---
title: "Reactions on two spheres in a channel"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:two-smooth-spheres-channel-reactions"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-smooth-spheres-channel-reactions
site: BCT Engineering
---

# Reactions on two spheres in a channel

Applied Mechanics · Chapter 2: Basic Concept in Statics and Static Equilibrium

- Asked in: 2073 Chaitra, 6 marks
- Video: [Sphere Problems in Mechanics / Ball and Roller Problems / Equilibrium in Engineering Mechanics / FBD](https://www.youtube.com/watch?v=yQjSbeKSDSM) — Excellent Ideas in Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-smooth-spheres-channel-reactions
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

Two smooth spheres of weight 200 N each are kept inside a channel as shown in the figure. The radius of each sphere is 20 cm. Determine reactions at contact surfaces. [Figure: spheres A (upper) and B (lower) in a channel of width 60 cm with floor and vertical walls; contact points labelled 1, 2, 3, 4.]

## Short answer

Geometry gives a 60° line of centres. For sphere A: R2 sin60°=200 N → R2=230.94 N, R1=R2 cos60°=115.47 N. For sphere B: R3=200+R2 sin60°=400 N, R4=R2 cos60°=115.47 N.

## Answer

**Reading of the figure:** two smooth spheres A (upper) and B (lower), each W = 200 N, radius 20 cm, in a channel 60 cm wide. A touches the left wall (contact 1) and B (contact 2). B touches the floor (contact 3) and the right wall (contact 4).

#### Geometry
Horizontal distance between centres = 60 − 20 − 20 = 20 cm. Distance between centres = 20 + 20 = 40 cm.
cos θ = 20/40, so θ = 60° (the line of centres makes 60° with the horizontal; A is up and to the left of B).

#### Sphere A (three forces: W, R₁ horizontal, R₂ along the line of centres)
```text
ΣFy = 0:  R2 sin60° − 200 = 0   ->  R2 = 200/0.8660 = 230.94 N
ΣFx = 0:  R1 − R2 cos60° = 0    ->  R1 = 230.94 × 0.5 = 115.47 N
```

#### Sphere B (W, R₂ from A, R₃ vertical, R₄ horizontal)
Sphere A pushes B down and to the right with 230.94 N.
```text
ΣFy = 0:  R3 − 200 − R2 sin60° = 0  ->  R3 = 200 + 200 = 400 N
ΣFx = 0:  R4 − R2 cos60° = 0        ->  R4 = 115.47 N
```
Check: the floor carries both weights, 400 N.

**Answer:** R₁ = 115.47 N (left wall on A), R₂ = 230.94 N (between A and B), R₃ = 400 N (floor on B), R₄ = 115.47 N (right wall on B).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The line of centres makes what angle with the horizontal?
   A. 60°
   B. 45°
   C. 90°
   D. 30°
   Answer: A. Horizontal offset 20 cm, centre distance 40 cm: cos θ = 0.5, θ = 60°. (Key point: Geometry)

2. The vertical equilibrium of A gives the contact force with B as R2 =
   A. 200 N
   B. 200 cos 60° = 100 N
   C. 400 N
   D. 200/sin 60° = 230.94 N
   Answer: D. Only R2's vertical component supports the 200 N weight. (Key point: Sphere A vertical)

3. The left wall reaction on A is:
   A. 0 N
   B. R2 cos 60° = 115.47 N
   C. 200 N
   D. 230.94 N
   Answer: B. The wall balances the horizontal component of R2. (Key point: Wall reaction)

4. The floor reaction on B is 400 N because:
   A. B is twice as heavy as A
   B. Friction acts upward
   C. The wall adds 200 N
   D. The floor supports the weight of both spheres
   Answer: D. R3 = 200 + R2 sin 60° = 200 + 200 = 400 N. (Key point: Floor reaction)

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