A uniform stick of length 1 meter and mass 200 gm is suspended from the ceiling through two vertical strings of equal length fixed at the ends. A small object of mass 20 gm is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.(adsbygoogle = window.adsbygoogle || []).push({});

The problem i attempted is basically from the chapter rotational motion.

Attempt:

I had the confusion regarding applying centre of mass to distribute both the masses ( of stick and object) at both the ends of the stick, thereby calculating resultant effective mass at both ends. e.g. as the stick is uniform, i distributed 200gm into 100 gm each and placed it at both the ends of the stick, secondly, i distributed the 20 gm mass into two parts such that its centre of mass is at 70 cm from the left end( as given in the problem), so by centre of mass formula i calculated it to be 6 gm and 14 gm to be placed at left and right ends respectively, so i came up with the answer as -: net force on right side=(1/100)*10+(14/1000)*10 = 1.14 Newtons. and on the left similarly, it comes to 1.06 Newtons.

But the answer is given as 1.12 N(right) and 1.04 N(left)

please help clarify my doubt ( or wrong solution)!

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# Tension in threads.

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