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SolutionsSet08

SolutionsSet08 - Homework Solutions Set 08 Problem 1 There...

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Homework Solutions Physics 121 Set 08 Spring 2008 - 1 - Problem 1 There are 5 forces acting on the climber. There are shown in the Figure below. d w h mg Nf Nh Fh Ff Since the system is in equilibrium, the net force in all directions must be zero. We immediately can conclude that the magnitude of the normal force associated with the feet is the same as the magnitude of the normal force associated with the hands: N f = N h The total friction force, due to the hands and the feet, is equal to f h + f f ! μ h N h + μ f N f = μ h + μ f ( ) N h Since the net force in the vertical direction must be zero, the magnitude of the friction force must be equal to the magnitude of the weight of the climber. We thus conclude that mg ! μ h + μ f ( ) N h or

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Homework Solutions Physics 121 Set 08 Spring 2008 - 2 - N h ! mg μ h + μ f ( ) The normal forces acting on the hands must be equal to the force with which the climbers pulls with his hands. The second requirement for equilibrium is the requirement that the net torque around any reference point is zero. Since we know the gravitational force, the normal force acting on the hands, and the friction force acting on the hands, we can evaluate the torque with respect to the feet of the climber: ! f = d + w ( ) mg " f h w " N h h = 0 Since the climber is pulling with the minimal force, the normal force will just be sufficient to supply the required friction force. We can thus rewrite the previous equation in the following way: d + w ( ) mg ! f h w = d + w ( ) mg ! μ h N h w ! N h h = d + w ( ) mg ! μ h w + h ( ) mg μ h + μ f ( ) = 0 The only unknown in this equation is the distance h , which we can now determine: h = μ h + μ f ( ) d + μ f w
Homework Solutions Physics 121 Set 08 Spring 2008 - 3 - Problem 2 Since the system is in equilibtrium, the net force on each “know” must be zero. Consider the left knot. The first condition for equilibrium tells us that F x !

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