Question

A block is placed on a wedge with coefficient of friction \(\mu = 0.5\). The wedge is accelerated horizontally towards the block. What is the minimum acceleration required so that the block does not slide down the wedge?

• A. \( g \)
• B. \( \frac{g}{2} \)
• C. \( \frac{g}{\sqrt{3}} \)
• D. \( \frac{g}{1 + \mu} \)

Hint:

When dealing with pseudo-forces, always draw the Free Body Diagram (FBD) from the perspective of the accelerating object. It turns a complex dynamic problem into a simpler static equilibrium problem!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem asks for the minimum acceleration of a wedge such that a block on its surface does not slide down. When the wedge is accelerated horizontally towards the block, the block experiences a pseudo force that increases the normal reaction and, subsequently, the maximum static friction available to prevent sliding.
Step 2: Key Formula or Approach:
In the frame of the accelerating wedge with angle $\theta$, the block experiences a pseudo force $ma$ horizontally.
Resolving forces perpendicular to the incline: \[ N = mg \cos\theta + ma \sin\theta \] Resolving forces parallel to the incline for the limiting case of sliding down: \[ mg \sin\theta = ma \cos\theta + \mu N \]
Step 3: Detailed Explanation:
Substituting $N$: \[ mg \sin\theta = ma \cos\theta + \mu(mg \cos\theta + ma \sin\theta) \] Solving for $a$: \[ mg \sin\theta - \mu mg \cos\theta = ma \cos\theta + \mu ma \sin\theta \] \[ g(\sin\theta - \mu \cos\theta) = a(\cos\theta + \mu \sin\theta) \] \[ a = g \frac{\sin\theta - \mu \cos\theta}{\cos\theta + \mu \sin\theta} = g \frac{\tan\theta - \mu}{1 + \mu \tan\theta} \] \[ a = g \frac{(4/3) - 0.5}{1 + 0.5(4/3)} = g \frac{4/3 - 1/2}{1 + 2/3} = g \frac{5/6}{5/3} = \frac{g}{2} \] This standard angle accurately matches Option (B).
Step 4: Final Answer:
The minimum acceleration required is $\frac{g}{2}$.