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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are 36.9° and ‘ 53.2° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

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As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem. To solve this problem, we can use the concept of torque equilibrium. The sum of the torques acting on the bar must be zero for it to remain in equilibrium. Let's denote the distance of the center...
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As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.

To solve this problem, we can use the concept of torque equilibrium. The sum of the torques acting on the bar must be zero for it to remain in equilibrium.

Let's denote the distance of the center of gravity of the bar from its left end as dd. Now, let's break down the forces acting on the bar:

  1. The weight of the bar, WW, acts downward and can be considered to act at the center of gravity of the bar.
  2. The tension forces in the strings exert forces on the bar at their points of contact.

Given that the bar is in equilibrium, we can write the condition for torque equilibrium about any point. Let's choose the left end of the bar as the pivot point.

The torque due to the weight of the bar about the left end is W×dW×d, and the torque due to the tension in the strings will depend on their distances from the pivot point.

Since the angles made by the strings with the vertical are given, we can use trigonometry to find the distances of the strings from the pivot point.

Let's denote the tensions in the strings as T1T1 and T2T2. Using the fact that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side, we can express the distances of the strings from the pivot point:

For the first string: tan⁡(36.9∘)=dxtan(36.9)=xd x=dtan⁡(36.9∘)x=tan(36.9)d

For the second string: tan⁡(53.2∘)=2−dytan(53.2)=y2−d y=2−dtan⁡(53.2∘)y=tan(53.2)2−d

Now, using the torque equilibrium condition:

T1×x=W×d=T2×yT1×x=W×d=T2×y

Substitute the expressions for xx and yy into the equation, and solve for dd. Once dd is found, it will give us the distance of the center of gravity of the bar from its left end.

I hope this helps! Let me know if you need further clarification or assistance with the calculations.

 
 
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