Friction and Tensile Forces in a Rope System

What is the maximum ratio of tensile forces in the two rope ends?

What factors need to be considered in determining the maximum ratio of tensile forces in a system involving a rope passing over a fixed sheave?

Maximum Ratio of Tensile Forces:

The maximum ratio of tensile forces in the two rope ends is 1.3.

To find the maximum ratio of tensile forces in the two rope ends, we need to consider the equilibrium condition of the system.

Let's denote the tension in the rope attached to the fixed load as T_load and the tension in the other end of the rope as T_force. The maximum ratio of these tensions occurs when the system is on the verge of motion or just about to move.

Considering the forces acting on the system, we have the following: 1. T_load: The tension in the rope due to the fixed load, acting downward. 2. T_force: The tension in the other end of the rope, acting upward. 3. F_friction: The frictional force acting between the rope and the sheave, opposing the motion.

Since the system is in equilibrium, the sum of the forces in the vertical direction should be zero: T_load - T_force - F_friction = 0

The frictional force can be calculated using the equation: F_friction = μ * N where μ is the coefficient of friction between the rope and the sheave, and N is the normal force acting on the rope due to the sheave.

In this case, the normal force N is equal to T_force, as the rope is passing over the sheave without any vertical displacement. Substituting the value of F_friction into the equilibrium equation: T_load - T_force - μ * T_force = 0 Rearranging the equation: T_load = (1 + μ) * T_force

Therefore, the maximum ratio of tensile forces in the two rope ends is given by: T_load / T_force = (1 + μ) Substituting the given coefficient of friction (μ = 0.30): T_load / T_force = (1 + 0.30) = 1.30

The maximum ratio of tensile forces = 1.30