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 $\alpha$ of the cart that will prevent the block from falling satisfies

• A. $\alpha>\frac{mg}{\mu}$
• B. $\alpha>\frac{g}{\mu m}$
• C. $\alpha\ge\frac{g}{\mu}$
• D. $\alpha$

Answer 1

The Correct Option is C

To solve this problem, we need to understand the conditions under which a block remains stationary relative to an accelerating cart. The block of mass $m$ is in contact with the cart, which is accelerating horizontally with acceleration $\alpha$. The key force at play here is the friction between the block and the cart, which provides the necessary force to keep the block from falling.

The static frictional force $f_s$ available is given by:

$$f_s \leq \mu N$$

where $N$ is the normal force exerted by the cart on the block and $\mu$ is the coefficient of static friction.

Since the block is not moving vertically, the force of gravity $mg$ acting downward must be balanced by the vertical component of the normal force. The normal force is equal to the horizontal mass times acceleration of the cart:

$$N = m \alpha$$

For the block to not fall, the vertical force exerted due to static friction must be equal to or greater than the gravitational force. Therefore,

$$f_s = \mu N \geq mg$$

Substituting the expression for $N$, we get:

$$\mu (m \alpha) \geq mg$$

After simplifying, we derive the necessary condition for the acceleration of the cart:

$$\alpha \geq \frac{g}{\mu}$$

This inequality ensures that the static friction is sufficient to counteract the gravitational pull on the block, preventing it from falling. This reasoning leads us to the correct answer:

• $\alpha \geq \frac{g}{\mu}$ - This is the correct condition ensuring the block remains stationary relative to the accelerating cart.