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Hint: Impending motion means the state of motion not long before the surface begins to slip. Here the static motion reaches the upper limit.
Complete step by step answer:The impending motion is mathematically given by:
$F = {\mu _s}N$
Where $F$ is the force, ${\mu _s}$ is the static friction and $N$ is the normal force.
The direction of the frictional force is consistently inverse to the forthcoming relative movement of the surfaces. In the event that the applied force ( $P$ ) rises then the frictional force ( $F$ ) additionally increments, until $F < {F_s}$ (limiting static frictional power). At the point when $F = {F_s}$ , the body has an unstable equilibrium and will move.
Body first remains in static condition, and then it goes through impending motion to change to motion.
No motion is the area up till the point the body starts slipping or begins approaching motion. As the system is in equilibrium the frictional force in this area is characterized by the conditions of equilibrium.
Here the frictional force is given by:
$F < {F_{\max }}$
Then comes impending motion where the body begins slipping. Here the static motion reaches the upper limit. The frictional force is given by:
$F = {F_{\max }} = {\mu _s}N$
Then the body goes into motion and begins moving in the direction the force is applied. Here the frictional force changes into kinetic friction.
The frictional force is given by:
$F = {F_{\max }} = {\mu _k}N$
According to question,
As the box is in static friction:
$
a \leqslant {\mu _s}g \\
{a_{\max }} = {\mu _s}g \\
\Rightarrow 0.15 \times 10 \\
\Rightarrow 1.5\,m{s^{ - 2}} \\
$
Hence, the maximum acceleration of the train in which a box lying on its floor will remain stationary is $1.5\,m{s^{ - 2}}$ .
Note:Here we have carefully observed the question whether the body is in static motion, kinetic motion or impending motion. Here the body is in static motion, so the given formula is applicable.
Complete step by step answer:The impending motion is mathematically given by:
$F = {\mu _s}N$
Where $F$ is the force, ${\mu _s}$ is the static friction and $N$ is the normal force.
The direction of the frictional force is consistently inverse to the forthcoming relative movement of the surfaces. In the event that the applied force ( $P$ ) rises then the frictional force ( $F$ ) additionally increments, until $F < {F_s}$ (limiting static frictional power). At the point when $F = {F_s}$ , the body has an unstable equilibrium and will move.
Body first remains in static condition, and then it goes through impending motion to change to motion.
No motion is the area up till the point the body starts slipping or begins approaching motion. As the system is in equilibrium the frictional force in this area is characterized by the conditions of equilibrium.
Here the frictional force is given by:
$F < {F_{\max }}$
Then comes impending motion where the body begins slipping. Here the static motion reaches the upper limit. The frictional force is given by:
$F = {F_{\max }} = {\mu _s}N$
Then the body goes into motion and begins moving in the direction the force is applied. Here the frictional force changes into kinetic friction.
The frictional force is given by:
$F = {F_{\max }} = {\mu _k}N$
According to question,
As the box is in static friction:
$
a \leqslant {\mu _s}g \\
{a_{\max }} = {\mu _s}g \\
\Rightarrow 0.15 \times 10 \\
\Rightarrow 1.5\,m{s^{ - 2}} \\
$
Hence, the maximum acceleration of the train in which a box lying on its floor will remain stationary is $1.5\,m{s^{ - 2}}$ .
Note:Here we have carefully observed the question whether the body is in static motion, kinetic motion or impending motion. Here the body is in static motion, so the given formula is applicable.
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