hw14 - TAM212 Introductory Dynamics Spring 2005 S....

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Unformatted text preview: TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 14 (2.125, 2.127) 2.125 (a) Total Work done = Change in kinetic energy W = ∆T Wspring + Wgravity on A + W f riction on B = T f inal − Tinitial = 0 − Tinitial A B = − Tinitial + Tinitial Let the distance by which A dropped (or B moved, or spring stretched) be y A . k δ (t1 )2 − δ (t2 )2 2 2 1 80 1 − − + yA = − 2 2 2 Wspring = = −40y2 + 40y A A Wg on A = m A y A = 161y A W f riction on B = µm B y A = 0.2(193)y A = 38.6y A On the right hand side RHS = − 1 163 2 1 193 2 (5) + (5) 2 32.2 2 32.2 Equating the two sides −40y2 + 162.4y A + 138 = 0 A y A = 4.78 f t 1 2 (b) If A stayed in its position, then the force in the cable holding it up will be 161 lb, which leaves A in equilibrium. Lets look at the free body diagram for B in such a state. 193 kδ f 161 B f 193 1 kδ f = 80(y A − ) 2 = 342 And this has to be balanced by the friction force and the weight of A. 342 = f + 161 f = 181 Lets check what is the maximum friction force which the weight of B will generate f MAX = 0.2(193) = 38.6 lb Since f > f MAX , the friction force along with the weight of A is not enough to keep the spring stretched in its current state. And so, A starts upward. 2.127 The length of the rope is a constant. Using this, (y A − K1 ) + π r1 + (y A − K1 − K2 ) + π r2 + (y B − K2 ) = constant ˙ ˙ ˙ ˙ Differentiating, 2 y A + y B = 0 y B = −2 y A So, B goes up twice as fast as A comes down. So A drops down y = 10 m when they meet. 2 K2 r 2 y A y B r1 K1 A B W = ∆T 1 1 m A v2 + m B v2 m A gy − m B gy = A B 2 2 15 2 5 15g(10) − 5g(20) = v A + (2v A )2 2 2 50(9.81) vA = 17.5 v A = 5.29 f t/sec (Down) v B = 10.26 f t/sec (Up) 3 ...
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hw14 - TAM212 Introductory Dynamics Spring 2005 S....

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