183_Dynamics 11ed Manual

183_Dynamics 11ed - Engineering Mechanics Dynamics Given 2 2 h sB Chapter 13 sA h = 2h sA vA vB =0 2 2 sA h 2 aB 2 vA sA aA sA vA 3 2 2 sA h 2

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

View Full Document Right Arrow Icon
Engineering Mechanics - Dynamics Chapter 13 Given hs B s A 2 h 2 + + 2 h = v B s A v A s A 2 h 2 + + 0 = a B v A 2 s A a A + s A 2 h 2 + + s A 2 v A 2 s A 2 h 2 + () 3 2 0 = TM g Ma B = T s B v B a B Find Ts B , v B , a B , = s B 1m = a B 2.203 m s 2 = v B 1.538 m s = T 1.802 kN = *Problem 13-48 Block B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the block’s vertical position y so that when is applied the block rises with a constant acceleration a B . Neglect the mass of the cord and pulleys. Solution: 2 F cos θ mg
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