---
title: "Force P to prevent block sliding down incline"
subject: "Applied Mechanics"
chapter: 5
key: "semester_1/applied_mechanics:block-50kg-force-p-prevent-sliding-down"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-50kg-force-p-prevent-sliding-down
site: BCT Engineering
---

# Force P to prevent block sliding down incline

Applied Mechanics · Chapter 5: Friction

- Asked in: 2067 Asoj, 6 marks
- Video: [Friction | friction engineering mechanics | Start motion up the incline | Keep it moving](https://www.youtube.com/watch?v=BgSCZ6Oh8eA) — KSG Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-50kg-force-p-prevent-sliding-down
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-5

## Question

Find the value of P to prevent the slide down of block as given below. [Figure: 50 kg block on a 30° incline, μ = 0.4, force P applied at 20° to the incline/horizontal as shown.]

## Short answer

Resolve forces parallel and perpendicular to the incline. Set friction F = 0.4 N acting up the slope to prevent downward slide. Equilibrium gives P (cos 20° +0.4 sin 20°) =490.5 (sin 30° -0.4 cos 30°). Solving yields P ≈70 N.

## Answer

**Reading of the figure:** a 50 kg block on a 30° incline, μ = 0.4. P pushes the block up the slope. Its line of action makes 20° with the incline surface, directed towards the surface (so P has a component along the slope, up, and a component pressing the block on to the plane). Take g = 9.81 m/s²: W = 50 × 9.81 = 490.5 N.

**Does it slide without P?** tan 30° = 0.577 > μ = 0.4, so (φ = 21.8° < 30°) the block slides down by itself. P is needed to hold it.

#### Impending motion down the slope (friction acts up the slope, F = μN)
Axes along and perpendicular to the incline.
```text
Perpendicular:  N = W cos30° + P sin20°
Along incline:  P cos20° + F = W sin30°,  F = 0.4 N
```
Substituting:
P cos20° + 0.4 (W cos30° + P sin20°) = W sin30°
P (cos20° + 0.4 sin20°) = W (sin30° − 0.4 cos30°)
P (0.9397 + 0.1368) = 490.5 (0.5 − 0.3464)
P × 1.0765 = 490.5 × 0.1536 = 75.35
P = **70.0 N**

Check: N = 490.5 × 0.8660 + 70.0 × 0.3420 = 448.7 N, F = 0.4 × 448.7 = 179.5 N, and P cos20° + F = 65.8 + 179.5 = 245.3 N = W sin30° ✓.

**Answer: P ≈ 70 N (minimum value to prevent the block from sliding down).** If the 20° is instead taken from the horizontal, the same method gives P ≈ 71 N.

## Check yourself

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

1. Why does the block slide down without P?
   A. Because P is absent
   B. tan 30° = 0.577 > mu = 0.4, so the incline is steeper than the friction angle
   C. Because mu is greater than tan 30°
   D. Because the mass is 50 kg
   Answer: B. Sliding occurs when tan(theta) > mu, i.e. incline angle exceeds phi = 21.8°. (Key point: Does block slide without P)

2. What is N for the block, with P pressing towards the plane at 20°?
   A. N = W cos30° + P sin20°
   B. N = W cos30° - P sin20°
   C. N = W sin30° + P cos20°
   D. N = W cos30°
   Answer: A. P has a component P sin20° pressing the block on to the plane, adding to W cos30°. (Key point: Normal reaction)

3. For impending slide down, which equation is used along the incline?
   A. P cos20° = W sin30° + 0.4 N
   B. P cos20° - 0.4 N = W sin30°
   C. P cos20° + 0.4 N = W sin30°
   D. P sin20° + 0.4 N = W sin30°
   Answer: C. Up the slope: P cos20° and friction (up the slope as the block tends to slide down); they balance W sin30°. (Key point: Equilibrium along incline)

4. What is the required P?
   A. About 245 N
   B. About 490 N
   C. About 180 N
   D. About 70 N
   Answer: D. P(1.0765) = 490.5 x 0.1536 = 75.35, so P = 70 N. (Key point: Value of P)

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