태권도 물리: Lesson 3 – energy and force, part 2

In lesson 1 we started with:

E=\frac{1}{2}mv^2+mg_y and F_{arm}=ma

  • Let m = mass of the fist and arm of a 45 kg (100 pounds) martial artist, approximately 10 kg (22 pounds).
  • Let v = velocity of the fist in the horizontal, or x direction only. Assume 5 \frac{m}{s}
    • Note: In lesson 2 we discovered that the actual velocity_{arm} = 4 \frac{m}{s}.
  • Inserting values we have E_{arm}=\frac{1}{2}(10kg)(4\frac{m}{s})^2-(4\frac{m}{s})(9.8\frac{m}{s^2})=80J-39.2J=40.8J and F_{arm}=(10kg)(9.8\frac{m}{s^2})=980N

This is about 220 pounds of force, if the entire arm is treated as a solid body. In actuality, the entire arm does not travel through the target. Assuming proper extension of the fist through the target zone, at most only 10% of the arm’s force will be delivered to the target.

  • Let l = length of fully extended arm, approximately 0.3 m
  • Let m = 10 kg, the mass of the fist and arm
  • Let v = 4 m/s, as calculated in lesson 2

F_i=\sum_{i=1}^{10 }m_i\frac{iv}{10t}=478.8N

Note that by treating the fist as one-tenth of the total arm, we find a more accurate assessment of the force delivered to target. In this case, one-tenth the size yields one-half the force, 108 pounds of force instead of 220 pounds.

Again, this assumes only linear motion. What about rotational motion enacted by twisting the fist during the punch? Let us save that for the next lesson.

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6 thoughts on “태권도 물리: Lesson 3 – energy and force, part 2

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  3. Pingback: 태권도 물리: Lesson 1 – energy and force | Aerospace Cubicle Engineer (ACE)

  4. Pingback: 태권도 물리: Lesson 2 – velocity and force | Aerospace Cubicle Engineer (ACE)

  5. Pingback: 태권도 물리: Lesson 3 – energy and force, part 2b | Aerospace Cubicle Engineer (ACE)

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