---
title: "Two blocks connected over a pulley"
subject: "Engineering Mechanics"
chapter: 2
key: "semester_3/engineering_mechanics:blocks-pulley-friction-acceleration"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/blocks-pulley-friction-acceleration
site: BCT Engineering
---

# Two blocks connected over a pulley

Engineering Mechanics · Chapter 2: Kinetics of Particles: Force, Mass and Acceleration

- Asked in: year not given, 8 marks
- Video: [In the figure two blocks are connected over a pulley](https://www.youtube.com/watch?v=kzco9H2y1-Q) — WNY Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/blocks-pulley-friction-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-2

## Question

Block A of mass 40 kg rests on a horizontal table whose coefficient of kinetic friction is 0.20. It is connected by a light inextensible string, passing over a smooth light pulley at the edge of the table, to block B of mass 25 kg hanging vertically. The system is released from rest. Find (a) the acceleration of the blocks, (b) the tension in the string, (c) the speed and distance moved after 2 s, and (d) the resultant load on the pulley support.

## Short answer

Using Newton's second law: T – μ m_A g = m_A a, m_B g – T = m_B a. Solve: a = (m_B g – μ m_A g)/(m_A+m_B) = 2.57 m/s², T = 181.1 N. After 2 s: v = 5.13 m/s, s = 5.13 m. Pulley load = √2 T = 256.1 N at 45° below horizontal toward A.

## Answer

**Given:** $m_A = 40$ kg, $m_B = 25$ kg, $\mu = 0.20$, $g = 9.81$ m/s². The string is inextensible, so both blocks have the same acceleration $a$. B moves down and A moves toward the pulley.

```text
   +-----+        ___
   |  A  |-------(   )     pulley
 ==+-----+==========\ \====
   friction <--      \ \ T
                      [B] (down, a)
```

#### Free-body equations
For A (horizontal): friction $F = \mu m_A g = 0.20 \times 40 \times 9.81 = 78.48$ N.

$$
T - 78.48 = 40a
$$

For B (vertical): $m_B g = 25 \times 9.81 = 245.25$ N.

$$
245.25 - T = 25a
$$

#### (a) Acceleration
Adding the two equations:

$$
245.25 - 78.48 = 65a \;\Rightarrow\; a = \frac{166.77}{65} = 2.566\ \text{m/s}^2
$$

#### (b) Tension

$$
T = 78.48 + 40(2.566) = 181.1\ \text{N}
$$

Check from B: $T = 25(9.81 - 2.566) = 181.1$ N. It agrees.

#### (c) Speed and distance after 2 s
Starting from rest with constant $a$:

$$
v = at = 2.566 \times 2 = 5.13\ \text{m/s}
\qquad
s = \tfrac12 a t^2 = 5.13\ \text{m}
$$

#### (d) Load on the pulley support
The two straight parts of the string pull on the pulley with force $T$ horizontally (toward A) and $T$ vertically downward (toward B). The pulley is light and smooth, so the load it passes to its support is the resultant of these two pulls:

$$
R = \sqrt{T^2 + T^2} = \sqrt{2}\,T = 256.1\ \text{N}
$$

This load acts at $45^\circ$ below the horizontal, directed toward the A side (down and back toward the table). The support pushes on the pulley with an equal and opposite force.

**Answer: a = 2.57 m/s²; T = 181.1 N; v = 5.13 m/s and s = 5.13 m at 2 s; pulley load = 256.1 N at 45° below the horizontal, toward A.**

## Check yourself

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

1. Which equation applies to block A on the table?
   A. T + 78.48 = 40a
   B. T - 78.48 = 40a
   C. T = 40a
   D. 78.48 - T = 40a
   Answer: B. Tension pulls A toward the pulley and kinetic friction mu m_A g = 78.48 N opposes it, so T - 78.48 = 40a. (Key point: Equation for A)

2. Adding the two equations eliminates T. The acceleration is:
   A. (245.25 - 78.48)/65 = 2.57 m/s^2
   B. 245.25/65 = 3.77 m/s^2
   C. (245.25 + 78.48)/65 = 4.98 m/s^2
   D. 166.77/40 = 4.17 m/s^2
   Answer: A. The driving force is m_B g - friction = 166.77 N acting on the total mass 65 kg. (Key point: Acceleration)

3. The tension in the string is:
   A. 245.25 N, equal to the weight of B
   B. 78.48 N, equal to the friction
   C. 181.1 N, which lies between the two
   D. 65 x 2.566 N
   Answer: C. T = 78.48 + 40(2.566) = 181.1 N, which also equals 25(9.81 - 2.566). It is less than B's weight because B accelerates. (Key point: Tension)

4. The load on the pulley support is the resultant of two forces T, one horizontal and one vertical. Its value is:
   A. 2T
   B. T
   C. sqrt(2) T = 256.1 N
   D. T/2
   Answer: C. Perpendicular forces of equal size T combine to sqrt(T^2 + T^2) = sqrt(2) T. (Key point: Pulley load)

## Related in Engineering Mechanics

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