Two blocks ($$m = 0.5$$ kg and $$M = 4.5$$ kg) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is $$\frac{3}{7}$$. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is $$N$$. (Round off to the Nearest Integer) [Take $$g$$ as 9.8 m s$$^{-2}$$]

We need to determine the maximum horizontal force $$F$$ that can be applied to the larger block so that both blocks continue to move together without relative slipping.

1. Identify the System Parameters

From the layout shown on page, we have:

• Mass of the smaller top block ($$m$$) = $$0.5\text{ kg}$$
• Mass of the larger bottom block ($$M$$) = $$4.5\text{ kg}$$
• Coefficient of static friction between the blocks ($$\mu_s$$) = $$\frac{3}{7}$$
• Acceleration due to gravity ($$g$$) = $$9.8\text{ m s}^{-2}$$

2. Analyze the Maximum Acceleration of the Top Block

When the horizontal force $$F$$ is applied to the bottom block $$M$$, the top block $$m$$ moves forward solely due to the static frictional force ($$f_s$$) acting between the two contact surfaces.

The maximum possible static frictional force that can act on block $$m$$ is:

$$f_{\text{max}} = \mu_s \cdot N = \mu_s \cdot mg$$

According to Newton's second law, this maximum frictional force dictates the maximum acceleration ($$a_{\text{max}}$$) the top block can experience without slipping:

$$m \cdot a_{\text{max}} = \mu_s \cdot mg$$

$$a_{\text{max}} = \mu_s \cdot g$$

Substitute the given values into the acceleration equation:

$$a_{\text{max}} = \frac{3}{7} \times 9.8 = 3 \times 1.4 = 4.2\text{ m s}^{-2}$$

3. Calculate the Maximum Applied Force ($$F$$)

For both blocks to move together as a single combined system without relative motion, the entire system of mass ($M + m$) must accelerate at this maximum rate ($a_{\text{max}}$):

$$F = (M + m) \cdot a_{\text{max}}$$

Substitute the masses and the calculated acceleration into the formula:

$$F = (4.5 + 0.5) \times 4.2$$

$$F = 5.0 \times 4.2 = 21\text{ N}$$

Conclusion

The maximum horizontal force that can be applied so that the blocks move together is 21 N.