# Toppling

Contents

## Summary

• When a body is suspended, the centre of mass is always positioned vertically below the point of suspension and this result is used to determine the angle of suspension.
• There are always two forces acting on that object when the object stands on a surface:
1.    Weight W
2.    The resultant of all the contact forces between that object and the surface.
• Equilibrium of an object is broken when the object is on the point of toppling.
• Toppling occurs when the weight force acts outside the base of the object.

When an object or particle stands on a surface, there are always two forces acting on that object:
i) The weight W of the object.
ii) The resultant of all the contact forces that are between that object and the surface and it usually acts through a point on both surfaces. The contact forces usually just consist of the frictional force parallel to the sliding surface and normal contact force which is perpendicular to the sliding surface.

Furthermore, from the article ”The Coefficient of Friction”, we can recall that for general equilibrium  $F\quad \le \quad \mu R$.  Also we discussed that limiting equilibrium is when frictional contact force is maximum, which happens when the object is at the point of sliding or moving.

Therefore, we can say that equilibrium of an object is broken when the object is on the point of toppling. Toppling occurs when the weight force acts outside the base of the object.

Lets now observe three situations:

From the second and third situation where the box is about to topple, we can see that the resultant force no longer acts in the same line as the weight ”W” and this produces an overall moment on the blocks, causing the box to topple.

#### Example #1

Q. A rectangular block of mass 3 kg is placed on a slope as shown in Fig 1. The angle alpha  $\alpha$  is gradually increased. Given that the coefficient of friction between the block and the slope is 0.6, find the angle at which the block topples.

Solution:

At the point where the block is about to topple about the edge E, the centre of mass is vertically above E. Therefore we calculate the angle  $\alpha$  using the equation:

$tan\quad \alpha \quad =\quad \frac { x }{ y }$

Where x = 0.4 m and y = 0.8 m

$tan\quad \alpha \quad =\quad \frac { 0.4 }{ 0.8 }$

$\alpha \quad =\quad arctan(0.5)$

$\alpha \quad =\quad 26.6$°       Ans

Hence, the angle at which the block topples is  $\alpha \quad =\quad 26.6$°.

#### Example #2

Q. A cubic block of mass 10 kg and side 12 cm, rests on a rough slope. The coefficient of friction between the block and the slope is 0.2. The slope is gradually increased. Will the block topple or slide?

Solution:

The block is on the point of toppling when its weight acts vertically through the its corner. This will occur when the angle reaches 45°.

The block is on the point of sliding when the frictional force up the slope is at its maximum possible value of R.

Resolving perpendicular to slope:

$R\quad =\quad 10g\quad cos\quad \alpha$

Resolving parallel to slope:

$uR\quad =\quad 10g\quad sin\quad \alpha$

$(0.2)10g\quad cos\quad \alpha \quad =\quad 10g\quad sin\quad \alpha$

$tan\quad \alpha \quad =\quad 0.2$

$\alpha \quad =\quad 11.3$°

The block will slide when the slope is at an angle of 11.3° and topple at an angle of 45°, as angle of sliding is less then the angle of toppling so the block will slide.