Three forces along the axes
- Input
- 50 N a 0°, 30 N a 90°, 20 N a 180°
- Expected output
- R = (30, 30) N; |R| = 30√2 ≈ 42,4264 N a 45°
Equilibrant: 225°.
resultant force calculator
Open the Resultant of forces mode, enter the magnitude and angle (from the x-axis, counterclockwise) of each force and pick the unit. The table shows Fx and Fy for each one and their sum, and the drawing chains the forces head to tail.
Equilibrant: 225°.
Equilibrant at 247.3801°.
Square diagonal.
No defined direction.
Approximate result.
The polygon closes.
Add or subtract component by component: A + B = (a₁ + b₁, a₂ + b₂) and, in 3D, also a₃ + b₃. For A = (3, 4) and B = (-1, 2): A + B = (2, 6) and A - B = (4, 2). Geometrically, the sum is the diagonal of the parallelogram built on A and B, or the vector that closes the chain when you place the tail of B on the tip of A.
From the positive x-axis, counterclockwise: 0° points right, 90° up, 180° left and 270° down. A 20 N force pointing left is 20 at 180°, never a negative magnitude.
The resultant is the single force that replaces the set. The equilibrant is the one that cancels the resultant, with the same magnitude and the opposite direction. If the resultant is already zero there is nothing to balance.
No. It treats concurrent forces, applied at the same point, and only adds vectors. Forces at different points also produce torque, which needs another calculation (force times lever arm).
(2, 6)
(4, 2)
(6, 8)
5
√5
≈ 2.2361
(3/5, 4/5)
≈ (0.6, 0.8)
(-√5/5, 2√5/5)
≈ (-0.4472, 0.8944)
53.1301°
5
63.4349°
Neither parallel nor orthogonal
10
10
5
(-1, 2)
√5
≈ 2.2361
(4, 2)
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