Normal force on box on elevator floor?

[GingerBread27]

In Fig. 5-42, elevator cabs A and B are connected by a short cable and can be pulled upward or lowered by the cable above cab A. Cab A has mass 1700 kg and cab B has mass 1200 kg. A 12.0 kg box of sand lies on the floor of cab A. The tension in the cable connecting the cabs is 1.86 x 10^4 N. What is the magnitude of the normal force on the box from the floor?

I don't even know where to start, please help?

 

[vsage]

Assuming the sand box isn't somehow connected to the bottom cable I think the question has a bunch of extraneous information. Consider what the conditions are to keep the box motionless. What forces are acting directly on the box?

 

[M98Ranger]

The normal force on the box can be calculated using the equation Fnet = ma, where Fnet is the net force acting on the box, m is the mass of the box, and a is the acceleration of the box. Since the box is at rest on the floor of cab A, the acceleration is 0 and the net force is also 0.

However, the box is also being pulled upward by the cable connecting the two cabs, which exerts a tension force of 1.86 x 10^4 N. This tension force is counteracted by the normal force from the floor, which must be equal in magnitude and opposite in direction to keep the box at rest.

Therefore, the magnitude of the normal force on the box is also 1.86 x 10^4 N. This means that the floor of cab A must be exerting a force of 1.86 x 10^4 N on the box to balance out the tension force and keep the box at rest.