A block of mass m1 = 1 kg and another mass m2 = 2 kg are placed together (see figure) on an inclined

Q: Why do we treat both blocks as a single system? Can't we solve them separately?

You absolutely can solve them separately! If you draw the free body diagram for $m_1$, you'll see a normal force $N_{12}$ pushing it up the incline. For $m_2$, that same $N_{12}$ pushes it down. When you add their equations of motion, $N_{12}$ cancels out, leaving you with the exact same system equation. Treating them as one system just saves you time and algebra!

Q: Wait, why is the normal force on $m_2$ only $m_2 g \cos\theta$? Doesn't $m_1$ press down on $m_2$?

Look closely at the diagram. The blocks are placed side-by-side along the incline, not stacked on top of each other vertically. $m_1$ is 'above' $m_2$ along the slope, so it pushes $m_2$ parallel to the incline, not perpendicular to it. Therefore, $m_1$ does not contribute to the normal force between $m_2$ and the plane.

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

This is a classic trap! $\mu N$ is the *maximum* possible static friction (limiting friction). Static friction is self-adjusting; it only pushes back as hard as it needs to. If the downward force is less than the maximum friction, the actual friction force will just perfectly balance the downward force to keep the net force zero.

Q: Why does the friction formula change when $\theta$ is $15^\circ$ or $20^\circ$?

Once the angle exceeds the critical tipping point ($11.5^\circ$), the downward force becomes stronger than the maximum static friction. The blocks break free and start sliding. When an object is sliding, we use kinetic friction, which is a fixed value given by $f_k = \mu_k N$. (Here, static and kinetic $\mu$ are the same).

Q: What if $m_1$ had friction too? How would the equation change?

If $m_1$ also had friction, say $\mu_1$, then both blocks would contribute to the resisting force. The maximum friction would be $f_{\text{max}} = \mu_1 m_1 g \cos\theta + \mu_2 m_2 g \cos\theta$. You would balance this against the same total downward force $(m_1 + m_2)g \sin\theta$.

Question

A block of mass $m_1 = 1	ext{ kg}$ and 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 the 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$]

\begin{tabular}{llll}
\hline
& List I & & List II 
\hline
P. & $	heta = 5^\circ$ & 1. & $m_2 g \sin 	heta$ 
Q. & $	heta = 10^\circ$ & 2. & $(m_1 + m_2) g \sin 	heta$ 
R. & $	heta = 15^\circ$ & 3. & $\mu m_2 g \cos 	heta$ 
S. & $	heta = 20^\circ$ & 4. & $\mu (m_1 + m_2) g \cos 	heta$ 
\hline
\end{tabular}

Options:

(A) P-1, Q-1, R-1, S-3

(B) P-2, Q-2, R-2, S-3

(C) P-2, Q-2, R-2, S-4

(D) P-2, Q-2, R-3, S-3

Sigma Percentile · Accepted Answer

Imagine you are standing on an inclined plane, holding two blocks. The top block, $m_1$, is on a perfectly smooth surface—zero friction! It is desperate to slide down. But it's blocked by $m_2$, which has a rough surface beneath it with a friction coefficient of $\mu = 0.3$. The question asks us to determine the nature and magnitude of the friction force on $m_2$ for various angles of inclination.

### Analyzing the Setup

Since $m_1$ has no friction, it will slide down and push against $m_2$. Because $m_1$ cannot pass through $m_2$, they will either remain at rest together or slide down together. This physical constraint allows us to treat both blocks as a single combined system of mass $(m_1 + m_2)$.

What forces are acting on this combined system? Gravity is pulling it straight down, but we only care about the components parallel and perpendicular to the incline. The component pulling the system down the incline is the sum of their individual sine components:
$$F_{	ext{down}} = m_1 g \sin	heta + m_2 g \sin	heta = (m_1 + m_2)g \sin	heta$$

### The Master Equation

Now, what is stopping this system from sliding? The only opposing force is the friction acting on $m_2$, because $m_1$ is on a frictionless surface. To prevent slipping, the maximum available static friction must be greater than or equal to the downward pulling force.

The maximum static friction (limiting friction) depends on the normal force acting on $m_2$. The normal force on $m_2$ is simply the perpendicular component of its own weight:
$$N_2 = m_2 g \cos	heta$$
Thus, the maximum friction available is:
$$f_{	ext{max}} = \mu N_2 = \mu m_2 g \cos	heta$$

For the system to remain at rest, we must have:
$$(m_1 + m_2)g \sin	heta \le \mu m_2 g \cos	heta$$

### The Critical Angle

Let's substitute the given values: $m_1 = 1	ext{ kg}$, $m_2 = 2	ext{ kg}$, and $\mu = 0.3$.
$$(1 + 2)g \sin	heta \le 0.3 	imes 2 	imes g \cos	heta$$
$$3 \sin	heta \le 0.6 \cos	heta$$

Dividing both sides by $3 \cos	heta$, we get:
$$	an	heta \le \frac{0.6}{3}$$
$$	an	heta \le 0.2$$

The problem kindly provides the approximation $	an(11.5^\circ) \approx 0.2$. This is our critical tipping point! As long as $	heta \le 11.5^\circ$, the blocks will not slip.

### The Dual Nature of Friction

This is where many students fall into a classic JEE trap. When the blocks are at rest ($	heta \le 11.5^\circ$), the friction is **static**. Static friction is smart; it only exerts as much force as necessary to maintain equilibrium. It does NOT equal $\mu N$ unless it is on the verge of slipping. Therefore, the static friction exactly balances the downward force:
$$f_{	ext{static}} = (m_1 + m_2)g \sin	heta$$

However, if the incline is steep enough ($	heta > 11.5^\circ$), the downward force overwhelms the maximum static friction, and the blocks begin to slide. Once sliding occurs, the friction becomes **kinetic**. Kinetic friction is constant and operates at its maximum value:
$$f_{	ext{kinetic}} = \mu m_2 g \cos	heta$$

### Final Calculation

Now we can easily match the lists.
- **P.** $	heta = 5^\circ$ and **Q.** $	heta = 10^\circ$: Both are less than $11.5^\circ$. The blocks are at rest. Friction is static, so $f = (m_1 + m_2)g \sin	heta$. This matches option 2.
- **R.** $	heta = 15^\circ$ and **S.** $	heta = 20^\circ$: Both are greater than $11.5^\circ$. The blocks are sliding. Friction is kinetic, so $f = \mu m_2 g \cos	heta$. This matches option 3.

The correct sequence is P-2, Q-2, R-3, S-3, which corresponds to option (d). A beautiful problem that tests your conceptual clarity on the self-adjusting nature of static friction!

Select Answer:

Visualized Solution

$System Setup$

$System Approach$

$Driving Force$

$Opposing Friction$

$Condition for Equilibrium$

$Substituting Values$

$Finding the Critical Angle$

$Static vs Kinetic Friction$

$Evaluating P and Q$

$Evaluating R and S$

$Final Answer$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation

The Critical Angle

The Dual Nature of Friction

Final Calculation

Similar Questions

List-I

List-II

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of $1 N$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (Take $g = 10 m/s^{2}$ )

A block of mass $m$ slides along a floor, while a force of magnitude $F$ is applied to it at an angle $θ$ as shown in figure. The coefficient of kinetic friction is $μ_{k}$ . Then, the block's acceleration $a$ is given by ( $g$ is acceleration due to gravity)

A block of mass $2 kg$ rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$ . The frictional force on the block is

A block rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$ . If the frictional force on the block is $10 N$ , the mass of the block (in $kg$ ) is ( $g = 10 m/s^{2}$ )

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Approach

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Driving Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Opposing Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Critical Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Static vs Kinetic Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating P and Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating R and S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer