Question

What is the minimum value of acceleration of wedge so that block start sliding? Given coefficient of friction between block and wedge is $\mu $ .

• A. $g\left(\frac{cos\theta + \mu \, sin\theta }{sin\theta - μcos \theta }\right)$
• B. $g\left(\frac{sin\theta + μcos\theta }{cos\theta - \mu \, sin\theta }\right)$
• C. $ \, g\left(\frac{sin\theta - μcos \theta }{sin\theta + μcos ⁡ \theta }\right)$
• D. $g\left(\frac{cos \, \theta - \mu \, sin\theta }{sin \, \theta + μcos \theta }\right)$

Answer 1

Answer: (B)

The free body diagram of the block is

The acceleration of the block is in the right direction so the block will experience a $pseudoforce$ in the left direction as shown in the free body diagram.
The block will start sliding on the wedge then friction force will act down the wedge.
The condition when block will slide on the wedge is
$macos\theta \geq mgsin\theta +\mu N$
For minimum acceleration
$macos\theta =mgsin\theta +\mu N.................\left(1\right)$
Now resolving the components of different $forces$ acting on the block
$N=mgcos\theta +masin\theta ...............\left(2\right)macos\theta =mgsin\theta +\mu N..............\left(3\right)$
On solving the equations $\left(2\right)$ and $\left(3\right)$
$\Rightarrow a\left(cos \theta - \mu sin \theta \right)=g\left(sin \theta + \mu cos \theta \right)\Rightarrow a=\frac{g \left(sin \theta + \mu cos \theta \right)}{\left(cos \theta - \mu sin \theta \right)}$