Find the magnitude and direction of the acceleration of a 500 kg elevator supported by a cable with a tension of 3000 N.

Question

Solve a numerical "elevator problem" for the tension of an elevator cable or the acceleration of an elevator.

Answer 1

Make a free-body diagram:

We are told tension is 3000N so, since I've chosen "+" as up, T = 3000

We are told the mass is 500 kg and W = mg so W = (9.81)(500) = 4905 N but, because we've chosen "+" as up, W = -4905

Summing the forces in the y-direction we get:

∑F_y = ma_y

3000 - 4905 = (500)a_y

-1905 = 500a_y

ay = -1905/500 = -3.8 m/s²

Since "-" is down, the acceleration is 3.8 m/s² in the downward direction.