Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass of 5.18 kg, the heavier box has a mass of 7.51 kg.
(a) Find the contact force between these boxes when a horizontal force of 3.9 N is applied to the light box.
To find the contact force between the two boxes, we will first calculate the acceleration of the system using Newton's second law of motion:
F_net = m * a
Where:
F_net is the net force applied to the system,
m is the total mass of the system, and
a is the acceleration of the system.
Since the force applied is acting on the light box, the net force is equal to the force applied:
F_net = 3.9 N
The total mass of the system is the sum of the masses of the two boxes:
m = 5.18 kg + 7.51 kg
Now we can find acceleration:
a = F_net / m
Substituting the values:
a = 3.9 N / (5.18 kg + 7.51 kg)
a ≈ 0.253 m/s²
Now that we have the acceleration, we can find the contact force between the two boxes. This force is equal in magnitude but opposite in direction to the force applied to the light box:
F_contact = -m * a
Substituting the values:
F_contact = -(5.18 kg + 7.51 kg) * 0.253 m/s²
F_contact ≈ -3.94 N
The contact force between the boxes is approximately -3.94 N, where the negative sign indicates that the force is acting in the opposite direction to the force applied to the light box.