How much more do heads-down tribes have to bear?

Classical Mechanics Level 5

In recent years, the popularity of smartphones has led to the emergence of "heads-down tribes". Recent studies have found that when playing with mobile phones, it is possible to bear 60 pounds of weight on the cervical spine, equivalent to hanging two large watermelons on the cervical spine, which is heavier than a seven-year-old child. When the body is upright, the pressure on the cervical spine is equal to the weight of the head; but when the head is down, the pressure on the cervical spine will change accordingly. Now the head and neck of the human body is simplified as the model shown in the figure; the center of gravity is at the head of point $P$ .The head is supported by $OP$ (which can be modeled as a light rod) , the cervical spine which can rotate around $O$ and the muscle pulling along the direction of $PQ$ . When the head is down, the angle between $OP$ and the vertical direction is $45°$ , and the angle between $PQ$ and the vertical direction is $60°$ . When the head is down, the pressure on the cervical vertebra is $n$ times about that of the cervical vertebra when it is upright .

Which of the following is closest to $n$ ?

3.3 2.0 2.8 4.2

1 solution

G Silb
Jul 11, 2019

Let $C$ be the compression along the spine, $W$ be the weight of the head, and $T$ be the tension in the muscle to hold the head upright.

Then $C=T\cos 15+W \cos 45$ [1]

Since the head isn't moving, the torque about the base of the spine is equal, so that

$T\cos 75 = W \cos 45$ [2]

Substituting [2] into [1],

$C=W \cos 45 [1+\tan 75]=3.3*W$

How much more do heads-down tribes have to bear?

1 solution

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