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PhysicsLaws of Motion
A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0·12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is :

(g = 10 m/s²)
Options
1
2·1
2
1·8
3
1·5
4
1·2
Correct Answer
Option 4 : 1·2 m s⁻²
Step-by-Step Solution
1

Understand the setup:

The box sits on the trolley floor. When the trolley accelerates, the only horizontal force available to accelerate the box along with it is static friction from the trolley floor.

2

Forces on the box (horizontal):

Normal force N = mg = 15 × 10 = 150 N

Maximum static friction: \(f_{max} = \mu_s \times N = 0.12 \times 150 = 18 \text{ N}\)

3

Apply Newton's Second Law to the box:

For the box to stay stationary relative to trolley, friction must provide its acceleration:

$$f = ma \implies a_{max} = \frac{f_{max}}{m} = \frac{\mu_s mg}{m} = \mu_s g$$

4

Calculate:

$$a_{max} = \mu_s \times g = 0.12 \times 10 = \mathbf{1.2 \text{ m/s}^2}$$

Note: mass of box (15 kg) cancels out and is irrelevant! ✓

Theory: Friction & Newton's Laws
1. What is Friction?

Friction is a contact force that opposes relative motion (or tendency of relative motion) between two surfaces in contact. It acts parallel to the surface of contact and arises due to microscopic interlocking of surface irregularities and intermolecular adhesive forces at contact points.

Friction is not always an undesirable force. In this problem, friction is the essential force that allows the box to accelerate with the trolley. Without friction, the box would remain stationary in the ground frame while the trolley slides under it. Walking, driving, and holding objects all depend on friction.

2. Types of Friction

Static Friction (fₛ): Acts when surfaces are not sliding. It is a self-adjusting force — it takes whatever value is needed to prevent relative motion, up to a maximum of μₛN. It is always equal and opposite to the applied force (or tendency-causing force) until the maximum is reached.

Kinetic (Sliding) Friction (fₖ): Acts when surfaces are sliding. Its magnitude is μₖN — constant regardless of speed or area. Since μₖ < μₛ always, kinetic friction is less than maximum static friction. This is why it is easier to keep an object sliding than to start the slide.

Rolling Friction: Acts on rolling objects. Much smaller than sliding friction — this is why wheels are used in vehicles. Rolling friction coefficient is very small (~ 0.001–0.005 for rubber on road).

3. Laws of Friction

📌 Law 1: Friction force is proportional to normal reaction: f = μN

📌 Law 2: Friction is independent of area of contact

📌 Law 3: Kinetic friction is independent of speed of sliding

📌 Law 4: μₛ > μₖ always (static > kinetic coefficient)

The coefficient of friction μ is dimensionless and depends on the nature of the two surfaces in contact. It does not depend on mass, area, or speed. Typical values: rubber on concrete ≈ 0.7, steel on steel ≈ 0.4, ice on ice ≈ 0.03.

4. Friction in Accelerating Systems

When an object rests on an accelerating surface, friction is the only horizontal force available to accelerate the object. The analysis is always from the ground (inertial) frame:

For object on accelerating surface:

\(f = ma\) (friction provides acceleration)

\(a_{max} = \mu_s g\) (when f = f_max)

Key insight: the maximum acceleration is μₛg, independent of mass. This is because both maximum friction force (μₛmg) and the required force (ma) are proportional to mass, so mass cancels.

5. Pseudo Force Analysis (Non-Inertial Frame)

In the reference frame of the accelerating trolley (non-inertial frame), a pseudo force acts on the box in the direction opposite to the trolley's acceleration. For the box to remain stationary in the trolley's frame, static friction must balance this pseudo force:

\(f_s \geq ma_{trolley}\)

\(\mu_s mg \geq ma_{trolley}\)

\(a_{trolley} \leq \mu_s g\)

Both the ground frame and trolley frame give the same condition: maximum acceleration = μₛg. Always choose the inertial (ground) frame for cleaner analysis in NEET.

6. Angle of Friction and Angle of Repose

The angle of friction (λ) is defined as tan λ = μ, where μ is the coefficient of friction. The resultant of normal force and friction force makes angle λ with the normal. The angle of repose (θ) is the maximum angle of inclination of a surface for which an object placed on it does not slide: tan θ = μₛ. Therefore angle of repose = angle of friction for static conditions.

7. Common NEET Friction Problems

📌 Object on trolley/conveyor belt: a_max = μg (this problem)

📌 Object on incline: N = mg cos θ, friction = μmg cos θ

📌 Two blocks, one on top of other: Friction between blocks limits relative motion

📌 Object being pushed horizontally: As normal force increases, friction increases

⚠️ Mass is irrelevant for maximum acceleration in this type of problem — always cancels out.

⚠️ Normal force = mg only when surface is horizontal and no vertical component of other forces acts.

⚠️ Static friction adjusts itself — it equals the required force up to μₛN, not always equal to μₛN.

⚠️ The 15 kg mass given in this question is a deliberate distractor — it does not affect the answer.

Frequently Asked Questions
1. Why does mass not matter in this problem?
a_max = μmg/m = μg. Both maximum friction (μmg) and required force (ma) are proportional to mass, so it cancels. The 15 kg is a deliberate distractor to mislead students who try to use it in their calculation.
2. What provides the force to accelerate the box?
Static friction from the trolley floor is the only horizontal force on the box. This friction acts forward (in the direction of trolley motion) on the box, accelerating it along with the trolley.
3. What is the friction force at maximum acceleration?
f = ma = 15 × 1.2 = 18 N. Or equivalently, f_max = μmg = 0.12 × 15 × 10 = 18 N. Both give the same answer, confirming that at maximum acceleration, friction has reached its limit.
4. What happens to the box if acceleration exceeds 1.2 m/s²?
The box starts to slide backward relative to the trolley. Kinetic friction now acts (slightly less than static friction maximum). The box lags behind the trolley and eventually falls off if the trolley continues to accelerate.
5. Is the normal force affected by horizontal acceleration?
No. For horizontal acceleration, the vertical forces are unchanged: N = mg. If the trolley were accelerating vertically (like an elevator), N would change and affect friction. But for horizontal motion, N = mg always.
6. What is the difference between μₛ and μₖ?
μₛ (static) is the coefficient for surfaces not sliding — it gives maximum static friction. μₖ (kinetic) is for sliding surfaces — always less than μₛ. In this problem, μ = 0.12 is given as static friction coefficient since the box is not sliding.
7. Can we solve this using pseudo force?
Yes. In the trolley's frame, pseudo force = ma backward. For equilibrium: μmg ≥ ma → a ≤ μg = 1.2 m/s². Both frames give the same maximum acceleration. Prefer ground frame for NEET as it is more straightforward.
8. What is the angle of repose for μ = 0.12?
Angle of repose θ = arctan(μ) = arctan(0.12) ≈ 6.84°. If the trolley floor were inclined at more than 6.84°, the box would slide even when the trolley is stationary. This is very gentle slope — the surfaces are quite slippery.
9. How does friction change during the acceleration?
Static friction self-adjusts. If trolley acceleration is 0.5 m/s², friction on box = 15 × 0.5 = 7.5 N (less than maximum 18 N). Only at the maximum acceleration of 1.2 m/s² does friction reach its full value of 18 N.
10. Does friction depend on area of contact?
No. Friction force f = μN is independent of area of contact. Whether the box rests on its large face or small face, the friction force is the same. This is Amontons' second law of friction — one of the most counterintuitive facts in mechanics.
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