126_PartUniversity Physics Solution

126_PartUniversity Physics Solution - 4-18 4.55. Chapter 4...

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4-18 Chapter 4 EVALUATE: The tension in the rope is not constant but increases from the bottom of the rope to the top. The tension at the top of the rope must accelerate the rope as well the 5.00-kg block. The tension at the top of the rope is less than F ; there must be a net upward force on the 6.00-kg block. 4.55. IDENTIFY: Apply m = F a ! ! to the barbell and to the athlete. Use the motion of the barbell to calculate its acceleration. SET UP: Let be upward. y + EXECUTE: (a) The free-body diagrams for the baseball and for the athlete are sketched in Figure 4.55. (b) The athlete&s weight is . The upward acceleration of the barbell is found from 2 (90.0 kg)(9.80 m/s ) 882 N mg == 2 1 00 2 y y yy v t a t −= + . 2 0 22 2( ) 2(0.600 m) 0.469 m/s (1.6 s) y yy a t = . The force needed to lift the barbell is given by lift barbell y Fw m a −= . The barbell&s mass is , so . 2 (490 N) (9 80 m s ) 50 0 kg /. / = . 2 lift barbell 490 N (50.0 kg)(0.469 m/s ) 490 N 23 N 513 N m a =+ = The athlete is not accelerating, so floor lift athlete 0 FF w . . floor lift athlete 513 N 882 N 1395 N w = + = EVALUATE: Since the athlete pushes upward on the barbell with a force greater than its weight the barbell pushes down on him and the normal force on the athlete is greater than the total weight, 1362 N, of the athlete plus barbell. Figure 4.55 4.56. IDENTIFY: Apply m = F a ! ! to the balloon and its passengers and cargo, both before and after objects are dropped overboard. SET UP: When the acceleration is downward take y + to be downward and when the acceleration is upward take to be upward. y + EXECUTE: (a) The free-body diagram for the descending balloon is given in Figure 4.56. L is the lift force. (b) y y Fm a ∑= gives and ( /3) Mg L M g 2 /3 L Mg = . (c) Now is upward, so , where m is the mass remaining. y + ( /2) Lm gm g 2/ 3 L Mg = , so . Mass 5 must be dropped overboard. 4 /9 mM = M EVALUATE: In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the total weight. Figure 4.56
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Newton&s Laws of Motion 4-19 4.57. IDENTIFY: Apply m = F a ! ! to the entire chain and to each link. SET UP: mass of one link. Let m = y + be upward. EXECUTE: (a) The free-body diagrams are sketched in Figure 4.57. is the force the top and middle links exert on each other. is the force the middle and bottom links exert on each other.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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126_PartUniversity Physics Solution - 4-18 4.55. Chapter 4...

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