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.

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

And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by

where m is the mass of the system.

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

To find m, we can use the weight of the system. In fact, the weight of an object is given by

where . Solving for m, and using W=5000 N, we find


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

and the acceleration has the same direction of the resultant force, so upward.

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n