The Problem

A cubical box with a side length of 6m is placed on a rough horizontal floor. We are given that the coefficient of friction between the floor and the cube is 0.2. We need to find the distance between the line of action of gravity and the line of action of the normal force.

Approach

To find the distance between the line of action of gravity and the line of action of the normal force, we need to understand the forces acting on the cube and analyze their directions and magnitudes.

Forces Acting on the Cube

1. Weight (mg): The weight of the cube acts vertically downwards from its center of mass. Its line of action is at the center of the cube.
2. Normal Force (N): The normal force acts perpendicularly to the surface of contact between the cube and the floor. Its line of action is at the point of contact between the cube and the floor.
3. Frictional Force (f): The frictional force acts parallel to the surface of contact between the cube and the floor. Its line of action is opposite to the impending motion of the cube.

Analyzing the Forces

Since the cube is placed on a rough horizontal floor, the normal force and the frictional force come into play.

The frictional force can be calculated using the formula f = μN, where μ is the coefficient of friction and N is the normal force.

Since the cube is at rest, the frictional force is equal to the maximum static frictional force, given by fs = μsN, where μs is the coefficient of static friction.

Let's calculate the normal force and the maximum static frictional force:

Normal force, N = weight of the cube = mg = (mass of the cube) * (acceleration due to gravity) = (density * volume of the cube) * g = ρ * a^3 * g

Maximum static frictional force, fs = μsN = μs * ρ * a^3 * g

Finding the Distance (x)

The distance (x) between the line of action of gravity and the line of action of the normal force can be found by considering the equilibrium of moments about the line of action of the normal force.

Since the cube is in equilibrium, the sum of moments about the line of action of the normal force must be zero.

The moment of the weight about the line of action of the normal force is equal to the moment of the frictional force about the same line.

Moment of the weight, Mw = (mg) * x = (ρ * a^3 * g) * x

Moment of the frictional force, Mf = (fs) * (a/2) = (μs * ρ * a^3 * g) * (a/2)

Setting Mw equal to Mf, we get:

(ρ * a^3 * g) * x = (μs * ρ * a^3 * g) * (a/2)

Simplifying the equation, we find:

x = (μs * a) / 2

Substituting the given values, we have:

x = (0.2 * 6) / 2 =

It will be 2.5