195_Dynamics 11ed Manual

# 195_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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

Engineering Mechanics - Dynamics Chapter 13 Given: v 50 m s = θ 15 deg = M 70 kg = g 9.815 m s 2 = Solution: + Σ F b = ma b ; N p sin () Mg 0 = N p M g sin () = N p 2.654 kN = Σ F n = ma n ; N P cos () M v 2 ρ = M v 2 N p cos () = 68.3 m = Problem 13-65 The man has weight W and lies against the cushion for which the coefficient of static friction is μ s . Determine the resultant normal and frictional forces the cushion exerts on him if, due to rotation about the z axis, he has constant speed v . Neglect the size of the man. Given: W 150 lb = s 0.5 = v 20 ft s = 60 deg = d 8f t = Solution:
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