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13.20 The system shown is at rest when a constant 30-lb force is applied to collar B. (a) If the force acts through the entir
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(13.20 0 Note: I ft = 0.3 m I lb =0.454 kg Given: 0) mars of collar, m = 18 16 me 18 x 0.454 = 8.172 kq solution: @ ft = 2x0.- (b) Distance a removed after which 30 16 so that . Vc to force to be Boo6m 0 V, 50 m | 30X0.454X9. → 6X0.454X g.81 N Veso →

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