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Example 4.7 Maximum Acceleration of a Train Solved Step by Step

 Maximum Acceleration of a Train Using Static Friction Explained

-  Dr.Sanjaykumar Pawar 

Physics diagram showing a box on the floor of an accelerating train with forces including static friction, normal reaction, weight, and acceleration labeled for educational purposes.
A box remains stationary inside an accelerating train due to static friction acting between the box and the train floor.


 

Example 4.7 - Maximum Acceleration of a Train Example 4.7 Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary,given that the co-efficient of static friction between the box and the train’s floor is 0.15.

Example 4.7

Question:
Determine the maximum acceleration of a 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.

Step 1: Understand the Concept

When the train accelerates, the box must also accelerate along with the train.

The force responsible for accelerating the box is the static friction between the box and the floor.

  • If friction is enough, the box remains stationary relative to the train.
  • If friction is not enough, the box starts sliding backward.

Therefore, the maximum acceleration depends on the maximum static friction available.

Step 2: Apply Newton's Second Law

According to Newton's Second Law:

F = ma

The only horizontal force acting on the box is static friction (fs).

ma = fs

Step 3: Write Maximum Static Friction

Maximum static friction is given by:

fs ≤ μsN

where:

  • μs = coefficient of static friction
  • N = normal reaction

Step 4: Find the Normal Reaction

Since the box is resting on a horizontal floor:

N = mg

Substitute N = mg into the friction formula:

fs ≤ μsmg

Step 5: Substitute into Newton's Law

From Newton's Law:

ma = fs

Using the maximum friction condition:

ma ≤ μsmg

Cancel mass (m) from both sides:

a ≤ μsg

Therefore:

amax = μsg

Step 6: Substitute the Given Values

Given:

μs = 0.15
g = 10 m/s²

Substitute into the formula:

amax = 0.15 × 10
amax = 1.5 m/s²
Final Answer:

amax = 1.5 m/s²

Quick Notes for Beginners

  • Static friction helps the box move with the train.
  • Maximum static friction = μsN.
  • For a horizontal surface, N = mg.
  • Using F = ma, we get:
ma = μsmg
a = μsg
a = 0.15 × 10 = 1.5 m/s²

Hence, the train can accelerate up to 1.5 m/s² without the box sliding.

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