Question

An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. Determine the magnitude and direction of its instantaneous acceleration.

Answer 1

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be

`F=T-W`
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by

`a= (F)/(m)`
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is

`F=T-W=6000 N-5000 N=1000 N`
To find m, we can use the weight of the system. In fact, the weight of an object is given by

`W=mg`
where
`g=9.81 m/s^2`. Solving for m, and using W=5000 N, we find

`m= (W)/(g)= (5000 N)/(9.81 m/s^2)=510 kg`

and at this point, we can calculate the acceleration of the system (elevator+people):

`a= (F)/(m)= (1000 N)/(510 kg)=1.96 m/s^2`
and the acceleration has the same direction of the resultant force, so upward.