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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?

Respuesta :

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

Answer:

[tex]T = 29430 N[/tex]

Explanation:

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

so we will have

[tex]T - mg = ma[/tex]

here T = tension in the supporting cable

mg = weight of elevator

now if the elevator is moving with steady speed

then we will have

[tex]T - mg = 0[/tex]

[tex]T = mg[/tex]

[tex]T = (3000)(9.81)[/tex]

[tex]T = 29430 N[/tex]