Question

A thin uniform equilateral plate rests in vertical plane with one of its ends 'A' on a rough horizontal floor, and other end 'C' on smooth vertical wall. The least angle (q) its base AC can make with horizontal will be –

• A. q = cot^–1$\left( 2\mu + \frac{1}{\sqrt{3}} \right)$
• B. q = tan^–1$\left( 2\mu + \frac{1}{\sqrt{3}} \right)$
• C. q = tan^–1$\left( 2\mu + \frac{1}{2\sqrt{3}} \right)$
• D. q = cot^–1$\left( 2\mu + \frac{1}{2\sqrt{3}} \right)$

Answer 1

Answer: (A)

q = cot^–1$\left( 2\mu + \frac{1}{\sqrt{3}} \right)$

Answer 2

N₁ = Mg ….(i)

µN₁ = N₂ ….(ii)

N₂ = µMg ….(iii)

Rotational equilibrium about A

Mg x cos (30⁰ + 0) – N₂ sinq = 0

Ž q = cot^–1 $\left( 2µ + \frac{1}{\sqrt{3}} \right)$