65. (a) To hold the crate at equilibrium in the final situation, GFmust have the same magnitude as the horizontal component of the rope’s tension Tsin θ, where is the angle between the rope (in the final position) and vertical: =FHGIKJ=°−sin....140012 019 5 But the vertical component of the tension supports against the weight: Tcos =mg.Thus, the tension isT= (230 kg)(9.80 m/s2)/cos 19.5° = 2391 N andF= (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.