Question

A board of mass M is placed on a rough inclined plane and a man of mass m walks down the board. If the coefficient of friction between the board and inclined plane is μ, the acceleration of the man, such that plank does not slip, is given by

• A. $a \leq \left( \frac { M - m } { m } \right) ( \cos \theta + \mu \sin \theta ) g$
• B. $a \geq \left( \frac { M + m } { m } \right) ( \sin \theta + \mu \cos \theta ) g$
• C. $a \leq \left( \frac { M + m } { m } \right) ( \sin \theta + \mu \cos \theta ) g$
• D. $a = \left( \frac { m } { M + m } \right) ( \sin \theta + \mu \cos \theta ) g$

Answer 1

Answer: (C)

$a \leq \left( \frac { M + m } { m } \right) ( \sin \theta + \mu \cos \theta ) g$

Answer 2

Let F₁ be the force between the man and the board and F₂ be the force of friction between the inclined plane and the board.

Here F₁ can have a value between Mg sin θ - μ (M + m) g cos θ and Mg sin θ + μ(M + m)g cos θ … (i)

Limiting value of F₂ = μN₂

= μ(M + m)g cos θ

the force equations are

F₁ + m sin θ = ma

i.e., F₁ = ma – mg sin θ

From (i)

Mg sin θ - μ(M + m)g cos θ ≤ F₁ ≤ Mg sin θ + μ (M + m)g cos θ

i.e., Mg sin θ - μ(M + m)g cosθ ≤ ma – mg sin θ ≤ mg sin θ + μ(M + m)g cos θ

i.e., $\left( \frac { M + m } { m } \right)$ (sin θ - μ cos θ)g ≤ a ≤ $\left( \frac { M + m } { m } \right)$ (sin θ + μ cos θ) g