1.A helicopter is lowering a man (weight= 882N) to a boat by means of a cable and harness. What force does the cable apply when the man is being accelerated downward at 1.1 m/s^2 ?
2.Elevators are designed with a maximum carrying capacity. More than this capacity could strain the motor or even break the supporting cable. If a 300kg elevator is designed to provide an upward acceleration of 2.5m/s^2 and the breaking strength of the supporting cable is 15000N and the average mass of a person is assumed to be 80kg, what is the maximum number of occupants for the elevator?
newton second law
2012-02-29 4:48 pm
回答 (2)
2012-02-29 5:04 pm
✔ 最佳答案
1) Let F be the force the cable is applying on the man, then:Mass of man = 88.2/9.8 = 90 kg
882 - F = 90 x 1.1
F = 783 N
2) Let n be the max. no. of person that the elevator can carry, then the mass of elevator + persons = 300 + 80n
Hence we have:
15000 - (300 + 80n) x 9.8 >= (300 + 80n) x 2.5
15000 >= (300 + 80n) x 12.3
15000 >= 3690 + 984n
984n <= 11310
n <= 11.5
Hence max. no. of person = 11
參考: 原創答案
2012-02-29 5:09 pm
1.By Newton's second law,
F=ma
R+W=ma
R-882=(882/10)(-1.1)
R=784.98N
2.Let b be the maximum number of occupants
R+W= ma
15000- (300+80b)(10) = (300+80b)(2.5)
15000-3000-800b = 750 +200b
1000b = 11250
b = 11.25
Thus,the maximum no. of occupants for the elevator is 11
F=ma
R+W=ma
R-882=(882/10)(-1.1)
R=784.98N
2.Let b be the maximum number of occupants
R+W= ma
15000- (300+80b)(10) = (300+80b)(2.5)
15000-3000-800b = 750 +200b
1000b = 11250
b = 11.25
Thus,the maximum no. of occupants for the elevator is 11
參考: f4 science student
收錄日期: 2021-04-20 12:17:08
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https://hk.answers.yahoo.com/question/index?qid=20120229000051KK00140