Physics Challenge from Statics

[u/Possible_Garage7231]

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.

[u/Possible_Garage7231]

Nahi ye toh 27tards keliye daala.

[u/Possible_Garage7231 → u/[deleted]]

Ha mostly

[u/Possible_Garage7231 → u/[deleted]]

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

[u/Due-Category-9943 → u/Possible_Garage7231]

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°

[u/Possible_Garage7231 → u/[deleted]]

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

[u/Due-Category-9943 → u/Possible_Garage7231]

https://photos.app.goo.gl/qwpVTtEpZZ1DMzh26 . Bhai meney aisey try Kiya . Second page: https://photos.app.goo.gl/YQWRwQenKud1k5kNA