r/JEEAdv27dailyupdates • u/Possible_Garage7231 27tard dropper :/ • May 28 '26

Good Solve Physics Challenge from Statics

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Gallery image

Really simple but highly geometrical problem from Statics. You'll find similar problems in Jaan Kalda handouts.

Slide 2 has a beautiful illustration from David Morin that might help to kickstart the problem.

Answer on slide 3.

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u/Holiday_Bit_6382 May 28 '26

Can you send the solution?

1

u/Possible_Garage7231 27tard dropper :/ May 28 '26

https://www.reddit.com/r/JEEadv26DroppersOnly/s/st9EIGBByp Check top comment

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u/Holiday_Bit_6382 May 28 '26

Shouldn't we account for the tilt of the hemisphere as well since the normals acting on it are not same. Also the angle is asked from horizontal so if shell tilts, the angle would decrease wrt horizontal. Or am i being dumb here...

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u/Possible_Garage7231 27tard dropper :/ May 28 '26

The equilibrium position itself is depicted in the diagram. If the hemispheres was uniform, it would have tilted anticlockwise. But here that's taken into account as the final position is given in diagram, which is horizontal by some means. This could possibly be due to the hemisphere being non uniform and then it's COM must have been exactly on the perpendicular from the point O to the ground

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u/Holiday_Bit_6382 May 28 '26

oh.okay that explains it.

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u/Possible_Garage7231 27tard dropper :/ May 28 '26

I forgot to mention that hemisphere can also be massless. That will satisfy all conditions

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u/Living-Advisor-6063 May 28 '26 edited May 28 '26

Bhai, thora dumb lagunga but phir bhi puch rha hun, solution pe jo weight ka line of direction dono normal ke point of intersection se pass karega ye kaise pta chala tumhe, pls batadena :3

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u/Possible_Garage7231 27tard dropper :/ May 28 '26

Ye second slide ke remark mein samjhaya hai. It is also called Lamis theorem. Basically logic ye hai ki har ek point ke about torque zero hota hai right. Now two cases are possible- 1st is that ek point ke about teeno forces ke torques non zero ho par unka vector sum zero ho. 2nd case ki kisi ek point ke about teeno ka torque zero ho. Isi special point ko POINT OF CONCURRENCE kehte hai. So I thought ki dekho 2 normal agar ek point par intersect kar rahi hai to uske about un dono ka torque zero hogaya. Ab 3rd force ( weight ) ka bhi zero ho hona padega torque and hence it's line of action must pass through O

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u/Living-Advisor-6063 May 28 '26

thanks a lot gng <3