ch5 - 5.8. IDENTIFY: Apply Newton's 1st law to the wrecking...

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5.8. IDENTIFY: Apply Newton’s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the cable. SET UP: The force diagram for the wrecking ball is sketched in Figure 5.8. Figure 5.8 EXECUTE: (a) y y Fm a = cos40 0 B Tm g °− = 2 4 (4090 kg)(9.80 m/s ) 5.23 10 N cos40 cos40 B mg T == = × °° (b) x x a = sin 40 0 BA TT = 4 sin 40 3.36 10 N AB = × E VALUATE : If the angle 40 is replaces by (cable B is vertical), then and ° 0 ° B g = 0. A T = 5.14.IDENTIFY: Apply m = F a G G to each block. 0 a = . SET UP: Take perpendicular to the incline and y + x + parallel to the incline. EXECUTE: The free-body diagrams for each block, A and B , are given in Figure 5.14. (a) For B , x x a = gives 1 sin 0 Tw α = and 1 sin = . (b) For block A , x x a = gives 12 sin 0 TT w −= and 2 2s i n = . (c) y y a = for each block gives cos nnw . (d) For 0 , and . For , 0 =→ nn w 90 ° 1 = , and . 2 2 T = w 0 nn EVALUATE: The two tensions are different but the two normal forces are the same. Figure 5.14a, b 5.21.IDENTIFY: Apply m F =a G G to each block. Each block has the same magnitude of acceleration a .
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SET UP: Assume the pulley is to the right of the 4.00 kg block. There is no friction force on the 4.00 kg block, the only force on it is the tension in the rope. The 4.00 kg block therefore accelerates to the right and the suspended block accelerates downward. Let x + be to the right for the 4.00 kg block, so for it , and let be downward for the suspended block, so for it x aa = y + y = . EXECUTE: (a) The free-body diagrams for each block are given in Figures 5.21a and b. (b) x x Fm a = applied to the 4.00 kg block gives (4.00 kg) Ta = and 2 10.0 N 2.50 m/s 4.00 kg 4.00 kg T a === . (c) y y a = applied to the suspended block gives mg T ma = and 22 10.0 N 1.37 kg 9.80 m/s 2.50 m/s T m ga == = −− . (d) The weight of the hanging block is . This is greater than the tension in the rope; . 2 (1.37 kg)(9.80 m/s ) 13.4 N mg g 0.75 Tm = EVALUATE: Since the hanging block accelerates downward, the net force on this block must be downward and the weight of the hanging block must be greater than the tension in the rope. Note that the blocks accelerate no matter how small m is. It is not necessary to have , and in fact in this problem m is less than 4.00 kg. 4.00 kg m > Figure 5.21a, b 5.24.IDENTIFY: Apply m F =a G G to the composite object of elevator plus student ( ) and also to the student ( tot 850 kg m = 550 N w = ). The elevator and the student have the same acceleration. SET UP: Let y + be upward. The free-body diagrams for the composite object and for the student are given in Figure 5.24a 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 mwg = = . EXECUTE: (a) y y a = applied to the student gives y nm gm a −= . 2 450 N 550 N 1.78 m/s 56.1 kg y g a m = . The elevator has a downward acceleration of . 2 1.78 m/s (b) 2 670 N 550 N 2.14 m/s 56.1 kg y a . (c) means . The student should worry; the elevator is in free-fall. 0 n = y a =− g (d) y y a = applied to the composite object gives .
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This note was uploaded on 04/29/2008 for the course PH 1110 taught by Professor Kiel during the Fall '08 term at WPI.

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ch5 - 5.8. IDENTIFY: Apply Newton's 1st law to the wrecking...

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