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PHYS310_homework_07_solutions

PHYS310_homework_07_solutions - Homework set 7 due...

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Homework set 7, due Monday, October 17, 11:30 am Following problems from Analytical Mechanics by Fowles & Cassiday 1. 5.1 (20 points) 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force acting on him is zero. The scale exerts an upward force, N r , whose value is equal to the scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus 0 0 N mg mA + - = r r r 0 5 0 4 4 g N mg mA N mg m N mg - - = - - = - = 5 5 4 4 W N mg W ′ = = = 150 . W lb ′ = (b) The acceleration is downward, in the same direction as g r 0 4 g N mg m - + = 3 4 4 W W W W ′ = - = 90 . W lb ′ = 2. 5.2 (20 points) (a) ( 29 cent F m r ϖ ϖ = - × × r r r r For r ϖ r r , 2 ˆ cent r F m r e ϖ = r 1 1 500 1000 s s ϖ π - - = = ( 29 2 6 2 ˆ 10 1000 5 5 cent r F e π π - = × × = r dynes outward (b) ( 29 2 2 4 1000 5 5.04 10 980 cent g F m r F mg π ϖ = = = × 3. 5.5 (20 points) 5.5 (a) f mg μ = - is the frictional force acting on the 0 A r (a) (b) N r mg r

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box, so 0 f mA ma - = r r r ( a r is the acceleration of the box relative to the truck. See Equation 5.1.4b.) Now, f
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