Question

A uniform rope has been placed on a sloping surface as shown in the figure. The vertical separation and horizontal separation between the end points of the rope are H and X respectively. The friction coefficient ($\mu$) is just good enough to prevent the rope from sliding down. Find the value of $\mu$.

• A. X/H
• B. H/X
• C. H*X
• D. H+X

Answer 1

Answer: (A)

X/H

Answer 2

Let $W$ be the weight of the rope, acting vertically downwards. Let $N_{res}$ be the resultant normal force exerted by the surface on the rope, acting perpendicular to the surface. Let $F_f$ be the resultant friction force exerted by the surface on the rope, acting parallel to the surface, opposing the downward motion.

Since the rope is just prevented from sliding down, it is in equilibrium. This means the net force acting on the rope is zero. The weight $W$ acts vertically downwards. The resultant of the normal force $N_{res}$ and the friction force $F_f$ must balance the weight $W$. Therefore, the resultant force $R = N_{res} + F_f$ must be acting vertically upwards.

The friction force $F_f$ is parallel to the surface, and the normal force $N_{res}$ is perpendicular to the surface. The resultant $R$ is the vector sum of $N_{res}$ and $F_f$. The angle between $N_{res}$ and $F_f$ is $90^\circ$.

Let $\theta$ be the angle the sloping surface makes with the horizontal. The normal force $N_{res}$ is perpendicular to the surface, so it makes an angle of $90^\circ - \theta$ with the horizontal. The friction force $F_f$ is parallel to the surface, so it makes an angle of $\theta$ with the horizontal.

The resultant force $R$ is vertical. Consider the angle between the resultant force $R$ and the normal force $N_{res}$. This angle is $90^\circ - \theta$. In the vector triangle formed by $N_{res}$, $F_f$, and $R$, we have: $\tan(\text{angle between } N_{res} \text{ and } R) = \frac{F_f}{N_{res}}$

Since the rope is on the verge of sliding, the static friction force is at its maximum value, $F_f = \mu N_{res}$. Therefore, $\tan(\text{angle between } N_{res} \text{ and } R) = \mu$.

We know that the angle between $N_{res}$ and $R$ is $90^\circ - \theta$. So, $\tan(90^\circ - \theta) = \mu$. This implies $\cot\theta = \mu$.

Now, we need to determine the effective angle $\theta$ of the slope from the given geometry. The vertical separation between the endpoints is $H$, and the horizontal separation is $X$. This implies that the tangent of the effective angle of the slope is given by the ratio of the vertical separation to the horizontal separation. $\tan\theta = \frac{H}{X}$

Substituting this into the equation for $\mu$: $\mu = \cot\theta = \frac{1}{\tan\theta} = \frac{1}{H/X} = \frac{X}{H}$

Thus, the value of the friction coefficient is $X/H$.