A block rests on a rough inclined plane making an angle of 30 degrees with the horizontal. The coeff

Q: Why didn't we use $\mu_s = 0.8$ in our calculation?

Because the block is not on the verge of slipping. The formula $f = \mu_s N$ only gives the maximum possible static friction, not the actual friction acting when the body is comfortably at rest.

Q: What if the block was on the verge of slipping?

If the problem stated the block was 'just about to slide', then we would use the limiting friction formula $f = \mu_s N$ to find the mass.

Q: How do we know for sure it's not on the verge of slipping?

If you calculate the maximum friction $\mu_s N$, it comes out to be $0.8 \times m \times 10 \times \cos 30^\circ \approx 13.85 \text{ N}$. Since the actual friction required to hold it is only $10 \text{ N}$, it is well below the maximum limit.

Q: Why is friction acting upwards along the incline?

Gravity is trying to pull the block down the incline. Friction always opposes the impending relative motion, so it must act upwards along the plane.

Q: What is the normal force in this scenario?

The normal force balances the perpendicular component of gravity. So, $N = mg \cos 30^\circ = 2 \times 10 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \text{ N}$.

Question

A block 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 $0.8$. If the frictional force on the block is $10	ext{ N}$, the mass of the block (in $	ext{kg}$) is ($g = 10	ext{ m/s}^2$)

Options:

(A) 2.0

(B) 4.0

(C) 1.6

(D) 2.5

Sigma Percentile · Accepted Answer

### The Illusion of Maximum Friction

Imagine a block resting peacefully on a rough inclined plane. The plane is tilted at an angle of $30^\circ$ to the horizontal. Gravity is relentlessly trying to pull the block down the slope, but friction is standing its ground, keeping the block perfectly still.

The problem gives us a very tempting piece of information: the coefficient of static friction, $\mu_s = 0.8$. For many students, seeing $\mu_s$ is like seeing a green light to immediately write down the formula $f = \mu_s N$. But beware—this is a classic trap designed to test your conceptual clarity!

### The Nature of Static Friction

Static friction is not a fixed value; it is a **smart, self-adjusting force**. It only works as hard as it needs to. The formula $f_{max} = \mu_s N$ only gives us the *maximum* possible value of static friction, which comes into play only when the object is on the absolute verge of slipping.

Since the problem simply states that the block "rests" on the plane, we cannot assume it is about to slip. Instead, we must rely on the fundamental condition of equilibrium: the net force must be zero.

### Balancing the Forces

Let's break down the forces acting on the block. Gravity ($mg$) acts straight down, but we can resolve it into two components:
1. Perpendicular to the plane: $mg \cos 30^\circ$ (balanced by the normal force $N$).
2. Parallel to the plane: $mg \sin 30^\circ$ (trying to pull the block down).

To keep the block at rest, the static friction $f$ acting up the incline must perfectly balance the downward pull of gravity. Therefore, our master equation is simply:
$$f = mg \sin 30^\circ$$

### The Final Calculation

We are given that the actual frictional force acting on the block is $10 	ext{ N}$. We also know that $g = 10 	ext{ m/s}^2$ and $\sin 30^\circ = 0.5$. Let's substitute these values into our equilibrium equation:
$$10 = m \cdot 10 \cdot \sin 30^\circ$$

$$10 = m \cdot 10 \cdot 0.5$$

$$10 = 5m$$

Solving for the mass $m$, we get:
$$m = 2.0 	ext{ kg}$$

And just like that, the problem unravels! The coefficient of friction $\mu_s = 0.8$ was nothing more than a red herring. By trusting the physical state of the system over blind formula application, we arrive smoothly at the correct answer.

Select Answer:

Visualized Solution

$Free Body Diagram$

$Resolving Gravity$

$The μ_{s} Trap$

$Equilibrium Condition$

$Substituting Values$

$Solving for m$

$Final Answer$

The Sigma Insight: Static and Kinetic Friction

The Illusion of Maximum Friction

The Nature of Static Friction

Balancing the Forces

The Final Calculation

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