- 2067 Asoj · 8 marks
Resultant of three forces on inclined block
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.
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.]
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.
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