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hw5_sol - Dynamics 487 Problem 4.67 The rope of mass m and...

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Dynamics 487 Problem 4.67 The rope of mass m and length l is released from rest with a very small amount of it hanging over the edge (i.e., s > 0 but it is very close to zero). Determine the speed of the rope as a function of s . Assume the surface is smooth and that there is no dissipation in the rope as it bends. Solution Since the system is conservative the work-energy principle is T 1 C V 1 D T 2 C V 2 ; (1) where position ± is taken to be when s ± 0 and position ² occurs at some point when s > 0 . With the datum as the table surface, it follows that V 1 D 0 and V 2 D W v . ² s=2/ ; (2) where W v refers to the weight of the vertical portion of the rope. The weight of the horizontal portion of the rope is W h , the constraint force perpendicular to the table is N and the rope’s mass is m ; it follows that W h1 D mg; W h2 D O.l ² s/=l ± mg; and W v2 D .s=l/ mg: (3) Applying Eqs. (3) to Eqs. (2) leads to the following potential energy terms: V 1 D 0 and V 2 D ² s 2 mg=2l: (4) The kinetic energies are given by T 1 D 0; T 2 D 1 2 O.l ² ± mv 2 h2 C 1 2 .s=l/ mv 2 D 1 2 mv 2 2 ; (5) where we note that v h2 D v D v 2 . Applying Eqs. (4) and (5) to Eq. (1), the answer for part (a) is 0 D mv 2 2 =2 ² s 2 mg=2l ) v 2 D s r g l , (6) for s > 0 .
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538 Solutions Manual Problems 5.23 through 5.25 Two persons A and B weighing 150 lb and 180 lb, respectively, jump off a Foating platform (in the same direction) with a velocity relative to the platform that is completely horizontal and with magnitude v 0 D 6 ft = s for both A and B . The Foating platform weighs 800 lb and assume that A , B , and the platform are initially at rest. Problem 5.23 Neglecting the water resistance to the horizontal motion of the platform, determine the speed of the platform after A and B jump at the same time. Problem 5.24 Neglecting the water resistance to the horizontal motion of the platform, and knowing that B jumps ±rst, determine the speed of the platform after both A and B have jumped. Problem 5.25 Neglecting the water resistance to the horizontal motion of the platform, and knowing that A jumps ±rst, determine the speed of the platform after both A and B have jumped. WARNING: the weight of A has been changed to 140 lb. Solution to 5.23 There are no external forces acting on the system. Therefore, the momentum is conserved in the x direction. Let ± and ² represent the time instants immediately before and after the jump. m A .v Ax / 1 C m B .v Bx / 1 C m P .v Px / 1 D m A .v Ax / 2 C m B .v / 2 C m P .v / 2 : (1) Recall that all masses are at rest at ± and that .v Ax / 2 D .v / 2 . Because the velocity v 0 is a relative velocity we must write ± v 0 D .v Ax / 2 ± .v / 2 ) .v Ax / 2 D .v / 2 ± v 0 : Eq. (1) becomes 0 D ± .v / 2 ± v 0 ² .m A C m B / C m P .v / 2 : Solve for .v / 2 .v / 2 D .m A C m B / v 0 m A C m B C m P D 1:71 ft = s : Solution to 5.24 There are no external forces acting on the system. Therefore, the momentum
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hw5_sol - Dynamics 487 Problem 4.67 The rope of mass m and...

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