DetetDete the maximum acceleration of the train in which a box is lying on its floor will remain stationary given that the coefficient of static friction between the box and the trains floor is 0.15
Take g = 10m/s2
DetetDete the maximum acceleration of the train in which a box is lying on its floor will remain stationary given that the coefficient of static friction between the box and the trains floor is 0.15
Take g = 10m/s2
For the box lying on the floor of train to remain stationary, the applied external force,F must be equal to the limiting value of static friction,fs as the acceleration of the box is caused due to the static friction.
But we know that fs ≤ μs N where μs is the coefficient of static friction between box and train and N the normal reaction.
The external force applied F= ma
Therefore we can write:
F = fs ≤ μs N
i.e. ma ≤ μs ×mg (Normal reaction N =mg)
i.e. a ≤ μs ×g
Therefore amax =μs ×g
amax = 0.15 × 9.8
(Given μs =0.15 and g =9.8 m/s2)
Therefore amax = 1.47 m/s2
Therefore the maximum accelaration of train so that a box lying on its floor will remain stationary =1.47 m/s2
Like if satisfied
For the box lying on the floor of train to remain stationary, the applied external force,F must be equal to the limiting value of static friction,fs as the acceleration of the box is caused due to the static friction.
But we know that fs ≤ μs N where μs is the coefficient of static friction between box and train and N the normal reaction.
The external force applied F= ma
Therefore we can write:
F = fs ≤ μs N
i.e. ma ≤ μs ×mg (Normal reaction N =mg)
i.e. a ≤ μs ×g
Therefore amax =μs ×g
amax = 0.15 × 9.8
(Given μs =0.15 and g =9.8 m/s2)
Therefore amax = 1.47 m/s2
Therefore the maximum accelaration of train so that a box lying on its floor will remain stationary =1.47 m/s2
Like if satisfied
