Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Three identical billiard balls each of mass 1 kg are placed as shown in the figure. The upper ball is smooth. Find minimum coefficient of static friction between lower balls and ground, so that balls remain stationary

Question Image

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

13

b

133

c

23

d

05

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Problem Statement

Three identical billiard balls, each of mass m = 1 kg, are placed in a triangular arrangement with two balls on the ground and one smooth ball resting on top. Find the minimum coefficient of static friction (μ) between the lower balls and the ground so that the balls remain stationary.

B C A mg N₁ N₂ 30° 2r

 

Solution Approach

Step 1: Geometry Analysis

For three identical spheres in contact with radius r:

  • Distance between centers of lower balls = 2r

  • The upper ball forms an equilateral triangle with the lower balls

  • The angle between vertical and the line joining centers = 30°

Step 2: Forces on Upper Ball A

The upper smooth ball experiences:

  • Weight: mg (downward)

  • Normal force N₁ from ball B (perpendicular to contact, at 30° from vertical)

  • Normal force N₂ from ball C (perpendicular to contact, at 30° from vertical)

Equilibrium conditions:

Horizontal: N₁ sin(30°) = N₂ sin(30°) → N₁ = N₂ = N
Vertical: 2N cos(30°) = mg
N = mg / (2 cos(30°)) = mg / √3

Step 3: Forces on Lower Ball B (or C)

Each lower ball experiences:

  • Weight: mg (downward)

  • Normal force from ball A: N (at 30° from vertical, pushing outward)

  • Normal force from ground: Nground (upward)

  • Friction force from ground: f (inward, preventing slip)

Equilibrium conditions:

Horizontal: f = N sin(30°) = (mg/√3) × (1/2) = mg/(2√3)
Vertical: Nground = mg + N cos(30°) = mg + (mg/√3) × (√3/2) = 3mg/2

Step 4: Minimum Coefficient of Friction

For the system to remain stationary:

f ≤ μ Nground
mg/(2√3) ≤ μ × (3mg/2)
μ ≥ mg/(2√3) × 2/(3mg) = 1/(3√3) = √3/9
Minimum Coefficient of Static Friction: μmin = √3/9 ≈ 0.192
 
Physical Interpretation: The friction must be sufficient to prevent the lower balls from sliding outward due to the outward horizontal component of the normal force from the upper ball. A coefficient of approximately 0.19 or higher is needed to keep the system stable.
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring