261_Dynamics 11ed Manual

261_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Engineering Mechanics - Dynamics Chapter 14 Solution: N c Mg cos θ () = 1 2 Mv A 2 Mga ( )sin + μ k N c a 1 2 B 2 = v B v A 2 2 ga ( + 2 k g cos a = v B 4.52 m s = * Problem 14-28 When the skier of weight W is at point A he has a speed v A . Determine his speed when he reaches point B on the smooth slope. For this distance the slope follows the cosine curve shown. Also, what is the normal force on his skis at B and his rate of increase in speed? Neglect friction and air resistance. Given: W 150 lb = v A 5 ft s = a 50 ft = b 100 ft = d 35 ft = Solution: yx a ( )cos π x b
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

Ask a homework question - tutors are online