Train Acceleration with Friction
Example 5.7
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train's floor is 0.15.
Box Status: Stationary
Given Values
Friction coefficient (μ) = 0.15
Gravity (g) = 10 m/s²
Maximum No-Slip Acceleration
amax = μg
= 0.15 × 10
= 1.5 m/s²
Current Condition
a ≤ μg
0.0 ≤ 1.5
Physics Explanation
The box remains stationary when the static friction can provide enough force to match the train's acceleration.
Static friction (fs) ≤ μN = μmg
Required force (F) = ma
For no slipping: fs ≥ F ⇒ μmg ≥ ma
⇒ a ≤ μg
Required force (F) = ma
For no slipping: fs ≥ F ⇒ μmg ≥ ma
⇒ a ≤ μg
In This Example:
amax = μg = 0.15 × 10 = 1.5 m/s²
If the train accelerates faster than 1.5 m/s², the box will begin to slip backward.
