Torque and EquilibriumHardQuestion 6157
Question
A ladder (mass M, length L) leans against smooth wall at angle θ. Friction at floor for equilibrium:
- AMg tanθ/2
- BMg/2tanθ
- CMg cosθ/2sinθCorrect
- DMg/2
Correct answer
Mg cosθ/2sinθ
Explanation
Taking torques about base: N_wall×L sinθ = Mg×(L/2)cosθ. f = N_wall = Mg cosθ/(2sinθ) = Mg cosθ/2sinθ.