Question

In about 1915, Henry Sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. 6-21). Sincosky’s mass was $\mathbf{79}\mathbf{}\mathbf{kg}$. If the coefficient of static friction between hand and rafter was $\mathbf{0}\mathbf{.}\mathbf{70}$, what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, Sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think Sincosky’s grip was remarkable, try to repeat his stunt)

Short answer

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

Step-by-step solution

Given

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

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

Determining the concept

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:

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{ma}$

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

Determining the least magnitude of the normal force

The relation between frictional force and the normal force is,

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

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,

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

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,

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

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