Physics-Tension Question

Question

An elevator weighing 4000 lbs falls with a downward acceleration of magnitude (1/2)g. What is the tension in the supporting cable?

So far, I used the formula Ft=Fg ma and got 58, 868N. Please show the steps to solving this question and the corrections, if any.

Answer 1

Ftotal = ma = T − mg

where m is the mass of the elevator,

a is the acceleration of the elevator,

T is the tension in the cable, and of course

g is the acceleration of the gravitational field at the earth's surface.

Therefore, T = ma + mg

Because the elevator mass is actually given as a weight and is in English units, presumably the result would also need to be in English units as well.  In this case, mg is given as 4000 lb and g in English units is 32.2 ft/sec².  The mass of the elevator is therefore

4000 lb / 32.2 ft/sec² = 124.2 slugs.  (If you aren't familiar with the slug as a unit of mass, 1 lb = 1 slug∙ft/sec² and a one slug mass weighs 32.2 lb in a standard gravitational field.)

One half g is an acceleration of 16.1 ft/sec², which is downward, hence is negative in the convention that upward accelerations are positive.

The tension on the cable is therefore

T = 124.2 slug ∙ (-16.1) ft/sec² + 4000 lb

= 2000 lb

Or you could convert to MKS units and convert back at the end.  In metric units, the tension will be 8898 N.