---
title: "Force for just motion of block B"
subject: "Applied Mechanics"
chapter: 5
key: "semester_1/applied_mechanics:block-b-just-motion-force-p-15deg"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-b-just-motion-force-p-15deg
site: BCT Engineering
---

# Force for just motion of block B

Applied Mechanics · Chapter 5: Friction

- Asked in: 2073 Chaitra, 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/block-b-just-motion-force-p-15deg
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-5

## Question

Determine the force required for just motion of the block B shown in the figure below. Take μs = 0.25 for all surfaces. [Figure: block A (20 N) on block B (50 N); A tied to a wall by a string at 15°; B pulled by force P to the left on the ground.]

## Short answer

From block A equilibrium, tension T = 4.851 N, normal N1 = 18.74 N, friction F1 = 4.69 N. For block B, ground normal N2 = 68.74 N, ground friction F2 = 17.19 N. The force P = F1 + F2 = 21.87 N, approximately 21.9 N.

## Answer

**Reading of the figure:** block A (20 N) sits on block B (50 N), which is on the ground. A is tied to a wall by a string at 15° to the horizontal (rising to the right). P pulls B to the left. μₛ = 0.25 for all surfaces. B slides to the left under A, which is held by the string.

#### Block A (held by the string)
Forces: weight 20 N, normal N₁ from B, friction F₁ = μN₁ from B (to the left, the direction of B's motion), tension T at 15°.
```text
ΣFy = 0:  N1 + T sin15° = 20      ->  N1 = 20 − 0.2588 T
ΣFx = 0:  T cos15° = F1 = 0.25 N1
          0.9659 T = 0.25 (20 − 0.2588 T)
          (0.9659 + 0.0647) T = 5
          T = 4.851 N
```
N₁ = 20 − 0.2588 × 4.851 = 18.74 N
F₁ = 0.25 × 18.74 = 4.69 N

#### Block B (just about to move to the left)
Forces: P to the left, weight 50 N, N₁ = 18.74 N pressing down from A, ground normal N₂, friction from A (F₁ = 4.69 N to the right) and ground friction F₂ (to the right).
```text
N2 = 50 + 18.74 = 68.74 N
F2 = 0.25 × 68.74 = 17.19 N
P = F1 + F2 = 4.69 + 17.19 = 21.87 N
```
**Answer: P ≈ 21.9 N.**

## Check yourself

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

1. Block A is held by the string, so friction F1 from B acts on A:
   A. Vertically
   B. Zero
   C. Along B's motion, balanced by T cos15°
   D. Opposite B's motion
   Answer: C. T cos15° = F1. (Key point: Held block)

2. N1 equals:
   A. 20
   B. T cos15°
   C. 20 - T sin15°
   D. 20 + T sin15°
   Answer: C. Vertical balance. (Key point: Normal N1)

3. N2 equals:
   A. 50 N
   B. 50 + N1 = 68.74 N
   C. 70 N
   D. 50 - N1
   Answer: B. B carries A's pressure. (Key point: Ground normal)

4. P equals:
   A. 50*0.25
   B. T
   C. F2 only
   D. F1 + F2 = 21.9 N
   Answer: D. Sum of frictions on B. (Key point: P)

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