SolutionsSet08

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 + f N f = h + 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 ! h + f ( ) N h or
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Homework Solutions Physics 121 Set 08 Spring 2008 - 2 - N h ! mg μ h + 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 ! h N h w ! N h h = d + w ( ) mg ! h w + h ( ) mg h + f ( ) = 0 The only unknown in this equation is the distance h , which we can now determine: h = h + f ( ) d + f w
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/20/2010 for the course PHY PHY 121 taught by Professor Wolfs during the Spring '08 term at Rochester.

Page1 / 10

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online