A 24.0 kg box (m1) rests on a table.(b) A 13.0 kg box (m2) is placed on top of the 24.0 kg box, as shown in Fig. 4-35.
Determine the normal force that the table exerts on the 24.0 kg box.
Determine the normal force that the 24.0 kg box exerts on the 13.0 kg box.
if both boxes are stacked, the force on the table is (m1+m2)g
The lower box exerts a force equal to the weight of the upper box, otherwise, the top box would be falling through the lower.
To determine the normal force that the table exerts on the 24.0 kg box, we need to consider the forces acting on the box in the vertical direction.
Firstly, we know that the weight of the box, which is the force of gravity acting on it, is given by the equation: Fg = m * g, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the 24.0 kg box is given as 24.0 kg, so we can calculate its weight as follows:
Fg1 = m1 * g = 24.0 kg * 9.8 m/s^2
= 235.2 N
Since the box is at rest on the table, the net vertical force acting on it must be zero. Therefore, the normal force exerted by the table on the 24.0 kg box is equal to the weight of the box, which in this case is 235.2 N.
To determine the normal force that the 24.0 kg box exerts on the 13.0 kg box, we need to apply Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
The normal force exerted by the 24.0 kg box on the 13.0 kg box is the same as the normal force exerted by the 13.0 kg box on the 24.0 kg box but in the opposite direction, according to Newton's third law.
Therefore, the normal force that the 24.0 kg box exerts on the 13.0 kg box is also 235.2 N.