A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough in

Q: Why can't we just use $T + f = mg \sin 60^\circ$ to find the answer?

That equation assumes the cylinder is in perfect equilibrium. You must always check if the surface can actually provide the required friction. If it can't, the equilibrium assumption is wrong!

Q: How do we know the direction of friction?

If the cylinder were to slide down, the contact point would tend to move down the incline. Friction opposes this relative motion, so it acts up the incline.

Q: Why is the torque equation $T \cdot R = f \cdot R$?

We are taking torques about the center of mass. The tension pulls tangentially at the top edge (distance R), creating a counter-clockwise torque. Friction acts tangentially at the bottom edge (distance R), creating a clockwise torque. Gravity and Normal force pass through the center, so they create zero torque.

Q: What happens since the required friction is greater than the maximum static friction?

The cylinder slips! It will accelerate down the incline while unwinding the string. The friction acting on it transitions to kinetic friction, which is equal to the limiting value $\mu_k N$.

Q: Why did we use $\mu_s$ for slipping friction?

In many physics problems, if the coefficient of kinetic friction ($\mu_k$) is not explicitly given, we approximate it using the coefficient of static friction ($\mu_s$).

Question

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, $\mu_s$, is 0.4)

Options:

(A) $\frac{7}{2} mg$

(B) $5 mg$

(C) $\frac{mg}{5}$

(D) $0$

Sigma Percentile · Accepted Answer

Imagine a solid cylinder resting on a $60^\circ$ incline, held back by a string wrapped around its outer edge. It looks like a classic equilibrium problem, right? But physics often hides subtle traps for the unwary. Let's dive into the forces at play and see why assuming equilibrium here might lead you astray.

### The Trap of Equilibrium
When we first look at this system, our instinct is to balance the forces and torques. Let's assume the cylinder is perfectly at rest.

For translational equilibrium along the incline, the upward forces (Tension $T$ and friction $f$) must balance the downward pull of gravity:
$$T + f = mg \sin 60^\circ$$

For rotational equilibrium about the center of mass, the counter-clockwise torque from the tension must balance the clockwise torque from friction:
$$T \cdot R = f \cdot R \implies T = f$$

Substituting $T = f$ into our first equation gives:
$$2f = mg \sin 60^\circ \implies f = \frac{\sqrt{3}}{4} mg \approx 0.433 mg$$

So, to keep the cylinder from slipping and rolling down, the surface needs to provide a frictional force of $0.433 mg$.

### The Reality Check
But can the surface actually provide this much friction? This is where we must check the physical limits of our system. The maximum static friction available is determined by the normal force and the coefficient of static friction $\mu_s$:
$$f_{	ext{max}} = \mu_s N$$

The normal force balances the perpendicular component of gravity:
$$N = mg \cos 60^\circ$$

Given $\mu_s = 0.4$, let's calculate the maximum available friction:
$$f_{	ext{max}} = 0.4 	imes mg \cos 60^\circ = 0.4 	imes mg 	imes 0.5 = 0.2 mg$$

### The Resolution
Look at those two numbers. The system **demands** $0.433 mg$ of friction to stay in equilibrium, but the surface can only **supply** a maximum of $0.2 mg$. The demand exceeds the supply!

Because the required friction is greater than the maximum possible static friction, our initial assumption of equilibrium is shattered. The cylinder will inevitably slip and accelerate down the incline.

When an object slips, the friction acting on it is the kinetic friction (which, in the absence of a separate $\mu_k$, we take as the limiting static friction). Therefore, the actual frictional force acting on the cylinder is simply the maximum available friction:
$$f = 0.2 mg = \frac{mg}{5}$$

This problem is a beautiful reminder: never blindly trust equilibrium equations without verifying that the physical constraints of the system can actually support them!

Select Answer:

Visualized Solution

$Free Body Diagram$

$Equilibrium Assumption$

$Equilibrium Equations$

$Required Friction$

$Maximum Static Friction$

$Calculating f_{max}$

$Conclusion$

$The Way Forward$

The Sigma Insight: Static and Kinetic Friction

The Trap of Equilibrium When we first look at this system, our instinct is to balance the forces and torques

The Reality Check But can the surface actually provide this much friction? This is where we must check the physical limits of our system

The Resolution Look at those two numbers

Similar Questions

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 ( $g = 10 m/s^{2}$ )

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 $g = 10 m/s^{2}$ )

Two masses $m_{1} = 5 kg$ and $m_{2} = 10 kg$ connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$ . The minimum weight $m$ that should be put on top of $m_{2}$ to stop the motion is

A block of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $μ$ . It is desired to make the body move by applying the minimum possible force $F$ . Find the magnitude of $F$ and the direction in which it has to be applied.

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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