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:
F = mg + ma = m(g + a)
F – force (Newtons, N)
g – acceleration due to gravity (10 m.s-2)
m – mass (kilograms, kg)
a – acceleration (metres per second per second, m.s-2)
To put it in words:
Force = Mass’s Weight + Mass’s Acceleration
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;
a = F/m – g = 1000/100 – 10 = 0 m.s-2
With 80 kg;
a = F/m – g = 1000/80 – 10 = 2.5 m.s-2
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.
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.
If the barbell in the above example weighed 0 kg the equation would give infinity as the acceleration, but in practice the force that the muscles produce at high speed is actually reduced (the force-velocity curve of a muscle, Fig (a)) – the available force produced by the muscles with 0 kg barbell goes into accelerating the mass of the muscles and arms.
The lifter’s max force of 1000 N would never be exceeded – faster moves would not involve a force higher than 1000 N.
Moving fast or slow should not be confused with cheat type lifts like the jerk or push press. This is why I chose the Bench Press as one cannot jump it up as in a say push press.
From a physiological perspective it seems that actin-myosin energy cycle times are limited therefore faster contractions necessarily involve fewer actin-myosin bonds at any one time so that faster contractions involve lower force (the force-velocity curve of a muscle).
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;
FT = FW + FA
FT – total force
FW – weight of mass
FA – force due to acceleration of mass
And knowing that;
½mv² = mgh (conservation of energy)
v² = 2gh (peak velocity at impact = velocity at start of deceleration at surface)
v² = 2as
a = (gh)/s
It can be deduced that as;
FT = mg + ma
FT = mg + (mgh)/s
FT = FW(1 + h/s)
h = s
then FT is twice the force of FW.
h/s is constant
then FT is constant.
h > 0
then FT is greater than FW.
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 + 0.3/0.005) = 4880 N
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 = s(FT/FW – 1)
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 = FW(1 + h/s)
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.