91影视

The angle of inclination of the head from the vertical is, $胃=40掳$.

The mass of the head is, m = 4.50 kg .

The distance of center of mass of the head from the pivot point P is, a = 11.0 cm .

The distance of the neck muscles from point P is, b = 1.50 cm .

When a force is applied to a body at a certain distance from the pivot point, the body starts to rotate about the pivot point. The rotational effect of the body is due to the 鈥榯orque鈥� produced on the body due to the force.

If the distance between the force applied and the pivot point increases, then the torque value increases; otherwise, the torque decreases.

The free-body diagram of the student鈥檚 head is given below,

Here, is the force applied by the neck muscles.

The weight of the student鈥檚 head acting downwards is given by,

$\begin{array}{r}W=mg\\ =4.50\mathrm{kg}脳9.81\mathrm{m}/{\mathrm{s}}^2\\ =44.145\mathrm{N}\end{array}$

The formula for the torque produced about the pivot point P is given by,

$\begin{array}{r}蟿=W脳a\sin 鈦�胃\\ =44.145\mathrm{N}脳0.11\mathrm{m}脳\sin 鈦�{40}^{鈭�}\\ =3.121\mathrm{N}鈰�\mathrm{m}\end{array}$

This torque produced is balanced by the neck muscles. Consider the force applied by the neck muscles is and it acts directly perpendicular to the angle $胃$of the neck.

So, balancing torques,

$\begin{array}{r}蟿=F脳b\\ F=\frac{蟿}{b}\end{array}$

Putting values,

$\begin{array}{r}F=\frac{3.121\mathrm{N}鈰�\mathrm{m}}{0.015\mathrm{m}}\\ =208.07\mathrm{N}\end{array}$

Hence, the tension in the neck muscles of the student鈥檚 head is 208.07 N .