A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force

Q: Why does the direction of friction change between the two cases?

Static friction is a smart force; it always opposes the *tendency* of relative motion. When a 2N force pulls down, the block wants to slide down, so friction acts up. When a 10N force pulls up, the block wants to slide up, so friction flips and acts down.

Q: Why do we use the maximum static friction $f = \mu N$ in both cases?

The problem uses the word 'maximum' force in both scenarios (maximum force down before it moves, and maximum force up before it moves). This implies the block is on the absolute verge of slipping in both cases, which is exactly when static friction reaches its limiting value, $\mu N$.

Q: Do we need to know the mass of the block to solve this?

No! As you can see in the final calculation, when we divide the two equations, the mass $m$ and acceleration due to gravity $g$ completely cancel out. The coefficient of friction depends only on the applied forces and the angle of the incline.

Q: What if the applied force was 5N up the incline?

If the force was 5N up, it would be between the two extremes. The block wouldn't be on the verge of slipping in either direction. The static friction would simply adjust itself to a value less than $\mu N$ to keep the block perfectly balanced.

Q: Why did we add and subtract the equations instead of substituting?

You can absolutely use substitution! However, adding and subtracting is a powerful algebraic trick here because the coefficients of $mg \sin 30^\circ$ and $\mu mg \cos 30^\circ$ are symmetric (+1 and -1). It makes the math much faster and less prone to errors.

Question

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force $2	ext{ N}$ down the inclined plane. The maximum external force up the inclined plane that does not move the block is $10	ext{ N}$. The coefficient of static friction between the block and the plane is (Take, $g = 10 	ext{ m/s}^2$)

Options:

(A) $\frac{2}{3}$

(B) $\frac{\sqrt{3}}{2}$

(C) $\frac{\sqrt{3}}{4}$

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### The Physics of the Inclined Plane

Imagine a block of mass $m$ resting on a rough inclined plane angled at $30^\circ$ to the horizontal.

Before we even look at the external forces, we must understand the fundamental forces at play. Gravity pulls the block straight down with a force of $mg$.

The inclined plane pushes back with a normal reaction force, $N$, which acts perpendicular to the surface.

To make our analysis simpler, we resolve the gravitational force into two components.

The component perpendicular to the plane is $mg \cos 30^\circ$, which perfectly balances the normal force $N$.

The component parallel to the plane is $mg \sin 30^\circ$, which constantly tries to pull the block down the incline.

### Scenario 1: The Downward Pull

The problem presents us with two distinct limiting cases. In the first case, a maximum force of $2	ext{ N}$ is applied down the incline, and the block remains at rest.

Because this is the **maximum** force before motion begins, the block is on the absolute verge of sliding down.

Static friction is a reactive force that always opposes the tendency of relative motion. Therefore, to prevent the block from sliding down, the static friction must act **upwards** along the incline at its maximum limiting value, $f = \mu N$.

We can now write our first force balance equation. The total downward force must equal the total upward force.

$$2 + mg \sin 30^\circ = \mu mg \cos 30^\circ$$

### Scenario 2: The Upward Pull

Now, let's visualize the second scenario. A maximum force of $10	ext{ N}$ is applied up the incline, and again, the block does not move.

In this situation, the block is on the verge of sliding **up** the incline.

Because the tendency of motion has reversed, the static friction must also flip its direction. It now acts **downwards** along the incline, still at its limiting value of $\mu N$.

Balancing the forces for this second state, the upward force equals the sum of the downward forces.

$$10 = mg \sin 30^\circ + \mu mg \cos 30^\circ$$

### The Mathematical Elegance

We have successfully translated the physics into a neat system of two linear equations:

$$ \mu mg \cos 30^\circ - mg \sin 30^\circ = 2 $$

$$ \mu mg \cos 30^\circ + mg \sin 30^\circ = 10 $$

While we could use substitution, there is a much more elegant algebraic trick. Notice how the $mg \sin 30^\circ$ terms have opposite signs.

If we simply add the two equations together, the sine terms cancel out beautifully.

$$ 2 \mu mg \cos 30^\circ = 12 \implies \mu mg \cos 30^\circ = 6 $$

Similarly, if we subtract the first equation from the second, the friction terms cancel out.

$$ 2 mg \sin 30^\circ = 8 \implies mg \sin 30^\circ = 4 $$

### The Final Calculation

We are now in the home stretch. We have isolated the two critical components of our system.

To find the coefficient of static friction, $\mu$, we divide our first result by our second result.

$$ \frac{\mu mg \cos 30^\circ}{mg \sin 30^\circ} = \frac{6}{4} $$

Notice how the mass $m$ and the acceleration due to gravity $g$ completely cancel out! This tells us that the coefficient of friction is independent of the block's mass.

$$ \mu \cot 30^\circ = \frac{3}{2} $$

We know from trigonometry that $\cot 30^\circ = \sqrt{3}$. Substituting this in, we get our final expression.

$$ \mu (\sqrt{3}) = \frac{3}{2} $$

$$ \mu = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} $$

**The coefficient of static friction between the block and the plane is exactly $\frac{\sqrt{3}}{2}$.**

Select Answer:

Visualized Solution

$Visual Anchor$

$Resolving Forces$

$Components of Gravity$

$Case 1: Downward Force$

$Equation for Case 1$

$Case 2: Upward Force$

$Equation for Case 2$

$Solving the Equations$

$Final Calculation$

The Sigma Insight: Static and Kinetic Friction

The Physics of the Inclined Plane

Scenario 1

Scenario 2

The Mathematical Elegance

The Final Calculation

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List-I

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A solid cylinder of mass $m$ is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is (The coefficient of static friction, $μ_{s}$ , is 0.4)

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Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Upward Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for Case 2

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

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