ch07-p065 - 65(a To hold the crate at equilibrium in the...

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65. (a) To hold the crate at equilibrium in the final situation, G F must have the same magnitude as the horizontal component of the rope’s tension T sin θ , where is the angle between the rope (in the final position) and vertical: = F H G I K J sin . . .. 1 400 12 0 19 5 But the vertical component of the tension supports against the weight: T cos = mg . Thus, the tension is T = (230 kg)(9.80 m/s 2 )/cos 19.5° = 2391 N and F = (2391 N) sin 19.5° = 797 N. An alternative approach based on drawing a vector triangle (of forces) in the final situation provides a quick solution. (b) Since there is no change in kinetic energy, the net work on it is zero.
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This note was uploaded on 03/19/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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