Question

A book of weight 20N is pressed between two hands and each hand exerts a force of 40 N. If the book just starts to slide down, then the coefficient of friction is:
A) 0.25
B) 0.2
C) 0.5.
D) 0.1.

Answer 1

Friction is defined as the resistance to the motion of the body the coefficient of friction which works on the body is kinematic coefficient of friction if the body is in motion and the friction is static coefficient of friction if the body is at rest.

Formula used: The formula of the friction force is given by,
${f_s} = \mu \cdot N$
Where $\mu$ is the coefficient of friction and $N$ is the normal reaction.
The formula of weight is given by,
$w = mg$
Where m is the mass of the body and g is the acceleration due to gravity.

Complete step-by-step answer:
It is given in the problem that a book is held by two hands which weighs 20N and the applied force by each hand is 40N and we need to find the value of force for the condition when the book starts to slip down to the ground.
Here the friction force will be equal to the weight of the book,
$\Rightarrow {f_s} = w$
As the formula of the friction is given by,
${f_s} = \mu \cdot N$
Where $\mu$ is the coefficient of friction and $N$ is the normal reaction.
Also the formula of weight is given by,
$w = mg$
Where m is the mass of the body and g is the acceleration due to gravity.
$\Rightarrow \mu \cdot N = mg$
As the force applied by each hand is equal and the weight of the body is 20N.
$\Rightarrow \mu \cdot \left( {2 \times 40} \right) = 20$
$\Rightarrow \mu = \dfrac{{20}}{{\left( {2 \times 40} \right)}}$
$\Rightarrow \mu = \dfrac{1}{4}$
$\Rightarrow \mu = 0 \cdot 25$
The coefficient of friction is equal to$\mu = 0 \cdot 25$.

The correct answer for this problem is option A.

Note: The formula of the friction is dependent on the coefficient of friction and normal reaction and the value of normal reaction is not always equal to the weight of the body; sometimes the applied force can also become the normal force as in this problem.