# An elevator weight 5000N is ascending with an acceleration of 3m/sec2. During this ascent its

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An elevator weight 5000N is ascending with an acceleration of 3m/sec2. During this ascent its operator whose weight is 700N is standing on the scale placed on the floor. What is the scale reading? What will be the total tension in the cable of the elevator during this motion?

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Given data, WE = 5000N, a = 3m/sec2, WO = 700N,

Let R = Reaction offered by floor on operator. This is also equal to the reading of scale.

T = total tension in the cable

Net upward force on operator

= Reaction offered by floor on operator – Weight of operator

= R – 700

But, Net force = ma

R – 700 = (700/9.81)X 3

R = 914.28N

Now for finding the total tension in the cable, Total weight of elevator is considered.

Net upward force on elevator and operator

= Total tension in the cable – Total weight of elevator and operator

= T – 5700

But net force = mass X acceleration

T – 5700 = (5700/9.81) X 3

T = 7445N