91Ó°ÊÓ

Mass,$\mathrm{m}=79\mathrm{kg}$

Coefficient of static friction,${\mathrm{μ}}_{\mathrm{s}}=0.70$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\underset{}{∑}\mathrm{ma}$

where, F is the net force, m is mass and a is an acceleration.

The relation between frictional force and the normal force is,

${\mathrm{f}}_{\mathrm{s}}={\mathrm{μ}}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}$

2 thumbs and 2 hands of finger are in touch with the rafter. So, there are 4 surface which are responsible for the frictional force. So,

${\mathrm{f}}_{\mathrm{s}}=4{\mathrm{μ}}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}$

Since, frictional force is in the opposite direction of the gravitational force and here rafter is not moving. So, the acceleration is zero. It gives,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{s}}={\mathrm{F}}_{\mathrm{g}} \\ =\mathrm{mg} \\ {\mathrm{F}}_{\mathrm{N}}=\frac{{\mathrm{f}}_{\mathrm{s}}}{4{\mathrm{μ}}_{\mathrm{s}}} \\ =\frac{\mathrm{mg}}{4{\mathrm{μ}}_{\mathrm{s}}} \\ \mathrm{Therefore},=2.8×{10}^2\mathrm{N} \end{gathered}$$

Hence, the least magnitude of the normal force on the rafter from each thumb or opposite fingers is$2.8×{10}^2\mathrm{N}$