---
title: "Resultant of three forces on inclined block"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:resultant-three-forces-600n-400n-300n"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-three-forces-600n-400n-300n
site: BCT Engineering
---

# Resultant of three forces on inclined block

Applied Mechanics · Chapter 3: Forces acting on particle and rigid body

- Asked in: 2067 Asoj, 8 marks
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-three-forces-600n-400n-300n
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Find the resultant of the given three forces. [Figure: forces on a block resting on a 25° incline: 600 N at 20° from the vertical, 400 N, and 300 N at 35° above the incline direction.]

## Short answer

Resolve each force into horizontal and vertical components. ΣFx = 896.15 N, ΣFy = 48.93 N. Resultant magnitude 897.5 N, direction 3.1° above the horizontal.

## Answer

**Reading of the figure:** a block on a 25° incline (rising to the right). Three forces act on the block:
- 600 N, directed towards the block, at 20° to the normal of the incline, on the left side of the normal. This makes it act at 45° below the horizontal (down and to the right).
- 400 N, directed up and to the right at 35° to the incline surface, so at 25° + 35° = 60° above the horizontal.
- 300 N, directed up along the incline, at 25° above the horizontal.

#### Components (x to the right, y upward)
| Force | Angle with +x | Fx (N) | Fy (N) |
|---|---|---|---|
| 600 N | −45° (315°) | 600 cos 45° = +424.26 | −600 sin 45° = −424.26 |
| 400 N | 60° | 400 cos 60° = +200.00 | 400 sin 60° = +346.41 |
| 300 N | 25° | 300 cos 25° = +271.89 | 300 sin 25° = +126.79 |
| **Sum** | | **ΣFx = 896.15** | **ΣFy = 48.93** |

#### Resultant
R = √(896.15² + 48.93²) = **897.5 N**
θ = tan⁻¹(48.93/896.15) = **3.1°** above the horizontal (towards the right).

Along and normal to the incline (if needed): along the incline (25°) R∥ = 896.15 cos 25° + 48.93 sin 25° = 832.8 N; normal R⊥ = −896.15 sin 25° + 48.93 cos 25° = −334.5 N.

**Answer: R ≈ 897.5 N acting at 3.1° above the horizontal.**

## Check yourself

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

1. Why is ΣFy small (48.93 N)?
   A. The 600 N downward component -424.26 nearly cancels 346.41 + 126.79
   B. The 300 N force has no vertical part
   C. All forces are horizontal
   D. The 400 N is downward
   Answer: A. Fy = -424.26 + 346.41 + 126.79. (Key point: Sum of components)

2. What is the resultant?
   A. 897.5 N at 3.1° above the horizontal
   B. 897.5 N at 86.9° above the horizontal
   C. 48.93 N horizontal
   D. 1300 N along the incline
   Answer: A. R = sqrt(896.15^2 + 48.93^2), tan(theta) = 48.93/896.15. (Key point: Resultant)

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