A stationary elevator and its contents have a combined mass of 3000Kg. The elevator is suspended by a single cable. (Assume three significant digits.)(adsbygoogle = window.adsbygoogle || []).push({});

Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest.

If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point?

If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point?

For a i did---F_g=mg = 3000*9.8=29400N (Down)

F_g=ă-Fă_T

F_T=29400 N (Up)

For c i did---Net Force downwards=mg-T

downward acceleration= Net force/m= (mg-T)/m

a = g-T/m

T= (g-a)m

= (9.8-3)3000= 20400N(Up)

Net Force acting upwards= mg= 3000*9.8= 29400N(Down)

how to do part b?what equation should i use?

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# Homework Help: Tension in elevator

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