Question

A rudimentary elevator is constructed by connected an elevator car of mass Me = 568 kg to a large counterweight of mass Me = 730 kg. A man of mass m = 90 kg steps on the elevator car and then the counterweight is released. a) Draw the free body diagrams for the elevator car, the counterweight, and the man. b) What is the acceleration of the counterweight (magnitude and direction)? c) What is the tension in the string? d) What is the apparent weight that the man experiences, i.e., what is the weight that would be measured by a scale placed beneath his feet? What mass would the scale read?

Answer 1

mc - 730kg , M = Grinen, Me=568kg 90 log FBD AT Mg meg mcg 6 Eauation of motion: To find a For elevator T-(metM) g = (Metm) a T= (568+90) teg xat (568x90) x 9.SIN T = 658 kg xa + 6448.UN For counter weight mag-tamaa ② substitute T in mag - 658kg xa-4 644g. UN = mea a (met 658kg) = (730x 9.81) N-64U8 UN a ip = 0.514 m/s2 Ang © Tension (T) = 0 T = Me (g-a) = 73 oleg (9.81-0.514) m/s2 IT = 678 6. UN JAng

& Apparent wt :-? N-mg = ma N = m(a+g) Na 90kg (0.519 + 9.81) m/ge N = 924.16 NAM Actual weight = mg = 90x9.8 = 882 N but scale will read 924 16 N ile man will feel more weight (weightfulnes),