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

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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
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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...
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## 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.

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