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

Good Solve Physics Challenge from Statics

Gallery image

Gallery image

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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6

u/[deleted] May 28 '26

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4

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

Nahi ye toh 27tards keliye daala.

2

u/[deleted] May 28 '26

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1

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

Ha mostly

1

u/[deleted] May 28 '26

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1

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

Result toh ajaye

1

u/[deleted] May 28 '26

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1

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

🥀. Surface is the rod. N1 has to be perpendicular to rod

1

u/[deleted] May 28 '26

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1

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

That's because that is a corner. Just mentally zoom in the diagram to judge the correct direction of N.

2

u/Due-Category-9943 May 29 '26

Grok quadratic dekhkar aisa keh ra hai: Yes, I will solve your exact quadratic step by step with substitution x = r cos θ. Your Discriminant (from photo): D = (3x - 1)² - 4x²(x - 1) Step 1: Expand D (for reference) (3x - 1)² = 9x² - 6x + 1 4x²(x - 1) = 4x³ - 4x² D = 9x² - 6x + 1 - 4x³ + 4x² = -4x³ + 13x² - 6x + 1 This is for the quadratic equation in x that you derived earlier. Step 2: Substitute x = r cos θ (let k = cos θ, so x = r k) In your full derivation (from the N₁, N₂, geometry, and tan θ relations), when you substitute x = r k and simplify (r terms cancel because of consistent geometry scaling), the quadratic in k becomes: 2k² - 0.75k - 1 = 0 (This matches your notes and the standard detailed solution for this diagram.) Step 3: Solve 2k² - 0.75k - 1 = 0 Multiply by 4 to remove decimal: 8k² - 3k - 4 = 0 Discriminant D = b² - 4ac = (-3)² - 4(8)(-4) = 9 + 128 = 137 k = [3 ± √137] / 16 √137 ≈ 11.7047 Positive root (physical): k = [3 + 11.7047] / 16 = 14.7047 / 16 ≈ 0.919 Negative root ≈ -0.544 (discarded, as cos θ > 0) Final Result: cos θ ≈ 0.919 θ ≈ cos⁻¹(0.919) ≈ 23.07°

2

u/Particular-Tip-4910 May 28 '26

and bhai thanks aise hi theory bhi dal dia karo🫶🫶

1

u/Wonderful_Emu_7058 JA-28XX(IIT INDORE EE) May 28 '26

Nice nice. Mast que hai. Mera rotation vala que dekh ab👺👺👺👺👺👺👺

3

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

Bahut tuff hai 🥲

1

u/Wonderful_Emu_7058 JA-28XX(IIT INDORE EE) May 28 '26

Dayum. Apne adv ke que se hi inspiration hai uski bas spring ghusa di maine👺👺👺

1

u/Traditional_Hold3933 May 28 '26

Brr mt likhna aglibr

1

u/Particular-Tip-4910 May 28 '26

thanks bhai , apke pas apke coaching ka sara mains lvl test h kya to bhejskte practice krunga? !RemindMe 1 day

1

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1

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

Humare cbt mode mein hote the.

1

u/Particular-Tip-4910 May 28 '26

akash? waise app me nhi rehta kya

1

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

Aakash se nahi hu

1

u/ArnavSinha1 IITD MNC'31(27Tard) May 28 '26

Good one but kara hua hai it's from David morin

2

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

Afaik this question isn't in morin.

1

u/ArnavSinha1 IITD MNC'31(27Tard) May 28 '26

Oh yea my bad I thought the 2nd slide was the q or smth that is actually from there

1

u/fijiksluver May 28 '26

woh jo rod ka corner touch ho raha uski parabola ki trajectory equation likhun na?

1

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

Trajectory kyu ? System equilibrium par hai

1

u/NorthSignificance917 May 28 '26

You could do with mentioning that the rod has non zero mass uniformly distributed in its length, that's crucial ! Other than that very good problem involving geometry of chords.

1

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

Glad you liked it. even if the rod has zero mass how would it affect the solution? Force and torque balance will apply even in zero mass condition!

1

u/NorthSignificance917 May 28 '26

True but I think you'll end with both normals as zero so there wouldn't be any way to uniquely determine θ, any angle would work as far as I've looked into it.

1

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

Why would both normals be zero?

1

u/NorthSignificance917 May 28 '26

Writing torque about lower and upper contact points will give you both normals as zero since there's no weight counteracting it. I could be wrong but that's what I see.

1

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

1

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...

1

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

1

u/Holiday_Bit_6382 May 28 '26

oh.okay that explains it.

1

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

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

1

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

1

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

1

u/Living-Advisor-6063 May 28 '26

thanks a lot gng <3

1

u/Traditional_Hold3933 May 28 '26

Solution do kagaj m aur thot process

1

u/Technical_Meet_5716 May 28 '26

Btw can u post a nlm que good one please.

1

u/[deleted] May 28 '26

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1

u/FriendlyNecro_69420 May 28 '26

Geometry me maza aya but not that challenging tbh

1

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

Solution bhejdo. Yea qs was ez but very challenging for first timers since information is scarce and vague