MythBusters knocking socks off (Season 8)

ewtkf3bD posted on Aug 17, 2016 at 08:41AM
Hitting Buster above his center of gravity causes his upper body to rotate "backwards," which is equivalent to the lower body rotating "forwards." Thus, the punch actually drives the feet deeper into the socks! (A backward blow, or a hit below the belt, creates a force in the right direction, but the right angle of the heel dampens it. Would it work anyway? Only one way to find out, I suppose.)

The sock-seperating force must, therefore, arise either from a linear acceleration of Buster's leg (i.e. from the "follow-through" of an ideal uppercut) or from torque as the leg is whipped.

In considering the linear case, assume an "ideal" uppercut that transmits all its force along the longitudinal axis of the body. By Newton's second law, the acceleration of the leg and body relative to the sock is given by:

a_body = F_body / m_body.

This is equal in magnitude (opposite in direction) to the acceleration of the sock relative to the leg:

a_sock = F_sock / m_sock.

Thus, considering only magnitudes, |F_body| = |F_sock| * (m_body / m_sock). Since the mass of the body is many, many times more than the mass of the sock, so must the force of the uppercut be many, many times more than the force needed to pull off the sock. Substituting the measured force of static friction on the sock for |F_sock| gives a lower bound for |F_body|.


For the other case (whipping), see my next post.

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hace más de un año ewtkf3bD said…
An object in circular motion about a center point feels an outward tug. In fact, this is due to the object's tendency to travel in a straight line -- it must be acted on by a force in the direction of the center of rotation in order to keep its circular path. The magnitude of this "centripetal acceleration" is given by a = v^2 / r, where v is the speed at which the object would fly away if allowed to do so, and r is the radius of the circle.

Let r be the distance from Buster's hip to his ankle. Let F be the force required to pull off his sock. Let m be the mass of the sock. Then, the sock will fly off when (v^2 / r) > (F / m), that is, when v > sqrt(r * F / m). It is clear from this inequality that shortening Buster's legs (decreasing r), using slicker socks (decreasing F), or using heavier socks (increasing m) result in a reduction in the minimum value of v.

Suppose, for instance, that r = 1 meter, F = 20 Newtons, and m = 100 grams. Then sqrt(r * F / m) = sqrt(200 m^2/s^2) = 14.1 m/s. The tangential velocity of Buster's foot must exceed 14.1 m/s in order to knock his socks off by the whipping action alone. This is equivalent to 14.1 / 3.14 = 4.49 revolution per second, or 269 rpm, about the hip joint....