A block of mass m is lying on the edge having inclination angle α= tan -1((1/5)). Wedge is moving with a constant acceleration, a=2 ms -2. The minimum value of coefficient of friction μ, so that m remains stationary with respect to wedge is

Question

A block of mass m is lying on the edge having inclination angle α= tan -1((1/5)). Wedge is moving with a constant acceleration, a=2 ms -2. The minimum value of coefficient of friction μ, so that m remains stationary with respect to wedge is (A) (2/a) (B) (5/12) (C) (1/5) (D) (2/5)

Tardigrade Admin · Accepted Answer

FBD of m in frame of wedge <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a95b2d81204221ea0207539cc263e0a2-.png" /> N=m g cos (α)-m a sin (α) Now, f=μ N= operatornamemacos α+ operatornamemg sin α ⇒ μ=(a cos α+g sin α/g cos α-a sin α) ⇒ μ=(a+g tan α/g-a tan α)=(6/12) ⇒ μ=(5/12)

Q. A block of mass $m$ is lying on the edge having inclination angle $α = tan^{- 1} (\frac{1}{5})$ . Wedge is moving with a constant acceleration, $a = 2 m s^{- 2}$ . The minimum value of coefficient of friction $μ$ , so that $m$ remains stationary with respect to wedge is

$\frac{2}{a}$

$\frac{5}{12}$

$\frac{1}{5}$

$\frac{2}{5}$

FBD of $m$ in frame of wedge

$N = m g cos (α) - ma sin (α)$
Now, $f = μ N = macos α + mg sin α$
$\Rightarrow μ = \frac{a c o s α + g s i n α}{g c o s α - a s i n α}$
$\Rightarrow μ = \frac{a + g t a n α}{g - a t a n α} = \frac{6}{12}$
$\Rightarrow μ = \frac{5}{12}$

Q. A block of mass m is lying on the edge having inclination angle α=tan−1(51​). Wedge is moving with a constant acceleration, a=2ms−2. The minimum value of coefficient of friction μ, so that m remains stationary with respect to wedge is

a2​

125​

51​

52​