---
title: "Acceleration of block A with blocks B and C"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:block-a-acceleration-blocks-abc"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-a-acceleration-blocks-abc
site: BCT Engineering
---

# Acceleration of block A with blocks B and C

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2074 Chaitra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-a-acceleration-blocks-abc
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Determine the acceleration of block A for the system, if the system starts from rest. Coefficient of friction between block A and table is 0.25 and that between blocks A and B is 0.35. Weights of blocks A, B and C are 100 N, 50 N and 500 N respectively. [Figure: block B sits on block A on a horizontal table; B is tied to the wall by a rope on the left; A is connected over a pulley at the table edge to hanging block C.]

## Short answer

Friction on A from B: 0.35×50 = 17.5 N; from table: 0.25×150 = 37.5 N; total 55 N. Dynamics: 500 – T = (500/9.81)a, T – 55 = (100/9.81)a. Adding gives a = (500−55)×9.81/600 = 7.28 m/s².

## Answer

#### Reading of the figure
Block B rests on block A, both on a horizontal table. B is tied by a rope to the wall on the left, so it cannot move. A is connected by a rope over a pulley at the table edge to the hanging block C. When released, C falls, A moves to the right and slides under the fixed block B. Weights: A = 100 N, B = 50 N, C = 500 N. μ(table–A) = 0.25 and μ(A–B) = 0.35.

```text
 wall |--rope--[B]
      |        [A]-----------o
                              |
                             [C]
```

#### Friction forces on A (A is sliding)
- Between A and B: N₁ = W_B = 50 N, so F₁ = 0.35 × 50 = 17.5 N
- Between table and A: N₂ = W_A + W_B = 150 N, so F₂ = 0.25 × 150 = 37.5 N
- Total friction on A opposing motion = 55 N

#### Equations of motion (a = acceleration of A and C)
- Block C: 500 − T = (500/9.81) a
- Block A: T − 17.5 − 37.5 = (100/9.81) a

Adding: 500 − 55 = (600/9.81) a, so a = 445 × 9.81/600 = **7.28 m/s²**

Tension in the rope AC: T = 500 − (500/9.81)(7.276) = **129.2 N**
(Rope force on B equals the friction 17.5 N.)

**Answer: acceleration of block A = 7.28 m/s² (to the right).** Driving force 500 N exceeds the friction of 55 N, so motion does start.

## Check yourself

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

1. Block B is tied to the wall. What does this mean for the motion?
   A. B moves with A
   B. B stays fixed while A slides beneath it
   C. B moves opposite to A
   D. B falls with C
   Answer: B. The rope keeps B at rest, so A slides under B and friction acts on A at both the A-B and table-A surfaces. (Key point: Reading the setup)

2. What normal reaction is used for the friction between the table and A?
   A. W_A = 100 N
   B. W_B = 50 N
   C. W_A + W_B = 150 N
   D. W_A + W_B + W_C = 650 N
   Answer: C. The table supports both A and the block B resting on it, so N = 150 N and F = 0.25 x 150 = 37.5 N. (Key point: Normal forces for friction)

3. What is the total friction opposing A's motion?
   A. 17.5 N + 37.5 N = 55 N
   B. 37.5 N only
   C. 0.35 x 100 = 35 N
   D. 0.25 x 650 = 162.5 N
   Answer: A. A-B friction is 0.35 x 50 = 17.5 N and table friction is 37.5 N, totalling 55 N. (Key point: Total friction on A)

4. Adding the equations for C and A eliminates T. What result follows?
   A. a = (500 - 55) x 9.81 / 600 = 7.28 m/s^2
   B. a = 500 x 9.81 / 600
   C. a = (500 - 55) x 9.81 / 100
   D. a = 55 x 9.81 / 600
   Answer: A. C: 500 - T = (500/g)a and A: T - 55 = (100/g)a. Adding gives 445 = (600/g)a, so a = 7.28 m/s^2. (Key point: Equations of motion)

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