Solutions06

# Solutions06 - Physics 140 Discussion 06 Solutions 5.3...

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Physics 140 - Discussion 06 - Solutions 5.3. I DENTIFY : 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. S ET U P : 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 = . = . E XECUTE : (a) From Figure 5.3a, 0 Σ = y F gives max b c 0 T m g + - = and 2 max (101 0 kg)(9 80 m/s ) 990 N. T = . . = From Figure 5.3b, 0 Σ = y F gives min b 0 T m g - = and 2 min (75 0 kg)(9 80 m/s ) 735 N. T = . . = (b) From Figure 5.3c, 0 Σ = y F gives b ( ) 0 - + = T m m g and 2 (19 5 kg 75 0 kg)(9 80 m/s ) 926 N. T = . + . . = E VALUATE : The tension in the chain increases linearly from the bottom to the top of the chain. Figure 5.3a–c __________________________________________________________________________________ 5.20. I DENTIFY : Apply m Σ = F a r r to the composite object of elevator plus student tot ( 850 kg) = m and also to the student ( 550 N). w = The elevator and the student have the same acceleration. S ET U P : Let y be upward. The free-body diagrams for the composite object and for the student are given in Figures 5.20a and b. T is the tension in the cable and n is the scale reading, the normal force the scale exerts on the student. The mass of the student is 56 1 kg. m w/g = = . E XECUTE : (a) y y F ma Σ = applied to the student gives . y n mg ma - = 2 450 N 550 N 1 78 m/s . 56 1 kg y n mg a m - - = = = - .

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