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Assignment 3 - PHYS 230 Fall 2006 Assignment 3 This...

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PHYS 230 Fall 2006 Assignment 3 This assignment is about forces and conservation of energy, with some applications of calculus that may be new to you. 1. [5 points.] A boat is secured to the dockside by a rope wrapped around a circular post, as shown. The rope subtends an angle θ on the post, as shown . . . of course, θ can be greater than 2 π . The coefficient of static friction between the rope and the post is μ 0 . 25. &% ’$ - @ @ @I T 0 T θ ¡ ¡ The owner of the boat is pulling on one end of the rope with a force T 0 . Show that, when θ 3 π , the force on the boat caused by the rope from the dockside is approximately 10 T 0 . This problem resembles the hanging rope example done in class – see the pdf on WebCT. We investigate the forces acting on a small piece of rope in contact with the post. The small piece has angular extent Δ θ . Therefore, its mass is Δ m = ρ Δ θ , where ρ is the mass per unit arc length. The forces are all “small” except T ( θ ) and T ( θ + Δ θ ) : they are the normal force Δ N and the friction force Δ F f . To fix ideas, imagine that T ( θ + Δ θ ) is greater than T ( θ ) , which means that Δ F f is towards the left. 1
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¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ D D D D D D D D D D D D Δ θ θ » » » » »9 X X X X Xz - 6 T ( θ ) T ( θ + Δ θ ) Δ F f Δ N The two tensions are directed tangentially at the two ends of the small piece of rope. The Normal force is directed radially outwards, and the friction force is also tangential – can you see why the exact spot at which it acts does not matter? Since the rope is in equilibrium, we must have: Δ N = T ( θ ) sin (Δ θ/ 2) + T ( θ + Δ θ ) sin (Δ θ/ 2) and T ( θ + Δ θ ) cos (Δ θ/ 2) - T ( θ ) cos (Δ θ/ 2) = Δ F f In the usual way, we write T ( θ θ ) as T ( θ )+Δ T T ( θ )+ dT Δ θ , write sin (Δ θ/ 2) Δ θ/ 2 , cos (Δ θ/ 2) 1 , and ignore products of small quantities. This gives: Δ N T ( θ θ and dT Δ θ Δ F f = μ Δ N Eliminating Δ N : dT μT ( θ ) We can now integrate: Z dT T = Z μdθ or ln T = μθ + constant In order that T = T 0 when θ = 0 , the constant of integration must be ln T 0 , so that T = T 0 e μθ When θ = 3 π , T 10 T 0 . 2
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2. [10 points.] Very similar to a problem we worked in class . . . but with an extra twist. A prisoner tries to escape to freedom by sliding down a rope. The top end of the rope is hung to a hook which can support a maximum vertical force of 600 N. The rope is 15 m long, and weighs 100 N. The prisoner weighs 700 N. Take g = 10 m/s 2 . (a) What is the smallest velocity with which the prisoner can reach the bottom of the rope? (He does not touch the ground.) Ans: approx. 9 m/s. The force that the prisoner exerts on the rope – and which the rope exerts on him – cannot be equal to his weight: if it were so, then the hook would give way. This is OK, however, because the prisoner is accelerating downwards, so that there must be a net force downwards on him.
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