Solution to Question 23
Let $\lambda$ be the mass per unit length of the rope. Let the length of the hanging portion $CD$ be $L$. According to the problem, the length of the portion on the floor $DE$ is $2L$.
System Equilibrium Analysis
The rope is being pulled slowly, meaning the system is in equilibrium. We analyze the forces on the distinct sections of the rope.
1. Section DE (On the floor)
- The rope lies horizontally.
- Normal reaction from the floor: $N = (\lambda \cdot 2L) g$.
- Friction force opposing motion: $f_k = \mu N = 2\mu \lambda L g$.
- Tension at point D ($T_D$) must overcome this friction to drag the rope.
- Equation: $T_D = 2\mu \lambda L g$.
2. Section CD (Hanging portion)
- This section hangs freely between the lift-off point C and the touch-down point D.
- Forces at D: Tension $T_D$ acts horizontally to the right.
- Forces at C: Tension $T_C$ acts tangentially to the cylinder. Since the arc BC subtends angle $\theta$ at the center (measured from the vertical), the tangent at C makes an angle $\theta$ with the horizontal.
- Weight: The total weight of this section is $W = \lambda L g$.
We resolve the tension at C into horizontal and vertical components:
- Horizontal Equilibrium: $T_C \cos \theta = T_D$
- Vertical Equilibrium: $T_C \sin \theta = W = \lambda L g$
Solving for Coefficient of Friction
Substitute $T_D = 2\mu \lambda L g$ into the horizontal equilibrium equation:
$$ T_C \cos \theta = 2\mu \lambda L g \quad \dots(1) $$From the vertical equilibrium equation:
$$ T_C \sin \theta = \lambda L g \quad \dots(2) $$Divide equation (2) by equation (1):
$$ \frac{T_C \sin \theta}{T_C \cos \theta} = \frac{\lambda L g}{2\mu \lambda L g} $$ $$ \tan \theta = \frac{1}{2\mu} $$ $$ \mu = \frac{1}{2 \tan \theta} = \frac{\cot \theta}{2} $$Numerical Substitution
We are given $\theta = \sin^{-1}(0.8)$.
- $\sin \theta = 0.8 = \frac{4}{5}$
- $\cos \theta = \sqrt{1 – 0.8^2} = 0.6 = \frac{3}{5}$
- $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{0.6}{0.8} = \frac{3}{4}$
Substituting this back into the expression for $\mu$:
$$ \mu = \frac{3/4}{2} = \frac{3}{8} $$