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Unformatted text preview: 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 on the post, as shown . . . of course, can be greater than 2 . The coefficient of static friction between the rope and the post is . 25. &% $ @ @ @I T T The owner of the boat is pulling on one end of the rope with a force T . Show that, when 3 , the force on the boat caused by the rope from the dockside is approximately 10 T . 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 . Therefore, its mass is m = , where is the mass per unit arc length. The forces are all small except T ( ) and T ( + ) : they are the normal force N and the friction force F f . To fix ideas, imagine that T ( + ) is greater than T ( ) , which means that F f is towards the left. 1 D D D D D D D D D D D D 9 X X X X Xz 6 T ( ) T ( + ) F f 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: N = T ( ) sin ( / 2) + T ( + ) sin ( / 2) and T ( + ) cos ( / 2) T ( ) cos ( / 2) = F f In the usual way, we write T ( + ) as T ( )+ T T ( )+ dT d , write sin ( / 2) / 2 , cos ( / 2) 1 , and ignore products of small quantities. This gives: N T ( ) and dT d F f = N Eliminating N : dT d T ( ) We can now integrate: Z dT T = Z d or ln T = + constant In order that T = T when = 0 , the constant of integration must be ln T , so that T = T e When = 3 , T 10 T . 2 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....
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This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.
 Fall '07
 Harris
 Conservation Of Energy, Energy, Force

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