136 - 2.136 Given Find Spring 2004 For the image shown F(t = 5 2t lb(where t is in seconds The coefficient of friction(static and kinetic = 0.2 The

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Spring, 2004 2.136 Given: For the image shown, F(t) = 5 + 2t lb (where t is in seconds). The coefficient of friction (static and kinetic) = 0.2. The box is at rest when t = 0. Its weight is 50-lb. Find: The velocity of the box when t = 10 s Solution: Use Impulse = change in momentum () 0 1 2 2 1 = = = = x m x x m dt i f t t & & & L L F I In the horizontal direction, ( ) ( ) x m f t F f t F x m F x & & & & = = = 2 / 3 30 cos Until the force F(t) is larger than the frictional force, 0 = x & & , i.e., 2 / 3 2 5 t f N + = µ This happens at t*, where t* is found by finding where ( ) 2 / 3 * 2 5 t N + = (1) We find N by looking in the y-direction: () ( ) t mg N mg N t mg N t F y m F y + + = = + + = + = = 5 . 2 0 2 / 2 5 0 30 sin 0 & & t N + = 5 . 52 (2) Using this in (1) ( ) 2 / 3 * 2 5 * 5 . 52 2 . 0 t t + = + , i.e., t*~4.03 s So, until 4.03s, the net force is zero, and after 4.03 s (once the block starts sliding,
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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