Question

A rope rests on two platforms, symmetrically both inclined as shown in figure. Coefficient of friction is 1 for the inclined surface.

• A. largest possible fractions of the rope that does not touch the platform is $\frac{n}{n+1}$ where n is (sin θ cos θ - sin$^2$θ)
• B. largest possible fractions of the rope that does not touch the platform is $\frac{n}{3+n}$ where n is (sin θ cos θ - sin$^2$θ)
• C. Decreasing θ would lead to smaller friction and bigger hanging fraction.
• D. Decreasing θ would result in smaller requirement of friction for equilibrium.

Answer 1

Answer: (A)

A

Answer 2

Let the total length of the rope be $L$. Let the length of the rope on each inclined platform be $L_p$. Let the total length of the rope that does not touch the platform (i.e., the hanging parts) be $L_h$. Due to symmetry, $L_h$ is composed of two equal hanging sections, one on each side. Let the length of the hanging part on one side be $L_{hang}$. So $L_h = 2 L_{hang}$. The total length of the rope is $L = 2L_p + L_h = 2L_p + 2L_{hang}$.

We want to find the largest possible fraction of the rope that does not touch the platform, which is $\frac{L_h}{L}$.

Consider the equilibrium of the rope segment on one inclined plane and the corresponding hanging part. Let the mass per unit length of the rope be $\lambda$. The mass of the rope segment on one inclined plane is $m_p = \lambda L_p$. The mass of the hanging rope on one side is $m_{hang} = \lambda L_{hang}$.

For the largest possible hanging fraction, the rope is on the verge of slipping. The hanging part pulls the rope down, and the component of the rope's weight on the incline also pulls it down. The friction force on the incline acts upwards to prevent slipping.

Forces acting on the rope segment on one inclined plane:

1. Weight component along the incline: $m_p g \sin\theta = \lambda L_p g \sin\theta$. This force acts down the incline.
2. Tension from the hanging part: The hanging part pulls the rope segment on the incline with a tension $T = m_{hang} g = \lambda L_{hang} g$. This force also acts down the incline.
3. Normal force: $N = m_p g \cos\theta = \lambda L_p g \cos\theta$. This force acts perpendicular to the incline.
4. Maximum static friction force: $f_{max} = \mu N = \mu (\lambda L_p g \cos\theta)$. This force acts up the incline to prevent slipping.

For equilibrium (or verge of slipping), the sum of forces along the incline must be zero: $T + m_p g \sin\theta = f_{max}$ $\lambda L_{hang} g + \lambda L_p g \sin\theta = \mu \lambda L_p g \cos\theta$

Divide by $\lambda g$: $L_{hang} + L_p \sin\theta = \mu L_p \cos\theta$ $L_{hang} = L_p (\mu \cos\theta - \sin\theta)$

Now, we need to find the fraction $\frac{L_h}{L} = \frac{2L_{hang}}{2L_p + 2L_{hang}} = \frac{L_{hang}}{L_p + L_{hang}}$.

Substitute the expression for $L_{hang}$: $\frac{L_h}{L} = \frac{L_p (\mu \cos\theta - \sin\theta)}{L_p + L_p (\mu \cos\theta - \sin\theta)}$ $\frac{L_h}{L} = \frac{\mu \cos\theta - \sin\theta}{1 + \mu \cos\theta - \sin\theta}$

Given that the coefficient of friction $\mu = 1$: $\frac{L_h}{L} = \frac{1 \cdot \cos\theta - \sin\theta}{1 + 1 \cdot \cos\theta - \sin\theta}$ $\frac{L_h}{L} = \frac{\cos\theta - \sin\theta}{1 + \cos\theta - \sin\theta}$

The derived fraction is of the form $\frac{X}{1+X}$ where $X = \cos\theta - \sin\theta$. Option (A) has the form $\frac{n}{n+1}$. Although the definition of $n$ in option (A) is given as $(\sin\theta \cos\theta - \sin^2\theta)$, which is inconsistent with our derived $X$, the functional form of option (A) matches the derived result. Options (C) and (D) are incorrect because decreasing $\theta$ increases the required friction for equilibrium (as $\cos\theta$ increases).