- First force F₁
- 30N
- Second force F₂
- 40N
- Angle between them θ
- 90°
50.0000N
Open with these values50.0000N
Result: 50.0000 NTwo forces at an angle combine into one, and its size sits between their difference and their sum. 30 N and 40 N give 70 N pointing the same way, 50 N at a right angle and 10 N head-on. The angle is the one enclosed between the two arrows, in degrees.
50.0000N
Open with these values70.0000N
Open with these values86.6025N
Open with these valuesR = √(F₁² + F₂² + 2 × F₁ × F₂ × cos θ)
| F₁, F₂, angle | What it is | Resultant |
|---|---|---|
| 0, 40, 90 | One force only | 40 |
| 3, 4, 90 | The 3-4-5 right triangle | 5 |
| 5, 5, 180 | Equal and opposite — they cancel | 0 |
| 7, 24, 90 | The 7-24-25 right triangle | 25 |
| 10, 10, 60 | Equal forces 60° apart | 17.3205 |
| 30, 40, 0 | Same direction — they add | 70 |
| 30, 40, 90 | At a right angle | 50 |
| 30, 40, 180 | Opposite — they subtract | 10 |
| 100, 50, 120 | A large and a small force, 120° apart | 86.6025 |
Use R = √(F₁² + F₂² + 2 × F₁ × F₂ × cos θ), where θ is the angle enclosed between the two forces. A 30 N and a 40 N force at 90° give √(900 + 1600) = 50 N.
The single force that has the same effect as two or more forces acting together. Instead of tracking each one separately you replace them with one equivalent arrow.
Because θ is the angle between the two force arrows, not the angle inside the triangle they form. Those two angles are supplementary, and swapping one for the other flips the sign — with the triangle angle the same formula carries a minus.
At 0° the cosine is 1 and the forces add: R = F₁ + F₂, the largest result possible. At 180° the cosine is −1 and they subtract: R = |F₁ − F₂|, the smallest. Everything else lies between those two.
Degrees, from 0 to 180, and the calculator converts internally. Reading 60 as radians instead of degrees would turn 17.32 N into 4.4 N, so the difference is not cosmetic.
Information, not professional advice.
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