Question

A force of 98N is required to just start moving a body of mass 100 kg over ice. The coefficient of static friction is $(g = 10m/{s^2})$
(A) $0.6$
(B) $0.4$
(C) $0.2$
(D) $0.1$

Answer 1

Hint :
In order to solve the above problem, first we draw a free body diagram of the body. And on balancing the vertical forces i.e., Normal and Weight, we get normal force.
After then using friction formula and balancing horizontal forces, we get coefficient of friction, where friction formula is given as
${f_L} = {\mu _S}N$
Where
${f_L} =$ Limiting friction
${\mu _S} =$ Coefficient of static friction
N $=$ Normal force

Complete step by step solution :
Given that
$m = 100kg$
So, the weight $=$ mg
$= 100 \times 10$ $[\because g = 10m/{s^2}]$
$W = 1000N$
Here weight is balance by normal force i.e., $W = mg = N$
From equation 1
N $=$ 1000 Newton …..(2)
If a body is just start moving then the friction is known as limiting friction and given as
${f_L} = {\mu _S}N$ …..(3)
Where
${\mu _S} =$ Coefficient of static friction
N $=$ Normal force
Here, given that the force is applying at the body to just start moving is 98 Newton.
So, from equation 3
$98 = {f_L} = {\mu _S}N$
${\mu _S} = \dfrac{{98}}{N}$
From equation 2
$\Rightarrow {\mu _S} = \dfrac{{98}}{{1000}}$
$\Rightarrow {\mu _S} = 0.098$
$\Rightarrow {\mu _S} \approx 0.1$
Hence, the coefficient of static friction ${\mu _S}$ is $0.1$.
So, option D is the correct answer.

Note :
Many times, students may get confused between static limiting and kinetic friction.
Static friction is a force that keeps an object at rest.
Limiting friction is the maximum friction that can be generated between 2 static surfaces is contact with each other. If it is required to just start the body to move.
Kinetic friction is defined as a force that acts between moving surfaces.