Question

State laws of static friction. A mass of $60kg$ is at rest on the platform. A coolie requires a force of $78.4N$ to just slide. Calculate coefficient of static friction.

Answer 1

Hint: The force of static friction depends on the coefficient of static friction(which is determined or a function of two materials) and the normal contact force for a system. On the flat surface of the platform, the normal force of the object is \[\eta {\text{ = }}mg\].

Formula Used:
\[{F_s}^{max\;} = {\mu _s}\eta \]
Where ${F_s}$ is the force of static friction, ${\mu _s}$ is the coefficient of static friction, $\eta $ is the normal force, $F_s^{\max }$ is the maximum force of static friction.

Complete step by step answer:
The relation between the limiting frictional force and normal reaction is governed by the law of limiting friction and is described as $F = \mu \eta $, $\mu $ is the coefficient of friction and is independent of the area of contact, $F$ is the frictional force and $\eta $ is the normal reaction.
Force required to just slide the object is \[78.4N\]
 \[\eta = m \times g\]
 denotes gravity and $m$ is the mass. The actual value of gravity is \[9.8{\text{ }}m/{s^2}\] on Earth.
\[{F_s}^{max\;} = 78.4\]
\[{F_s}^{max\;} = {\mu _s}mg\]
$\Rightarrow {\mu _s} = \dfrac{{78.4}}{{60 \times 9.8}}$
$\Rightarrow {\mu _s} = \dfrac{{78.4}}{{588}}$
$\Rightarrow {\mu _s} = \dfrac{{784 \times {{10}^{ - 1}}}}{{588}}$
$\Rightarrow {\mu _s} = 1.33 \times {10^{ - 1}}$
$\Rightarrow {\mu _s} = 0.133$
The coefficient of static friction is \[0.133\].

Note: The specific manifestation of friction which attempts to resist efforts to move an object that is currently at rest with respect to a surface. Static friction attempts to keep objects in contact from sliding along the surface of the platform. When the net force attempting to create sliding motion exceeds a certain limiting value proportional to the normal force exerted by the platform on the object, Static friction will not be able to stop motion. The coefficient of static friction is assigned the Greek letter "mu" \[\left( \mu \right),\] with a subscript "S". This implies a limit to the strength of the static friction force. The limiting behavior is well approximated by the mathematical expression:
Force of static friction \[ \leqslant \] (coefficient of static friction) maximum force of static friction $ = $ (coefficient of static friction) (normal force)
\[{F_s} \leqslant {\mu _s}\eta \] and \[{F_s}^{max\;} = {\mu _s}\eta \].