A block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction

Question

A block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $a$ of the cart that will prevent the block from falling satisfies:

Accepted Answer

1. **Forces Analysis:** Consider the block of mass $m$ against the vertical surface of the accelerating cart. 
   - **Horizontal Direction:** The cart accelerates with acceleration $a$. The normal reaction $N$ provided by the vertical surface of the cart on the block causes this acceleration. According to Newton's Second Law: $N = ma$.
   - **Vertical Direction:** The weight $mg$ acts downwards. To prevent the block from falling, the static friction force $f_s$ acting upwards must balance the weight. Thus, $f_s = mg$.
2. **Friction Condition:** For the block not to slip, the required static friction must be less than or equal to the limiting value ($f_s \le f_{s,max}$). The limiting friction is given by $f_{s,max} = \mu N$ [NCERT Class 11, Physics Part I, Section 5.9 Friction].
3. **Derivation:**
   $$ f_s \le \mu N $$
   Substitute $f_s = mg$ and $N = ma$:
   $$ mg \le \mu (ma) $$
   Cancel the mass $m$ (since $m > 0$):
   $$ g \le \mu a $$
   $$ a \ge \frac{g}{\mu} $$
   Thus, the minimum acceleration required is $g/\mu$.

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