A regular hexagon of mass m is placed on a sufficiently rough surface. A horizontal force F is applied to topple the hexagon. Minimum value of F is?

Steps to Solve

1. Determine the center of mass and pivot point

For a regular hexagon, the center of mass is at its geometric center. When a horizontal force ( F ) is applied to the hexagon, it will pivot about one of its vertices.

2. Identify the forces acting on the hexagon

The forces acting on the hexagon are:

• The gravitational force ( F_g = mg ) acting downwards at the center of mass.
• The applied horizontal force ( F ).
• The normal force ( N ) acting upwards at the point of contact.

3. Apply the conditions for toppling

For the hexagon to begin toppling, the torque due to the applied force must equal the torque due to the gravitational force about the pivot point.

4. Calculate the torques

Let the distance from the pivot point to the center of mass be ( d ).

• The torque due to the gravitational force is:

$$ \tau_g = mg \cdot d $$

• The torque due to the applied force is:

$$ \tau_F = F \cdot \frac{L}{2} $$

where ( L ) is the length of a side of the hexagon.

5. Set the torques equal to establish equilibrium

Set the two torque expressions equal to each other for toppling:

$$ F \cdot \frac{L}{2} = mg \cdot d $$

6. Solve for the applied force ( F )

Rearranging the equation gives:

$$ F = \frac{2mg \cdot d}{L} $$

We need to determine ( d ) for a regular hexagon.

For a regular hexagon with side length ( L ), the distance ( d ) from the center to the vertex is:

$$ d = \frac{L \sqrt{3}}{2} $$

Substituting ( d ) into the force equation yields:

$$ F = \frac{2mg \cdot \frac{L \sqrt{3}}{2}}{L} = mg \sqrt{3} $$