---
title: "Force P to move block A (blocks A and B)"
subject: "Applied Mechanics"
chapter: 5
key: "semester_1/applied_mechanics:two-blocks-force-p-move-block-a-30deg"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-blocks-force-p-move-block-a-30deg
site: BCT Engineering
---

# Force P to move block A (blocks A and B)

Applied Mechanics · Chapter 5: Friction

- Asked in: 2073 Shrawan, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-blocks-force-p-move-block-a-30deg
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-5

## Question

Two blocks A and B of 40 N and 20 N respectively are in equilibrium position as shown in the figure below. Calculate the force P required to move block A. Take μ = 0.3 for all surfaces. [Figure: block B resting on block A, B tied to a wall by a string inclined at 30°; block A on the ground pulled by force P to the left.]

## Short answer

For block B, string tension T = 5.905 N, normal from A N1 = 17.05 N, friction F1 = 0.3 N1 = 5.11 N. On block A, ground normal N2 = 40 + 17.05 = 57.05 N, ground friction F2 = 17.11 N. P = F1 + F2 = 22.23 N, so force required is about 22.2 N to the left.

## Answer

**Reading of the figure:** block B (20 N) rests on block A (40 N), which is on the ground. B is tied to a wall by a string inclined at 30° to the horizontal (rising towards the wall, on the right). P pulls A to the left. μ = 0.3 for all surfaces. When A moves left, B tends to move with it, but the string holds B, so B stays at rest and A slides under it.

#### Block B (equilibrium, friction from A at the limit)
Forces: weight 20 N, normal N₁ from A, friction F₁ = μN₁ from A (acting to the left, the direction of A's motion), string tension T (up and to the right at 30°).
```text
ΣFy = 0:  N1 + T sin30° − 20 = 0     ->  N1 = 20 − 0.5 T
ΣFx = 0:  T cos30° − F1 = 0, F1 = 0.3 N1
          0.866 T = 0.3 (20 − 0.5 T)
          0.866 T + 0.15 T = 6
          T = 5.905 N
```
N₁ = 20 − 0.5 × 5.905 = 17.05 N
F₁ = 0.3 × 17.05 = 5.11 N

#### Block A (moving to the left, so all frictions act to the right)
Forces: P to the left, weight 40 N, normal pressure N₁ = 17.05 N from B (downward), ground normal N₂, friction from B (F₁ = 5.11 N), ground friction F₂ = 0.3 N₂.
```text
ΣFy = 0:  N2 = 40 + 17.05 = 57.05 N
F2 = 0.3 × 57.05 = 17.11 N
ΣFx = 0:  P = F1 + F2 = 5.11 + 17.11 = 22.23 N
```
**Answer: P ≈ 22.2 N (to the left).**

## Check yourself

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

1. When P pulls A, what happens to B?
   A. B moves with A
   B. B stays at rest held by the string while A slides under it
   C. B slides on the string
   D. B tips
   Answer: B. The string holds B. (Key point: Which moves)

2. N1 on B is:
   A. 20 - T sin30°
   B. 20 + T sin30°
   C. 20 cos30°
   D. 20
   Answer: A. Tension's vertical component reduces N1. (Key point: Normal on B)

3. N2 on A is:
   A. 40 - N1
   B. 40 + N1
   C. 40 + 20
   D. 40
   Answer: B. A carries its own weight plus N1 from B. (Key point: Ground normal)

4. P equals:
   A. F2 only
   B. F1 + F2 = 22.2 N
   C. F1 - F2
   D. T
   Answer: B. Both frictions oppose A. (Key point: Force P)

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