and parallel components using trigonometric functions such as sine, cosine, and tangent. This process is crucial for analyzing the equilibrium of suspended objects where the angles at which support forces (like string tensions) act must be related to the vertical and horizontal directions to correctly set up the equations of equilibrium.

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