Question

Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary. It is given that the coefficient of static friction between the box and the floor of the train is $0.13.{\text{ }}\left( {g = 9.8m/{s^2}} \right)$.

Answer 1

Hint
When the train starts to move, the box on the floor of the train experiences a force of static friction. As long as the static friction force is equal to the force with which the box is moving along with the train, the box remains stationary with respect to the train.
We use the formula of static friction and formula of force to solve this question. For the box to be at rest the force with which the box is moving along with the train should be equal to or lesser than the force due to static friction. So, we take the force of static friction equal to force applied on the box due to the acceleration of the train and find the acceleration from this equation.
- Static friction ${f_s} = {\mu _s}{F_N}$
- Force acting on the box $F = ma$
Here, Coefficient of static friction is represented by ${\mu _s}$, Mass of box is represented by $m$, Acceleration due to gravity is represented by $g$, and Acceleration of the train is represented by $a$

Complete step by step answer
When the train is in motion, the box which is on the floor of the train moves along with the train. The force acting on the box due to the acceleration of the train is given by $F = ma$
There is also a force of static friction acting on the box. The product of the coefficient of static friction and the normal force on the box gives us the force of static friction.
$ \Rightarrow {f_s} = {\mu _s}{F_N} $
$ \Rightarrow {f_s} = {\mu _s} \times m \times g $
For the box to remain stationary the force experienced on the box must be lesser or equal to the force of static friction.
$ \Rightarrow {f_s} \geqslant F $
$ \Rightarrow {\mu _s}{F_N} \geqslant ma $
$ \Rightarrow {\mu _s} \times m \times g \geqslant ma $
Coefficient of the friction is given, ${\mu _s} = 0.13$
Acceleration due to gravity$g = 9.8m/{s^2}$
Substituting the above values,
$ \Rightarrow m.a = {\mu _s}mg $
$ \Rightarrow a = {\mu _s}g = 0.13 \times 9.8 $
$ \Rightarrow a = 1.274m/{s^2} $
Hence the box remains stationary as long as the acceleration of the train is lesser than or equal to $[a = 1.274m/{s^2}$
The maximum acceleration of the train for which the box remains stationary is $[a = 1.274m/{s^2}$

Note
Students might get confused on seeing the term static friction. The formula for static friction is the same as that of friction force. The coefficient of static friction is the maximum ratio of applied force to normal force with no motion.