Force components and resultants
A force can be split into perpendicular components or combined with other forces into a single resultant. Nearly every statics problem uses these two skills.
Rectangular components
A force F at angle \(\theta\) above the positive x-axis has components \(F_x = F\cos\theta\) and \(F_y = F\sin\theta\). A 500 N force at 30° has \(F_x = 500\cos 30° \approx 433\) N and \(F_y = 500\sin 30° = 250\) N. Signs show direction: components pointing left or down are negative.
Magnitude and direction from components
Going the other way, \(F = \sqrt{F_x^2 + F_y^2}\) and \(\theta = \tan^{-1}\frac{F_y}{F_x}\), checking the quadrant from the signs. Components of 300 N and 400 N combine to 500 N at about 53.1° above the x-axis.
Slopes instead of angles
Directions are often given as a slope triangle, such as 3 : 4 : 5. A 250 N force along a 3-4-5 slope, rising 3 for every 4 across, has components 250 x 4/5 = 200 N horizontally and 250 x 3/5 = 150 N vertically, with no trigonometry needed.
Adding forces by components
To find the resultant of several forces, add all x-components and all y-components separately, then combine. Forces of 200 N at 0° and 300 N at 60° give Rx = 200 + 150 = 350 N and Ry = 259.8 N, so R is about 436 N at about 36.6°.
The parallelogram law
Two forces add as the diagonal of the parallelogram they form, or equivalently by placing them head to tail in a triangle. The law of cosines then gives the resultant's magnitude: with 200 N and 300 N at 60° apart, R squared = 200 squared + 300 squared + 2(200)(300)cos 60° = 190,000, so R is about 436 N, matching the component method.
Resolving along chosen axes
Axes do not have to be horizontal and vertical. For a block on a 25° slope, it is easier to resolve its weight along and perpendicular to the slope: the component along the slope is W sin 25° and the component into the slope is W cos 25°.
Unit vectors
Unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point along x and y with magnitude 1. A force can be written \(\mathbf{F} = 433\mathbf{i} + 250\mathbf{j}\) N. Adding vectors is then just adding coefficients, which keeps bookkeeping tidy in larger problems.
Worked example
Two tugboats pull a ship: one with 40 kN along the ship's axis, the other with 30 kN at 90° to it. The resultant is the square root of (1,600 + 900) = 50 kN, directed at tan inverse (30/40), about 36.9°, from the ship's axis. To pull straight along the axis, the second tug would need to change its angle.
Key takeaways
A force F at angle theta has components F cos theta and F sin theta, with signs giving direction. Magnitude is the square root of the sum of squared components, and direction is tan inverse of Fy over Fx, checked by quadrant. Slope triangles such as 3-4-5 give components directly. Add forces by summing components, or use the parallelogram law and law of cosines. Choose axes, such as along a slope, that simplify the problem. Unit vectors i and j keep vector arithmetic organised.
Flashcard
F cos theta
Check your understanding
What are the components of a 500 N force at 30 degrees above the x-axis?