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An elevator of mass 3000 kg is ascending at a steady speed of 2.0 m/s. What is the

force in the cable supporting the elevator?

Verified Answer :

alekos

F = mg
we know the mass and g = 9.8 m/s/s

Answer:

\(T = 29430 N\)

Explanation:

When the elevator is moving at uniform speed then we can say that acceleration of the elevator is zero

so we will have

\(T - mg = ma\)

here T = tension in the supporting cable

mg = weight of elevator

now if the elevator is moving with steady speed

then we will have

\(T - mg = 0\)

\(T = mg\)

\(T = (3000)(9.81)\)

\(T = 29430 N\)

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