Forces Acting on Boxes in an Accelerating Elevator

When considering a scenario where three boxes are stacked on top of one another within an elevator that is accelerating upwards, it's important to analyze the forces at play. This article dives into the detailed process of applying Newtons Second Law of Motion to calculate the forces contributing to the motion of each individual box. We will focus on two key points: first, how the top box exerts a force on the middle box, and second, how the bottom box exerts a force on the middle box.

Definitions and Assumptions

To begin, we define some essential variables and make some assumptions. Let us denote the masses of the top, middle, and bottom boxes as m1, m2, and m3, respectively. Additionally, let a represent the upward acceleration of the elevator, and g represents the acceleration due to gravity, which is approximately 9.81 m/s2.

Force on the Top Box Acting on the Middle Box

To find the force that the top box (m1) exerts on the middle box (m2), we need to consider the forces acting on the top box.

Weight of the top box: The weight W1 of the top box is given by:

W1 m1g

The net force Fnet1 on the top box, due to the upward acceleration of the elevator, can be expressed as:

Fnet1 m1a

The normal force N12 that the top box exerts on the middle box is the sum of the weight of the top box and the force due to the upward acceleration:

N12 W1 Fnet1 m1g m1a m1g m1a

Thus, the force that the top box acts on the middle box is:

N12 m1g m1a m1g(1 a/g)

Force on the Bottom Box Acting on the Middle Box

To find the force that the bottom box (m3) exerts on the middle box (m2), we need to consider the forces acting on the middle box.

Weight of the middle box: The weight W2 of the middle box is given by:

W2 m2g

The net force Fnet2 acting on the middle box, due to its acceleration upwards, can be expressed as:

Fnet2 m2a

The normal force N23 that the bottom box exerts on the middle box is equal to the weight of the middle box plus the net force acting on it:

N23 W2 Fnet2 m2g m2a m2g m2a

Thus, the force that the bottom box acts on the middle box is:

N23 m2g m2a m2g(1 a/g)

Summary of Forces

Force that the top box exerts on the middle box: N12 m1g m1a m1g(1 a/g) Force that the bottom box exerts on the middle box: N23 m2g m2a m2g(1 a/g)

These equations clearly illustrate how the upward acceleration of the elevator affects the forces between the boxes, providing a comprehensive understanding of the dynamics at play in this scenario.