638_Dynamics 11ed Manual

638_Dynamics 11ed Manual - is then placed against the wall,...

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

View Full Document Right Arrow Icon
Engineering Mechanics - Dynamics Chapter 19 t 1 3s = W S 180 lb = k A 1.25 ft = r i 1.5ft = r o 2.75 ft = g 32.2 ft s 2 = Solution: (a) System as a whole 0 W B r i t 1 + W S g k A 2 ω W B g r i () r i + = W B r i t 1 g W S k A 2 W B r i 2 + = 2.48 rad s = (b) Parts separately Guesses T 1lb = 1 rad s = Given 0 Tt 1 r i + W S g k A 2 = Tt 1 W B t 1 W B g r i () = T Find T , () = T 4.81 lb = 2.48 rad s = Problem 19-9 The disk has mass M and is originally spinning at the end of the strut with angular velocity . If it
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: is then placed against the wall, for which the coefficient of kinetic friction is k determine the time required for the motion to stop. What is the force in strut BC during this time? Given: M 20 kg = 60 deg = 60 rad s = r 150 mm = k 0.3 = g 9.81 m s 2 = 636...
View Full Document

Ask a homework question - tutors are online