Question

The mass of a lift is 500 kg what will be the tension on its cable when it is going up with an acceleration of 20 m/s^2? (g=9.8 m/s^2)

Answer 1

Let the tension in the cable be T

The equation of motion for the lift is:

T−mg=ma where m=mass of the lift, a= acceleration of the lift)

⇒T=m(g+a)

⇒T=500(10+2)=500×12=6000N.

Answer 2

Mass of the lift(m) = 500 kg

Acceleration of the lift(a) =20 m/s²

g = 9.8 m/s²

Force ‘ma’ is in direction of motion i.e. downwards. Also mg is being applied in the downward direction. The direction of tension is always along the string and away from the body i.e. in upward direction.

So on balancing the forces

We get an equation

T=mg+ma

T=m(g+a)

T=500(9.8+20) →500×29.8 → 14900N