A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The

Q: Wait, why isn't the answer $0.7 \times 9.8 \times \sqrt{3}$ N?

That is the *maximum* possible static friction (limiting friction). Static friction is lazy; it only works as hard as it needs to. If the force pushing the block down is only $9.8$ N, friction will only push back with $9.8$ N to keep it at rest.

Q: Why do we use $\cos 30^\circ$ for the normal force?

Gravity acts straight down, but the surface is tilted. The normal force only balances the component of gravity that presses *perpendicularly* into the surface. By geometry, this component is $mg \cos 30^\circ$.

Q: What if the driving force was greater than the maximum static friction?

Then the block would start sliding! At that point, static friction breaks, and *kinetic* friction takes over. The frictional force would then be calculated using the coefficient of kinetic friction ($f_k = \mu_k N$).

Q: Does the mass of the block affect whether it slides?

Interestingly, no! If you set $mg \sin\theta > \mu_s mg \cos\theta$, the mass $m$ and gravity $g$ cancel out. The condition for sliding depends entirely on the angle $\theta$ and the coefficient of friction $\mu_s$ (specifically, if $\tan\theta > \mu_s$).

Q: Why is static friction called "self-adjusting"?

Imagine pushing a heavy fridge. If you push with 10 N, it doesn't move, so friction is 10 N. Push with 50 N, it still doesn't move, friction is 50 N. It adjusts to exactly match your applied force, up to its breaking point.

Question

A block of mass $2	ext{ kg}$ rests on a rough inclined plane making an angle of $30^\circ$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

Options:

(A) $9.8	ext{ N}$

(B) $0.7 	imes 9.8 	imes \sqrt{3}	ext{ N}$

(C) $9.8 	imes \sqrt{3}	ext{ N}$

(D) $0.7 	imes 9.8	ext{ N}$

Sigma Percentile · Accepted Answer

### The Deceptive Grip: Mastering Static Friction on an Incline

Imagine you are standing on a steep, rough mountain slope, trying to hold a heavy boulder in place. You can feel the exact moment when the boulder wants to slip, and you have to push back just enough to keep it steady.

This physical intuition is exactly what is being tested in this classic JEE problem. It looks like a straightforward plug-and-chug calculation, but it hides one of the most beautiful and deceptive concepts in Newtonian mechanics: the self-adjusting nature of static friction.

Let's break down the physics and uncover the trap that catches so many students off guard.

### Analyzing the Setup

We are given a block of mass $m = 2 	ext{ kg}$ resting on an inclined plane. The plane is tilted at an angle of $	heta = 30^\circ$ with the horizontal.

The surface is rough, and the coefficient of static friction between the block and the plane is given as $\mu_s = 0.7$. Our ultimate goal is to determine the exact magnitude of the frictional force acting on the block.

To solve this, we must first understand the invisible tug-of-war happening on that inclined plane. Gravity is relentlessly trying to pull the block down, while the microscopic roughness of the surface is fighting to keep it anchored.

### Resolving the Forces

Gravity acts straight down towards the center of the Earth with a force of $mg$. However, because the block is on a tilted surface, this gravitational force doesn't act entirely along the direction of potential motion.

We must resolve gravity into two perpendicular components. The first component acts perpendicular to the inclined plane, pressing the block into the surface. This is balanced by the Normal reaction force $N$. By simple geometry, this component is:

$$N = mg \cos 30^\circ$$

The second component acts parallel to the inclined plane. This is the "driving force" that actively tries to slide the block down the ramp. We can calculate it as:

$$F_{	ext{down}} = mg \sin 30^\circ$$

### The Friction Arsenal

Now, let's talk about friction. Because the block is initially at rest, we are dealing with **static friction**.

Static friction is not a fixed value; it is a highly intelligent, self-adjusting force. It will only exert exactly as much force as is necessary to counteract the driving force and prevent motion.

However, static friction has a breaking point. It can only grow up to a certain maximum limit, known as the **limiting friction**. If the driving force exceeds this limit, the static friction "breaks," the block starts to slide, and kinetic friction takes over. The formula for this maximum limit is:

$$f_{s,	ext{max}} = \mu_s N = \mu_s mg \cos 30^\circ$$

### The Master Calculation

To find out what friction is actually doing, we must first calculate the driving force and the maximum possible static friction, and then compare them.

Let's calculate the driving force pulling the block down:

$$F_{	ext{down}} = 2 	imes 9.8 	imes \sin 30^\circ$$

Since $\sin 30^\circ = \frac{1}{2}$, the calculation simplifies beautifully:

$$F_{	ext{down}} = 2 	imes 9.8 	imes \frac{1}{2} = 9.8 	ext{ N}$$

Next, let's calculate the absolute maximum resistance the surface can offer:

$$f_{s,	ext{max}} = 0.7 	imes 2 	imes 9.8 	imes \cos 30^\circ$$

Substituting $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we get:

$$f_{s,	ext{max}} = 0.7 	imes 19.6 	imes \frac{\sqrt{3}}{2} \approx 11.88 	ext{ N}$$

### The Final Verdict

Here is where the magic happens. We compare the two values we just found.

The driving force trying to pull the block down is **$9.8 	ext{ N}$**. The maximum frictional force the surface *could* provide is **$11.88 	ext{ N}$**.

Because the driving force is strictly less than the maximum limiting friction ($9.8 < 11.88$), the block does not have enough energy to break the static grip. It remains perfectly at rest.

This is the trap! Many students calculate the maximum friction ($11.88 	ext{ N}$) and immediately select it as the answer. But remember, static friction is lazy and self-adjusting. It doesn't need to use its full strength of $11.88 	ext{ N}$. It only needs to perfectly balance the driving force to keep the net force at zero.

Therefore, the actual static friction force adjusting itself to stop the block is exactly equal to the driving force:

$$f = F_{	ext{down}} = 9.8 	ext{ N}$$

The correct answer is indeed option (a). By understanding the physical reality behind the equations, you can easily sidestep the mathematical traps and arrive at the elegant truth.

Animated Solution for Physics - Laws of Motion: 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

Select Answer:

Visualized Solution

The Inclined Plane Setup

Forces Acting on the Block

Resolving Gravity

The Role of Friction

Calculating the Driving Force

Evaluating the Driving Force

Calculating Maximum Static Friction

Evaluating Maximum Static Friction

The Crucial Comparison

The Self-Adjusting Nature of Friction

The Sigma Insight: Static and Kinetic Friction

The Deceptive Grip

Analyzing the Setup

Resolving the Forces

The Friction Arsenal

The Master Calculation

The Final Verdict

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