Splitting an arrow in two
Tonight's idea
A vector is an arrow: a size and a direction (you met them in grade 10). A pull of 10 newtons (N, the unit of force) at 30° above the horizontal is a vector.
Any arrow is the sum of a horizontal arrow and a vertical arrow. Finding them is called resolving, and the two parts are the components. For an arrow of size \(F\) at angle \(\theta\), measured anticlockwise from the positive \(x\)-direction:
horizontal part \(= F\cos\theta\), vertical part \(= F\sin\theta\).
This is the unit circle scaled by \(F\), so it works for every angle. At 120° the horizontal part is negative: it points left.
Why bother? Arrows in different directions are awkward to add. Components are just numbers: horizontal parts add, vertical parts add, and the two never mix. So resolve each one, add the parts, then recombine: for total parts \(x\) and \(y\), the size is \(\sqrt{x^2 + y^2}\) and the angle satisfies \(\tan\theta = y/x\). Sketch the result to pick the right quadrant.
Watch one: two pulls on one ring
A ring is pulled by 10 N at 30° and by 6 N at 120°. Resolve both:
\[10\cos 30° = 8.66,\quad 10\sin 30° = 5.00; \qquad 6\cos 120° = -3.00,\quad 6\sin 120° = 5.20.\]
Add: horizontal \(8.66 - 3.00 = 5.66\), vertical \(5.00 + 5.20 = 10.20\). Recombine: size \(\sqrt{5.66^2 + 10.20^2} = 11.66\) N, at angle \(\tan^{-1}(10.20/5.66) = 61.0°\).
A check: 30° and 120° are at right angles, so Pythagoras gives the size directly, \(\sqrt{10^2 + 6^2} = \sqrt{136} = 11.66\). The two methods agree.
Your turn
- A rope pulls a sledge with 20 N at 35° above the horizontal. Find the horizontal and vertical components, to 2 decimal places. Which part actually moves the sledge forward?
- A vector has horizontal part \(-5\) and vertical part \(12\). Find its size and its angle, measured anticlockwise from the positive \(x\)-direction. (Sketch it first.)
- A boat steers at 12 km/h in the direction 60° (anticlockwise from east). A current pushes it due east at 5 km/h. Find the boat's actual speed and direction.
- A lamp weighing 40 N hangs at rest from two wires, each at 30° to the ceiling. At rest, all the forces add to zero. Each wire pulls with the same force \(T\). Find \(T\). Then find \(T\) if the wires are pulled almost flat, at 5° to the ceiling.
Stretch: tilt the axes
Components need two perpendicular directions, not necessarily horizontal and vertical. A 500 N crate sits on a ramp tilted at 20°. Resolve its weight, which points straight down, into a part along the ramp and a part pressing into the ramp. What force along the ramp holds the crate still, if there is no friction?
Answers
- Horizontal \(20\cos 35° = 16.38\) N, vertical \(20\sin 35° = 11.47\) N. Only the horizontal part moves the sledge forward; the vertical part lifts it slightly.
- Size \(\sqrt{25 + 144} = 13\). The arrow points left and up, in the second quadrant. \(\tan^{-1}(12/5) = 67.4°\), so the angle is \(180° - 67.4° = 112.6°\). (A calculator fed \(12/(-5)\) says \(-67.4°\), which is the opposite arrow.)
- Boat: \(12\cos 60° = 6\) east, \(12\sin 60° = 10.39\) north. Current: 5 east, 0 north. Total: 11 east, 10.39 north. Speed \(\sqrt{11^2 + 10.39^2} = 15.13\) km/h, direction \(\tan^{-1}(10.39/11) = 43.4°\) anticlockwise from east.
- The horizontal parts of the two tensions are equal and opposite, so they cancel. The vertical parts must hold up the weight: \(2T\sin 30° = 40\), so \(T = 40\) N. At 5°: \(T = 20/\sin 5° = 229.5\) N. A nearly flat wire has almost no vertical part, so it must pull very hard. This is why a washing line can never be pulled perfectly straight.
Stretch. The weight makes an angle of 20° with the direction straight into the ramp. So the part pressing into the ramp is \(500\cos 20° = 469.8\) N and the part along the ramp is \(500\sin 20° = 171.0\) N, pointing downhill. A force of 171.0 N up the ramp holds the crate. Check: \(469.8^2 + 171.0^2 = 500^2\).