A stick AB of mass M is tied to one end of a light string OA. A horizontal force F = Mg is applied at end B of the stick and it remains in equilibrium in the position shown. If α = tan⁻¹(n), the value of n is ...
Physics · Mixed Chapter
JEE Mains

Question

A stick AB of mass M is tied to one end of a light string OA. A horizontal force F = Mg is applied at end B of the stick and it remains in equilibrium in the position shown. If α = tan⁻¹(n), the value of n is ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Take moments about A so that tension in the string has no moment. The horizontal force F at B has moment FL cosα and the rod's weight Mg acts at its midpoint giving opposite moment (Mg)(L/2) sinα. Step 2: For equilibrium FL cosα=(MgL/2) sinα, or tanα=2F/(Mg). Step 3: Since F=Mg, tanα=2. Therefore α=tan⁻¹n gives n=2.