Vectors and Forces

Vector Components

splitting a force into across and up

A force along a slope splits into a horizontal part and a vertical part using a 3-4-5 triangle, with no trigonometry.

Example

A force pointing along a slope can be split into a horizontal part and a vertical part using a 3-4-5 triangle, no trigonometry needed. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A pull along a slope

A rope pulls with a force of 5 newtons along a slope that rises 3 for every 4 across. We want the across part and the up part separately.

F=5 N,slope 3:4F = 5\ \text{N}, \quad \text{slope } 3:4
A force along a rising slopeA single force arrow pointing up and to the right along a steadily rising slope.OF

The 3-4-5 triangle

The slope 3 up and 4 across has length 5, because 3 and 4 are the legs of a right triangle, so the across fraction is 4 over 5 and the up fraction is 3 over 5.

32+42=9+16=25=523^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}

The across part

Multiply the force by the across fraction: 5 times 4 over 5 is 4 newtons.

Fx=5 N⋅45=4 NF_x = 5\ \text{N} \cdot \frac{4}{5} = \hl{4}\ \text{N}

The up part

Multiply the force by the up fraction: 5 times 3 over 5 is 3 newtons.

Fy=5 N⋅35=3 NF_y = 5\ \text{N} \cdot \frac{3}{5} = \hl{3}\ \text{N}
The force split into across and up partsThe original force arrow with its horizontal across part and vertical up part drawn from the same point.OFFxFy

Same direction, larger force, larger parts

Keep the same 3 up for 4 across direction. Bigger force makes both component parts grow in the same ratio.

FFxFy5 N4 N3 N10 N8 N6 N15 N12 N9 N\begin{array}{c|c|c}F & F_x & F_y \\ \hline 5\ \text{N} & 4\ \text{N} & 3\ \text{N} \\ 10\ \text{N} & 8\ \text{N} & 6\ \text{N} \\ 15\ \text{N} & 12\ \text{N} & 9\ \text{N}\end{array}
mechanics A 3-4-5 slope keeps the components exact whole numbers, so the across and up parts can be read off directly.