YF_ISM_05

# YF_ISM_05 - APPLYING NEWTON'S LAWS 5 F y 5.1 IDENTIFY a = 0...

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5-1 A PPLYING N EWTON S L AWS 5.1. IDENTIFY: 0 a = for each object. Apply yy Fm a = to each weight and to the pulley. SET UP: Take y + upward. The pulley has negligible mass. Let r T be the tension in the rope and let c T be the tension in the chain. EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a. a = gives r 25.0 N Tw == . (b) The free-body diagram for the pulley is given in Figure 5.1b. cr 2 50.0 N TT . EVALUATE: The tension is the same at all points along the rope. Figure 5.1a, b 5.2. IDENTIFY: Apply m = F a G G to each weight. SET UP: Two forces act on each mass : w down and () = up. EXECUTE: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w . EVALUATE: The tension is the same in all three cases. 5.3. IDENTIFY: Both objects are at rest and 0 a = . Apply Newton’s first law to the appropriate object. The maximum tension max T is at the top of the chain and the minimum tension is at the bottom of the chain. SET UP: Let y + be upward. For the maximum tension take the object to be the chain plus the ball. For the minimum tension take the object to be the ball. For the tension T three-fourths of the way up from the bottom of the chain, take the chain below this point plus the ball to be the object. The free-body diagrams in each of these three cases are sketched in Figures 5.3a, 5.3b and 5.3c. b+c 75.0 kg 26.0 kg 101.0 kg m =+= . b 75.0 kg m = . m is the mass of three-fourths of the chain: 3 4 (26.0 kg) 19.5 kg m . EXECUTE: (a) From Figure 5.3a, 0 y F = gives max b+c 0 Tm g −= and 2 max (101.0 kg)(9.80 m/s ) 990 N T . From Figure 5.3b, 0 y F = gives min b 0 g and 2 min (75.0 kg)(9.80 m/s ) 735 N T . (b) From Figure 5.3c, 0 y F = gives b 0 Tmm g −+ = and 2 (19.5 kg 75.0 kg)(9.80 m/s ) 926 N T =+ = . 5

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5-2 Chapter 5 EVALUATE: The tension in the chain increases linearly from the bottom to the top of the chain. Figure 5.3a–c 5.4. IDENTIFY: Apply Newton’s 1st law to the person. Each half of the rope exerts a force on him, directed along the rope and equal to the tension T in the rope. SET UP: (a) The force diagram for the person is given in Figure 5.4 1 T and 2 T are the tensions in each half of the rope. Figure 5.4 EXECUTE: 0 x F = 21 cos cos 0 TT θθ −= This says that 12 TTT == (The tension is the same on both sides of the person.) 0 y F = sin sin 0 TT m g +− = But , so 2s i n Tm g θ = 2 (90.0 kg)(9.80 m/s ) 2540 N 2sin 2sin10.0 mg T = ° (b) The relation i n g = still applies but now we are given that 4 2.50 10 N T (the breaking strength) and are asked to find . 2 4 (90.0 kg)(9.80 m/s ) sin 0.01764, 22 ( 2 . 5 0 1 0 N ) mg T = × 1.01 . EVALUATE: /(2sin ) g = says that / 2 g = when 90 (rope is vertical). T →∞ when 0 since the upward component of the tension becomes a smaller fraction of the tension. 5.5. IDENTIFY: Apply m = F a G G to the frame. SET UP: Let w be the weight of the frame. Since the two wires make the same angle with the vertical, the tension is the same in each wire. 0.75 Tw = .
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## This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_05 - APPLYING NEWTON'S LAWS 5 F y 5.1 IDENTIFY a = 0...

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