A non-uniform bar of weight W is suspended at rest by two...

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 $2 \mathrm{~m}$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

Key Concepts

Force Decomposition

Force decomposition involves breaking down a force into its perpendicular components, typically along horizontal and vertical axes. This method simplifies the analysis of systems involving oblique forces, such as those acting at an angle in a suspended system, by enabling the separate consideration of each directional component in equilibrium equations.

Center of Gravity

The center of gravity is the specific point where the entire weight of an object can be considered to act. For non-uniform bodies, this point is not necessarily at the geometric center, making its determination essential for assessing how gravitational forces contribute to rotational dynamics.

Torque and Moments

Torque, or moment of force, is a measure of the rotational tendency of a force about a pivot point. It is calculated as the product of the force magnitude and the perpendicular distance from the pivot. Balancing torques is critical when analyzing systems in rotational equilibrium.

Static Equilibrium

Static equilibrium refers to the state where an object remains at rest because the sum of the forces and the sum of the moments (torques) acting on it are zero. This principle is fundamental in determining conditions under which a system, such as a suspended bar, does not accelerate linearly or rotationally.

Transcript

00:01 In this problem, a non -uniform bar of length 2 meter have weight w and its center of gravity is at distance d from the left end.

00:13 So we have to calculate what is the value of d if the system is in equilibrium.

00:20 So first of all, we will draw what are the forces here acting.

00:24 So if we go to draw the tension force, so in left end, let us say tension force is t1.

00:30 And in right end the tension forces t2 therefore this tensions can be resolved into its horizontal and vertical component so let us say this is horizontal component and this one is vertical component therefore this angle will be 36 .9 degree so this angle is 36 .9 degree therefore vertical component will be t1 cost 36 .9 and horizontal component will be t1 sine 36 .9 degree now similarly if we resolve the t2 components so its vertical component and horizontal component will be first we two we have to find the angle so this angle is 53 .1 degree so this angle will be 53 .1 degree therefore its vertical component will be t2 cost 53 .1 degree and horizontal component will be t2 cause t2 sign 53 .1 degree.

01:40 Now the system is in equilibrium it means that all the horizontal and vertical forces must be balanced.

01:46 Therefore what are the horizontal forces? horizontal forces t1 sign 36 .9 and t2 sign 53 .1 so both of these must be equal.

01:56 Therefore since system is in equilibrium therefore t1 sine 36 .9 is equals to t2 sine 53 .1 degree.

02:22 Therefore t1 is equal to sine 53 .1 divided by sine 36 .9 into t2.

02:34 So t1 is equal to 4 upon 3t2.

02:39 So this is the first relation that we have got from the equilibrium.

02:46 Now, we have to find the distance d.

02:52 For that, we can use the rotational equilibrium...