SME
Statics · Problem type
Resultant of three forces by components
Use this for “Forces of 200 N at 15°, 300 N at 90° and 100 N at 225° act at a point. Find the resultant.”
Diagram
F₁ = 200 N at α₁ = 15° from +x: (193.2, 51.76) N.
F₂ = 300 N at α₂ = 90° from +x: (0, 300) N.
F₃ = 100 N at α₃ = 225° from +x: (−70.71, −70.71) N.
ΣFₓ = 122.5 N and ΣFᵧ = 281.1 N.
R
=
√
122.5
2
+
281.1
2
=
306.6
N
at
66.45°
from
+x.
This is an example. Type your own values to start a new problem.
Rₓ (N)
=
F₁ (N)
×
cos(
α₁
°)
+
F₂ (N)
×
cos(
α₂
°)
+
F₃ (N)
×
cos(
α₃
°)
Rᵧ (N)
=
F₁ (N)
×
sin(
α₁
°)
+
F₂ (N)
×
sin(
α₂
°)
+
F₃ (N)
×
sin(
α₃
°)
R (N)
=
(
Rₓ (N)
)
2
+
(
Rᵧ (N)
)
2
β
Resultant direction
worked out (not typed) · °
Formulas
Rₓ
=
F₁
×
cos(
α₁
°)
+
F₂
×
cos(
α₂
°)
+
F₃
×
cos(
α₃
°)
122.4745 = 200 × cos(15°) + 300 × cos(90°) + 100 × cos(225°)
Rᵧ
=
F₁
×
sin(
α₁
°)
+
F₂
×
sin(
α₂
°)
+
F₃
×
sin(
α₃
°)
281.0531 = 200 × sin(15°) + 300 × sin(90°) + 100 × sin(225°)
R
=
√
Rₓ² +
Rᵧ
2
306.5793
=
√
122.4745
2
+
281.0531
2
β
=
atan2(Rᵧ,
Rₓ)
66.4539 = atan2(281.0531, 122.4745)
Assumptions
•
Every force acts at one point, so the forces add as vectors.
•
Each angle α is measured counterclockwise from the positive x-axis.
•
The direction β comes from atan2(Rᵧ, Rₓ), so it lands in the right quadrant.
Step-by-step
WE KNOW
F₁ = 200 N, α₁ = 15°, F₂ = 300 N, α₂ = 90°, F₃ = 100 N, α₃ = 225°
FIND
resultant x-part (Rₓ), resultant y-part (Rᵧ), resultant size (R), resultant direction (β)
Step 1 · Find resultant x-part (Rₓ)
Use Rₓ = F₁ × cos(α₁°) + F₂ × cos(α₂°) + F₃ × cos(α₃°)
Add each force’s x-part, F cos α, signs and all: a force pointing left adds a negative part.
Rₓ = 200 × cos(15°) + 300 × cos(90°) + 100 × cos(225°)
Rₓ = 193.1852 + 0 + (−70.7107)
Rₓ = 122.4745 N
Step 2 · Find resultant y-part (Rᵧ)
Use Rᵧ = F₁ × sin(α₁°) + F₂ × sin(α₂°) + F₃ × sin(α₃°)
Add each force’s y-part, F sin α, signs and all: a force pointing down adds a negative part.
Rᵧ = 200 × sin(15°) + 300 × sin(90°) + 100 × sin(225°)
Rᵧ = 51.7638 + 300 + (−70.7107)
Rᵧ = 281.0531 N
Step 3 · Find resultant size (R)
Use R = √(Rₓ² + Rᵧ²)
The two parts are at right angles, so the size is the hypotenuse of their right triangle.
R
=
√
122.4745
2
+
281.0531
2
R
=
√
15,000 + 78,990.9
R
=
306.5793
N
Step 4 · Find resultant direction (β)
Use β = atan2(Rᵧ, Rₓ)
The angle from the positive x-axis: tan⁻¹ of Rᵧ over Rₓ, turned into the quadrant the parts point to.
β
=
atan2(281.0531,
122.4745)
(122.4745, 281.0531) is in quadrant I: tan⁻¹ gives the angle as it is
β
=
tan⁻¹(
281.0531
122.4745
)
=
66.4539°
β
=
66.4539°
Check
Put the numbers back in. Both sides should match.
122.4745 = 200 × cos(15°) + 300 × cos(90°) + 100 × cos(225°) ✓
281.0531 = 200 × sin(15°) + 300 × sin(90°) + 100 × sin(225°) ✓
306.5793
=
√
122.4745
2
+
281.0531
2
✓
66.4539 = atan2(281.0531, 122.4745) ✓
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