Why does ground reaction force decrease at the start of a vertical jump?

Question

I don't understand why reaction force would decrease at the start of a countermovement jump. I've tried looking at this diagram and considering moments and how lowering the position of the center of gravity may affect them, but I don't seem to understand why this would be. I'm still in high school,so a simple explanation would help :)

Answer 1

I don't understand why reaction force would decrease at the start of a countermovement jump.

The reaction force changes because of changes in the vertical acceleration of the center of gravity of the persons body throughout the counter movement, as follows.

At position A, before the person body starts dropping, the net force on the person is zero, so that $N=mg$, where $N$ is the upward ground reaction force and $mg$ is the force of gravity acting downward on the body.

In positions A through B the center of gravity of the body drops giving it a net downward acceleration of $a$ as the upper part of the body drops while the arms remain essentially vertical. Or, $$F_{net}=ma=mg-N$$

Where $F_{net}$ is the net downward force on the body, $a$ is vertical acceleration of the center of gravity and is positive downward, and $N$ is the ground reaction force. For $a$ to be positive, $N<mg$, which is why the reaction force is shown going down.

In the positions B through C a combination of two things are happening. The upper part of the body is still dropping but at the same time we see the persons arms are swinging (accelerating) upward, counteracting the continued downward acceleration of upper body moving downward until they eventually cancel each other at position C, where once again $N=mg$.

In positions C through E the upward acceleration of the arms is causing a net acceleration upward. For that to happen, $N>mg$, which is the reason the GRF goes positive

In positions E through F the arms are swinging (accelerating) downward while the upper body is accelerating upwards, the net result being a slight decrease in the GRF followed by an increase at the point of launch. I do not understand, however, why there is negative GRF at position G as the person is no longer in contact with the ground (?).

You can simulate some of this by standing on bathroom scale. The scale reading reflects the GRF. At position A while you are still (no acceleration of your center of gravity), the GRF will simply equal your weight. Now try the following while observing the scale to simulate the effects of dropping and raising you body.

1. Suddenly allow your body to drop by bending your knees, but keep your arms vertical downward. You will observe the scale momentarily deflecting downward (giving you a momentarily lower "weight"), reflecting a drop in the GRF.

2. Now, starting in a squatting position with your hands hanging vertically downward, quickly stand up. You will observe the scale momentarily deflecting upward (giving you a momentarily higher "weight"), reflecting an increase in the GRF.

To simulate the effects of swinging your arms:

1. Standing straight up, suddenly raise your hands over your head while observing the scale. You will observe the scale momentarily deflecting upward (giving you a momentarily higher "weight"), reflecting an increase in the GRF.

2. Again standing straight up starting with your hands over your head, drop them suddenly to your sides. You will observe the scale momentarily deflecting downward (giving you a momentarily lower "weight"), reflecting a drop in the GRF.

Hope this helps.

Answer 2

The decrease in reaction force is not exactly due to lowering of center mass, is is due to acceleration of center of mass. Earlier the reaction force was balancing gravity hence there was no acceleration. Now, because the reaction force is less than gravity, he accelerates downward, and while jumping up, because the reaction is more than gravitational force, he accelerates upwards.