# Force

#### Dropping a 100 kg ball creates large forces – this doesn’t mean the body applies large forces to the ball.

Moving fast (or for that matter just moving) is not very ‘dangerous’.

Here is an equation that every barbell man should know:

where;

##### a – acceleration (metres per second per second, m.s-2)

To put it in words:

## Application

Assume that a lifter is doing a Rack bench from the chest. He applies 1000 N of force to the bar. Swap the equation around:

With 100 kg;

With 80 kg;

With 50 kg;

##### a = F/m – g = 1000/50 – 10 = 10 m.s-2

Thus the lighter the weight the greater the acceleration that can be applied to the weight.

## Discussion

It is very important to note, in the above example, that the acceleration is a result of the changing mass. People often just think of the equation as;

##### F = ma

and reduce this further to;

##### F ∝ a

and thereby conclude that high acceleration equals high force. This leads to the ridiculous conclusion that somehow the force is created from nowhere by acceleration. In fact, the force is due to the muscle and can be no different from that created by the muscle.

# Impact

## Shot Put

Imagine a shot put that is dropped onto a deformable surface. The distance it is dropped is h, the distance it decelerates on the deformable surface is s.

Restating the equation from above to;

Where;

##### FA – force due to acceleration of mass

And knowing that;

##### a = (gh)/s

It can be deduced that as;

Thus if;

##### h = s

then FT is twice the force of FW.

And if;

##### h/s is constant

then FT is constant.

And if;

##### h > 0

then FT is greater than FW.

## Examples

Assume that an 8 kg shot is dropped from 0.3 m on to concrete which deforms by 0.005 m (g = 10 m.s-2).

Assume that an 8 kg shot is dropped from 0.3 m on to sand which deforms by 0.2 m (g = 10 m.s-2).

##### FT = 80(1 + 0.3/0.2) = 200 N

Let’s apply the equation to a 90 kg man jerking a 30 kg barbell; he is measured as generating 3500 N of ground reaction force. Assume he dips by a mere 0.05 metres and then straightens;

##### h = 0.05(3500/1200 – 1) = 0.0958 m

So the man and barbell jumps above the ground by 0.0958 metres after the jerk. 3500 N is the ground reaction force, much of the force is simply created by the man’s accelerating body mass. The force of the bar on the man during the jerk is;

##### FT = 300(1 + 0.0958/0.05) = 875 N

The remaining 2625 N is due to the man’s accelerating body mass – without the bar he would generate 2625 N ground reaction force anyway.

The man and barbell might be visualised as a stack of bricks where the lowest brick gets the greatest force from the weight and acceleration.

It is the man’s muscles that are the source of the force – he determines h and s.

These figures would be mirrored on the descent if the man decided to apply the same forces.