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b) When the rope starts to brake the fall at t = to the rope is stretched resulting in a force F. Let y = y(t) be the length of the rope stretched at time t. Formulate an initial value problem that describes the fall for t > to = 0.

b) When the rope starts to brake the fall at t = to the rope is stretched resulting in a force F. Let y = y(t) be the length of the rope stretched at time t. Formulate an initial value problem that describes the fall for t > to = 0.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question b)

We are going to create a model for the forces acting when a climber with mass m falls on a rope. To simplify the problem, we will I assume that there is no external and internal friction associated with the rope, and that there is no slack in the rope when the climber falls a height h/2 above the last belay, see Figure 1. The rope has length L (in meters m) when the fall occurs. All motion in the model is vertical (along the y-axis), and we will assume that the positive direction is downward. Figure 1: Climber with mass m falling freely for a length h until the elastic rope starts to brake the fall at y = 0 and t = to = 0. At this time, the climber has speed vo. The rope has length L, is fixed at point S, and is without slack when the climber falls. σ = Eε Here, tension o is defined as force per area: a) The climber is in free fall until the rope starts to catch the fall. Set up the model that applies to free fall and derive the expression for the speed vo the climber has when the rope starts to brake the fall. In this subquestion and in subquestion d) you can assume that the speed of the climber is only a function of y such that: dy v(y). d²y dv dt² dy = V dt b) When the rope starts to brake the fall at t = to the rope is stretched resulting in a force F. Let y = y(t) be the length of the rope stretched at time t. Formulate an initial value problem that describes the fall for t > to = 0. F A and & is the deformation (strain) of the rope given by the relative stretch c) In the problem you have found in b) the force from the rope that brakes the fall is included. We will assume that this follows Hook's law. Using the terminology from elasticity theory, we can express Hook's law by the tension (stress) o on the rope: y(t) - F₂ (y) 9 W² 0 = ε = (1 - cos(wt)) + Y L EA L L Elastic modulus E is a material constant and A is the cross section of the rope. We will assume that EA is constant. Show that the tensile forceF, in the model with these assumptions becomes Y (1) and that the solution of the initial value problem from b) given the force (4) becomes: √2gh W -sin(wt) (2) (3) h y = 0 (4) (5)
Transcribed Image Text:We are going to create a model for the forces acting when a climber with mass m falls on a rope. To simplify the problem, we will I assume that there is no external and internal friction associated with the rope, and that there is no slack in the rope when the climber falls a height h/2 above the last belay, see Figure 1. The rope has length L (in meters m) when the fall occurs. All motion in the model is vertical (along the y-axis), and we will assume that the positive direction is downward. Figure 1: Climber with mass m falling freely for a length h until the elastic rope starts to brake the fall at y = 0 and t = to = 0. At this time, the climber has speed vo. The rope has length L, is fixed at point S, and is without slack when the climber falls. σ = Eε Here, tension o is defined as force per area: a) The climber is in free fall until the rope starts to catch the fall. Set up the model that applies to free fall and derive the expression for the speed vo the climber has when the rope starts to brake the fall. In this subquestion and in subquestion d) you can assume that the speed of the climber is only a function of y such that: dy v(y). d²y dv dt² dy = V dt b) When the rope starts to brake the fall at t = to the rope is stretched resulting in a force F. Let y = y(t) be the length of the rope stretched at time t. Formulate an initial value problem that describes the fall for t > to = 0. F A and & is the deformation (strain) of the rope given by the relative stretch c) In the problem you have found in b) the force from the rope that brakes the fall is included. We will assume that this follows Hook's law. Using the terminology from elasticity theory, we can express Hook's law by the tension (stress) o on the rope: y(t) - F₂ (y) 9 W² 0 = ε = (1 - cos(wt)) + Y L EA L L Elastic modulus E is a material constant and A is the cross section of the rope. We will assume that EA is constant. Show that the tensile forceF, in the model with these assumptions becomes Y (1) and that the solution of the initial value problem from b) given the force (4) becomes: √2gh W -sin(wt) (2) (3) h y = 0 (4) (5)
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