284_Dynamics 11ed Manual

284_Dynamics 11ed Manual - ( ) 2 = T 2 1 2 W g v 2 2 = V 2...

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Engineering Mechanics - Dynamics Chapter 14 Solution: T 1 1 2 Mv 0 2 = V 1 0 = T 2 0 = V 2 1 2 k A k B + () d 2 = 1 2 Mv 0 2 0 + 0 1 2 k A k B + () d 2 + = d M k A k B + v 0 = d 0.73 m = * Problem 14-68 A block of weight W rests on the smooth semicylindrical surface. An elastic cord having a stiffness k is attached to the block at B and to the base of the semicylinder at point C . If the block is released from rest at A ( θ = 0 ° ), determine the unstretched length of the cord so the block begins to leave the semicylinder at the instant = 2 . Neglect the size of the block. Given: W 2lb = a 1.5 ft = k 2 lb ft = g 9.81 m s 2 = 2 45 deg = Solution: T 1 0 = V 1 1 2 k π a
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Unformatted text preview: ( ) 2 = T 2 1 2 W g v 2 2 = V 2 1 2 k 2 ( ) a 2 W a ( )sin 2 ( ) + = Guess 1 ft = v 2 1 ft s = Given W sin 2 ( ) W g v 2 2 a = 1 2 k a ( ) 2 + 1 2 W g v 2 2 1 2 k 2 ( ) a 2 W a ( )sin 2 ( ) + + = v 2 Find v 2 , ( ) = v 2 5.843 ft s = 2.77 ft = 282...
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