# Calculating the force required to topple an object?

1. Dec 1, 2014

### paulw99

Is there a formula that calculates the force required to "push over" or topple an object? I know all the details of the object (length, width, height, material, etc) and the object will be resting on a flat surface. If anyone can point me in the right direction I'd be very grateful.

Note: This is not a homework/school related question.

2. Dec 1, 2014

### ShayanJ

How much do you know about rotational dynamics?

3. Dec 1, 2014

Nothing.

4. Dec 1, 2014

### CWatters

There isn't a simple equation that covers every situation.

Consider a huge rock in the shape of a cube 10m x 10m x 10m. It would take a considerable force to lift one side just 1 mm off the ground.... but once you have lifted it to the point where it's balanced on an edge the force required to topple it over could be very small. The weight of a small dust particle landing on one side might be enough to topple it. In practice it would depend on what the rock was resting on. If it was on soft ground it might sink in and be quite difficult to topple. If it was on a hard surface it might be easy. Which force did you want to know, the initial force required to lift an edge or the final force required to topple it? The initial force is reasonably easy to calculate. It's also reasonably easy to calculate the force needed to lift it at any angle in between.

The above assumes you are lifting it slowly. The situation gets more complicated if you lift the edge of an object quickly. For example suppose you put some explosive under one edge. An explosion would apply a force for a very short period of time. It might only be 1/10th of a second but it might be long enough to get the thing rotating at some velocity about the opposite edge. The question then becomes will it keep rotating or slow, stop and fall back down? As the block rotates the centre of mass rises and that takes energy (=mgh). If you know the initial velocity you might be able to calculate if the explosion gave the block enough energy to lift the centre of mass to the point where it would continue to topple over. If not, then it would fail to reach that height and fall back down.

5. Dec 1, 2014

### CWatters

PS: If the force is limited to a horizontal push on the side of the object then you need to know about the friction between the object and the ground. For example will it topple over or just slide along the ground? In the real world it's not always easy to be sure what will happen because the friction changes as the object is lifted. eg It might start to topple, then as friction reduces it might start to slide and fall back down. Things get easier if the object is up against a step or similar that stops it sliding.

6. Dec 3, 2014

### Staff: Mentor

Generally, you need to look at tilting the object so that its centre of mass moves to be directly above the supporting edge the block is rotating about.Then the tiniest extra force is enough to make it overbalance---and it obligingly completes the rotation by itself (with the assistance of gravity).

7. Apr 24, 2015

### Venugopalan CV

Consider a cubic object of size L x B x H and density 'D'. Weight of the object W=LxBxHxD. Since the object is assumed to be homogenous, the Centre of Gravity will be at the centre with co-ordinates from one of teh edges as L/2, B/2 and H/2. The minimum force to topple it will be when the force is applied on the top edge horizontally when you are looking on the side containing the smaller base of B. The object will tend to topple about the diagonally opposite edge. The stabilising effect due to the weight of the object about this edge will be WxB/2 and the destabilising moment due to applied force F will be FxH. When (FxH) exceeds (WxB/2), the object will start to topple. The actual force at th time of toppling will be less since the moment arms change during the toppling process.