hw15 - (5 2 t-10 dt = 50 32 2 v f-0 f f t 2-5 t)f f 10 2 5...

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

View Full Document Right Arrow Icon
TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 15 (2.135, 2.139) 2.135 The maximum friction force generated is f max = μ N = 0 . 2(50) = 10 lb . The box doesn’t start moving till this force is matched and exceeded. F ( t ) = 10 5 + 2 t = 10 t = 2 . 5 sec Using Euler’s Frst law of motion Z 10 2 . 5 ( F ( t ) - f max ) dt = m ( v f - v i ) Z 10 2 . 5
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (5 + 2 t-10) dt = 50 32 . 2 ( v f-0) f f ( t 2-5 t )f f 10 2 . 5 = 50 32 . 2 v f v f = 36 . 2 f t / sec 2.139 Using conservation of momentum ( m 1 + m 2 ) v f = m 1 v 1 + m 2 v 2 50 g v f = 20 g (1) + v f = . 4 / ; f t / sec 1...
View Full Document

This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online