# Punch Physics

## Impulse of a Taekwondo Punch

For example, consider the amount of force needed to break a board supported at both ends struck in the center by a force F. The downward force will be shared by the two supports, so each will supply an upward force of F/2. Imagine that the board is deflected downwards by the force, as shown in green. In this case the top surface of the board will be in compression, and the bottom surface will be in tension.This force will produce a torque about an axis through the middle of the board. To understand this, consider the forces on the left half of the board that are due to the right half. The top of the board is in compression, so the right half is pushing to the left. The bottom of the board is in tension, so here the right half is pulling to the right. These forces are indicated by the red horizontal arrows. This means the forces will produce a counter-clockwise torque about the center. The board will break if this torque is greater than:
tmax = 1/6 Wh2sb

where W is the width of the board, h is the thickness, and sb is the modulus of rupture of wood.Notice that the upward force from the left support will tend to produce a torque in the opposite direction from the torque due to the stresses. The magnitude of this torque is simply r*F (since r and F are perpendicular in this case), or F/2 * L/2. If this torque is less than tmax given above, then the board will not break. Thus, the minimum force needed to break the board can be calculated by setting these two torques equal to each other:

FL/4 = 1/6 Wh2sbF = 2/3 sb Wh2/L