If a 6.0 kg object is sitting on top of a 7.0 kg object on a level surface, state the magnitude and direction of the forces exerted by the 7.0 kg object.
According to Newton's Third Law of Motion (Newton III), whenever an object is "sitting on a surface" there are two forces at play. One is the gravitational force acting on the object pulling it toward the center of the Earth (its weight). The other is a Normal force which is created by the surface and prevents the object from passing through the surface. These forces are equal in magnitude, but opposite in direction as preducted by Newton III.
The 6.0 kg object has a weight of 6.0 kg X (-9.8 m/s^2) = (-58.8 N)
Or one would say, "59 Newtons downward." (Rounded to two significant digits.)
The 7.0 kg object is in the way and prevents the 6.0 kg object from actually moving in the downward direction. Therefore the 7.0 kg object is applying an equal and opposite Normal force against the bottom of the 6.0 kg object. Therefore, one of the forces that the 7.0 kg object is exerting is "59 Newtons upward Normal force against the bottom of the 6.0 kg object."
The 7.0 kg object also has its own weight which is pulling it toward the center of the Earth. The weight of the 7.0 kg object is given by
7.0 kg X(-9.8 m/s^2) = -68.6 N
However, the 7.0 kg object is also holding up the 58.8 N object as well. Therefore, the 7.0 kg object is exerting a downward force on the level surface which is equal to the sum of the two weights:
-68.6 N + (-58.8 N) = -127.4 N or -130 N to two dsignificant digits. We could then say that the 7.0 kg object is exerting a downward force of 130 N.