A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together on an inclined plane with angl

Q: Why do we assume $m_1$ and $m_2$ move together?

If they were separate, $m_1$ (with no friction) would have an acceleration of $g \sin \theta$, while $m_2$ (with friction) would have a smaller acceleration of $g \sin \theta - \mu g \cos \theta$. Since $m_1$ is behind $m_2$ and wants to move faster, it will push into $m_2$, forcing them to move together with a common acceleration.

Q: Why doesn't $m_1$ contribute to the normal force for friction?

The blocks are placed side-by-side along the incline, not stacked on top of each other vertically relative to the plane. The normal force from the incline on $m_2$ only depends on the mass of $m_2$ itself ($m_2 g \cos \theta$).

Q: Why is the static friction equal to $(m_1 + m_2)g \sin \theta$ and not $\mu m_2 g \cos \theta$?

Static friction is a 'smart' force; it only applies as much force as necessary to prevent motion, up to its maximum limit. If the system is stationary, the friction must exactly balance the total downward driving force, which is $(m_1 + m_2)g \sin \theta$.

Q: What happens if the angle is exactly $11.5^\circ$?

At exactly the critical angle, the system is on the verge of slipping. The static friction reaches its absolute maximum value, $\mu m_2 g \cos \theta$, which in this specific case is exactly equal to the driving force $(m_1 + m_2)g \sin \theta$.

Q: Why is the kinetic friction the same formula as the maximum static friction?

The problem states that the coefficient of static and dynamic (kinetic) friction are equal ($\mu = 0.3$). Therefore, once the object starts moving, the kinetic friction is constant and equal to $\mu N$.

Question

A block of mass $m_1 = 1	ext{ kg}$ another mass $m_2 = 2	ext{kg}$, are placed together (see figure) on an inclined plane with angle of inclination $	heta$. Various values of $	heta$ are given in List I. The coefficient of friction between the block $m_1$ and the plane is always zero. The coefficient of static and dynamic friction between the block $m_2$ and the plane are equal to $\mu = 0.3$. In List II expressions for the friction on block $m_2$ are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by $g$.
[useful information : $	an (5.5^\circ) \approx 0.1$ ; $	an(11.5^\circ) \approx 0.2$ ; $	an (16.5^\circ) \approx 0.3$]

List-I:

P. $	heta = 5^\circ$

Q. $	heta = 10^\circ$

R. $	heta = 15^\circ$

S. $	heta = 20^\circ$

List-II:

1. $m_2 g \sin 	heta$

2. $(m_1 + m_2)g \sin 	heta$

3. $\mu m_2 g \cos 	heta$

4. $\mu(m_1 + m_2)g \cos 	heta$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Blocks

Imagine you are standing on an inclined plane, watching two blocks, $m_1$ and $m_2$, placed side-by-side.

The upper block, $m_1$, is on a perfectly smooth surface. It has absolutely zero friction holding it back. Naturally, it wants to slide down the incline as fast as gravity will allow.

However, right below it is $m_2$. This block is on a rough patch, experiencing friction. Because $m_1$ wants to accelerate faster than $m_2$, it will inevitably crash into $m_2$ and push it.

This is our first major breakthrough: **the two blocks will move together as a single, combined system**.

### Analyzing the Forces

Now that we know they move together, let's look at the forces pulling them down.

The total driving force is simply the component of their combined weight acting parallel to the incline. This is given by:
$$F_{	ext{drive}} = (m_1 + m_2)g \sin 	heta$$

But what about the resistance? Friction is the only thing fighting this downward slide, and it **only acts on $m_2$**.

The normal force pressing $m_2$ into the incline is $m_2 g \cos 	heta$. Therefore, the absolute maximum friction—the limiting friction—that the surface can muster is:
$$f_{	ext{max}} = \mu m_2 g \cos 	heta$$

### The Tipping Point: Condition for Slipping

For the blocks to actually break free and slide down, the driving force must overpower the maximum possible friction.

Mathematically, the system slips if:
$$(m_1 + m_2)g \sin 	heta > \mu m_2 g \cos 	heta$$

Let's rearrange this to find the critical angle. Dividing both sides by $\cos 	heta$ and the masses, we get:
$$	an 	heta > \frac{\mu m_2}{m_1 + m_2}$$

This is a beautiful, elegant condition. It tells us exactly when the system will yield to gravity.

### Calculating the Critical Angle

Let's plug in the numbers given in the problem. We have $m_1 = 1	ext{ kg}$, $m_2 = 2	ext{ kg}$, and $\mu = 0.3$.

Substituting these values into our condition:
$$	an 	heta > \frac{0.3 	imes 2}{1 + 2}$$

$$	an 	heta > \frac{0.6}{3}$$

$$	an 	heta > 0.2$$

Now, look at the "useful information" provided at the end of the question. It states that $	an(11.5^\circ) \approx 0.2$.

This is our magic number! The critical angle is **$11.5^\circ$**. If the incline is steeper than this, the blocks slip. If it's shallower, they stay put.

### The Verdict: Matching the Cases

Armed with our critical angle, evaluating the cases in List I becomes a breeze.

**Cases P ($5^\circ$) and Q ($10^\circ$):**
Both of these angles are less than $11.5^\circ$. The incline isn't steep enough. The system remains perfectly stationary.

In this static state, friction is a "smart" force. It doesn't need to use its maximum power; it only uses exactly what is needed to balance the driving force. 
$$f_{	ext{static}} = (m_1 + m_2)g \sin 	heta$$
This perfectly matches option (2) in List II.

**Cases R ($15^\circ$) and S ($20^\circ$):**
These angles are greater than $11.5^\circ$. Gravity wins! The system slips down the incline.

Once slipping occurs, kinetic friction takes over. Since the problem states that static and kinetic friction coefficients are equal, the friction acts at its maximum constant value:
$$f_{	ext{kinetic}} = \mu m_2 g \cos 	heta$$
This perfectly matches option (3) in List II.

### Final Conclusion

Bringing it all together, P and Q map to 2, while R and S map to 3.

The correct matrix match is **P-2, Q-2, R-3, S-3**.

This problem is a fantastic reminder to always check the physical state of the system before blindly applying friction formulas!

List-I

List-II

Select Matching Pairs:

Visualized Solution

System Analysis: $m_{1}$ and $m_{2}$

Limiting Friction on $m_{2}$

Condition for Slipping

Critical Angle Calculation

Finding the Critical Angle $θ_{c}$

Cases P and Q: $θ < 11. 5^{\circ}$

Cases R and S: $θ > 11. 5^{\circ}$

Final Matrix Match

The Sigma Insight: Static and Kinetic Friction

The Setup

Analyzing the Forces

The Tipping Point

Calculating the Critical Angle

The Verdict

Final Conclusion

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