Precalculus · N.VM.A.3

Forces and velocities as vectors

A plane flies through air that is itself moving, and several ropes pull on one load at once. What single motion, or single pull, do they add up to?

You need

Adding and subtracting vectorsPrecalculus Magnitude and direction of a vectorPrecalculus

Where you are 3 in Vector addition and scalar multiplication

Trigonometry7

Polar coordinates and vectors10

Vector addition and scalar multiplication3

Adding and subtracting vectors

Forces and velocities as vectors

Scaling a vector

When you reach for it

Some quantities have a size and a direction. A velocity is a speed together with a direction of motion, and a force is a push or a pull of a certain strength in a certain direction. Both are vectors, and when two or more of them act together their combined effect is their sum, found exactly as in adding vectors.

Three friends meet the idea in three ways. Aigul flies a seaplane between lakes; her instruments show which way the nose points and how fast the plane moves through the air, but the air itself is moving. Baurzhan takes a motorboat across a river whose water is flowing downstream. Dauren rigs ropes and cables: several pulls on one ring, a chandelier hung from two cables, a cart held on a ramp. In each case the question is the same: what single vector has the effect of all of them together, and, when something stays put, what must the forces be for that sum to be zero?

The procedure

Velocities add

A plane’s air velocity \mathbf{a} is its motion through the air: its magnitude is the airspeed and its direction is the heading, the way the nose points. The wind \mathbf{w} is the motion of the air over the ground. The plane’s motion over the ground is

\mathbf{g} = \mathbf{a} + \mathbf{w}

whose magnitude is the ground speed and whose direction is the true course. The reason is the triangle rule. In one hour the plane moves \mathbf{a} through the air while the air carries everything in it by \mathbf{w}, so the plane ends up \mathbf{a} followed by \mathbf{w} from where it started, tip to tail. A boat works the same way: its velocity through the water plus the current’s velocity gives its velocity over the riverbed.

Bearings in, bearings out

Navigators give directions as bearings, measured clockwise from north, as in navigation with bearings. With x pointing east and y pointing north, a vector of magnitude d on bearing \theta is

\langle d\sin\theta,\ d\cos\theta \rangle

To turn a sum back into a bearing, find its magnitude and its direction angle \alpha (counterclockwise from east) as in magnitude and direction, then use \theta = 90^\circ - \alpha, adding 360^\circ if that is negative.

A wind is named by where it comes from: a north wind blows from north toward south. A wind from bearing 300^\circ therefore blows toward 300^\circ - 180^\circ = 120^\circ, and that second bearing is the direction of \mathbf{w}.

Forces add

Forces acting on one object combine by vector addition. Their sum

\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \cdots + \mathbf{F}_n

is the resultant, the single force with the same effect as all of them. Resolve each force into \langle \|\mathbf{F}\|\cos\alpha,\ \|\mathbf{F}\|\sin\alpha \rangle, add the components, and convert back if a size and direction are wanted.

The object is in equilibrium when the forces on it add to the zero vector. The force \mathbf{E} = -\mathbf{R}, the equilibrant, is the one extra force that brings a set of forces into equilibrium: \mathbf{R} + (-\mathbf{R}) = \mathbf{0}. It has the same magnitude as \mathbf{R} and the opposite direction, so its direction angle is \alpha + 180^\circ (or \alpha - 180^\circ).

At rest: two equations

For an object held at rest, the forces on it are in equilibrium. A vector is zero exactly when both of its components are zero, so one vector equation becomes two number equations:

\begin{aligned} \text{sum of the } x\text{-components} &= 0 \\ \text{sum of the } y\text{-components} &= 0 \end{aligned}

When every force’s direction is known, these two equations can be solved for two unknown magnitudes. The physical facts in each problem (a weight pulls straight down, a cable pulls along itself, a smooth ramp pushes straight out from its surface) are stated as given; the work here is the vector algebra. Forces are measured in newtons, written \text{N}.

Worked examples

Example 1

Ground speed and true course

Aigul’s seaplane has heading 050^\circ and airspeed 160 miles per hour. The wind blows from 300^\circ at 30 miles per hour. Find her ground speed and true course, to the nearest tenth.

Solution

The wind comes from 300^\circ, so it blows toward 120^\circ. Resolve both velocities with \langle d\sin\theta, d\cos\theta \rangle:

\begin{aligned} \mathbf{a} &= \langle 160\sin 50^\circ,\ 160\cos 50^\circ \rangle \\ &\approx \langle 122.57, 102.85 \rangle \\ \mathbf{w} &= \langle 30\sin 120^\circ,\ 30\cos 120^\circ \rangle \\ &\approx \langle 25.98, -15.00 \rangle \end{aligned}

Adding components (Figure 1),

\mathbf{g} = \mathbf{a} + \mathbf{w} \approx \langle 148.55, 87.85 \rangle

Figure 1Example 1. The air velocity \mathbf{a} and the wind \mathbf{w} placed tip to tail. Their sum \mathbf{g} is the motion over the ground.

Ground speed: \displaystyle \|\mathbf{g}\| \approx \sqrt{148.55^2 + 87.85^2} \approx 172.6 miles per hour. Both components are positive, so \displaystyle \alpha = \arctan\frac{87.85}{148.55} \approx 30.6^\circ (inverse tangent), and the true course is 90^\circ - 30.6^\circ = 59.4^\circ, that is, 059.4^\circ. The wind turned her 9.4^\circ clockwise from her heading, and its eastward part sped her up.

Check with the triangle in Figure 1. At the head of \mathbf{a} the angle between -\mathbf{a} and \mathbf{w} is 180^\circ - (120^\circ - 50^\circ) = 110^\circ, so the law of cosines gives \|\mathbf{g}\|^2 = 160^2 + 30^2 - 2(160)(30)\cos 110^\circ, which is about 29{,}783.4, and \displaystyle \sqrt{29{,}783.4} \approx 172.6.

Example 2

Crossing a river

Baurzhan’s motorboat moves at 2.5 meters per second through still water. A river 120 meters wide, with straight parallel banks, flows at 1.5 meters per second. (a) He points the boat straight across. How long does the crossing take, where does he land, and what is his actual speed? (b) Which way must he point to land directly opposite his start, and how long does that crossing take?

Solution

Put x downstream and y straight across. The current is \mathbf{c} = \langle 1.5, 0 \rangle.

(a) The boat’s velocity through the water is \mathbf{b} = \langle 0, 2.5 \rangle, so over the riverbed he moves at \mathbf{b} + \mathbf{c} = \langle 1.5, 2.5 \rangle. Only the y-component carries him across, so the crossing takes \displaystyle \frac{120}{2.5} = 48 seconds. In those 48 seconds the x-component carries him 1.5 \cdot 48 = 72 meters downstream (Figure 2). His actual speed is \displaystyle \sqrt{1.5^2 + 2.5^2} = \sqrt{8.5} \approx 2.92 meters per second, along a line \displaystyle \arctan\frac{1.5}{2.5} \approx 31.0^\circ downstream of straight across.

Figure 2Example 2(a). Pointed straight across, the boat is carried 72 meters downstream of C while it crosses.

(b) Point the boat upstream at an angle \varphi from straight across, so \mathbf{b} = \langle -2.5\sin\varphi,\ 2.5\cos\varphi \rangle. To go straight across, the downstream component of \mathbf{b} + \mathbf{c} must be zero:

-2.5\sin\varphi + 1.5 = 0, \qquad \sin\varphi = 0.6

so \varphi = \arcsin 0.6 \approx 36.9^\circ upstream of straight across. Since \varphi is acute, \displaystyle \cos\varphi = \sqrt{1 - 0.36} = 0.8, the across speed is 2.5 \cdot 0.8 = 2 meters per second, and the crossing takes \displaystyle \frac{120}{2} = 60 seconds. Going straight costs 12 seconds, because part of the boat’s speed is spent canceling the current.

Example 3

Three pulls on one ring

In a three-way tug-of-war, three ropes are tied to one steel ring. With x east and y north, Aigul pulls with 200 N at direction angle 20^\circ, Baurzhan with 150 N at 110^\circ, and Dauren with 120 N at 230^\circ. Find the resultant, and the fourth pull that would hold the ring in equilibrium.

Solution

Resolve each pull with \langle \|\mathbf{F}\|\cos\alpha, \|\mathbf{F}\|\sin\alpha \rangle:

\begin{aligned} \mathbf{F}_1 &\approx \langle 187.94, 68.40 \rangle \\ \mathbf{F}_2 &\approx \langle -51.30, 140.95 \rangle \\ \mathbf{F}_3 &\approx \langle -77.13, -91.93 \rangle \end{aligned}

Adding, \mathbf{R} \approx \langle 59.50, 117.43 \rangle. Its magnitude is \displaystyle \sqrt{59.50^2 + 117.43^2} \approx 131.6 N, and both components are positive, so its direction angle is \displaystyle \arctan\frac{117.43}{59.50} \approx 63.1^\circ. The ring is pulled mostly northward, in a direction between Aigul’s rope and Baurzhan’s.

The equilibrant is -\mathbf{R} \approx \langle -59.50, -117.43 \rangle: 131.6 N at 63.1^\circ + 180^\circ = 243.1^\circ. Added to the three pulls it gives the zero vector.

Example 4

A chandelier on two cables

Dauren hangs a chandelier of weight 500 N from a knot held by two cables. The left cable makes 30^\circ with the horizontal and the right cable 45^\circ (Figure 3). Each cable pulls on the knot along itself, away from the knot. Find the tension (the size of the pull) in each cable.

Solution

Figure 3Example 4. The three forces on the knot. The knot is at rest, so they add to the zero vector.

Let T_1 and T_2 be the unknown tensions. The left cable points at direction angle 180^\circ - 30^\circ = 150^\circ and the right one at 45^\circ:

\begin{aligned} \mathbf{T}_1 &= \left\langle -\tfrac{\sqrt{3}}{2}T_1,\ \tfrac{1}{2}T_1 \right\rangle \\ \mathbf{T}_2 &= \left\langle \tfrac{\sqrt{2}}{2}T_2,\ \tfrac{\sqrt{2}}{2}T_2 \right\rangle \\ \mathbf{W} &= \langle 0, -500 \rangle \end{aligned}

The knot is at rest, so both component sums are zero:

\begin{aligned} -\tfrac{\sqrt{3}}{2}T_1 + \tfrac{\sqrt{2}}{2}T_2 &= 0 \\ \tfrac{1}{2}T_1 + \tfrac{\sqrt{2}}{2}T_2 &= 500 \end{aligned}

The first equation says \displaystyle \frac{\sqrt{2}}{2}T_2 = \frac{\sqrt{3}}{2}T_1. Substituting into the second, \displaystyle \frac{1 + \sqrt{3}}{2}T_1 = 500, so

T_1 = \frac{1000}{1 + \sqrt{3}} = 500(\sqrt{3} - 1) \approx 366.0 \text{ N}

after rationalizing the denominator. Then \displaystyle T_2 = \frac{\sqrt{3}}{\sqrt{2}}T_1 = 250(3\sqrt{2} - \sqrt{6}) \approx 448.3 N. The steeper cable carries more. The two tensions add to about 814 N, more than the weight, because the cables also pull sideways against each other.

Check. Upward parts: 0.5(366.0) + 0.7071(448.3) \approx 183.0 + 317.0 = 500.

Example 5

A cart held on a ramp

Dauren holds a loaded cart of weight 800 N at rest on a smooth ramp tilted 25^\circ, using a rope parallel to the ramp. The ramp pushes on the cart straight out from its surface. Find the tension in the rope and the size of the ramp’s push.

Solution

With x horizontal and y up, the rope pulls up the ramp at direction angle 25^\circ, and the ramp’s push is perpendicular to the ramp, at 25^\circ + 90^\circ = 115^\circ. Call their sizes T and P. Since \cos 115^\circ = -\sin 25^\circ and \sin 115^\circ = \cos 25^\circ, the equilibrium equations are

\begin{aligned} T\cos 25^\circ - P\sin 25^\circ &= 0 \\ T\sin 25^\circ + P\cos 25^\circ - 800 &= 0 \end{aligned}

The first gives T = P\tan 25^\circ. Substituting into the second,

P\,\frac{\sin^2 25^\circ + \cos^2 25^\circ}{\cos 25^\circ} = 800

and the Pythagorean identity turns the numerator into 1. So P = 800\cos 25^\circ \approx 725.0 N and T = P\tan 25^\circ = 800\sin 25^\circ \approx 338.1 N. The same algebra with any tilt \beta gives T = 800\sin\beta: nothing on a level floor, the whole weight on a vertical wall.

Example 6

The heading for a wanted course

Aigul wants a true course of due east, 090^\circ, at airspeed 150 miles per hour, with the wind blowing from 210^\circ at 25 miles per hour. Find the heading she must fly and her ground speed, to the nearest tenth.

Solution

The wind blows toward 030^\circ, so

\begin{aligned} \mathbf{w} &= \langle 25\sin 30^\circ,\ 25\cos 30^\circ \rangle \\ &= \langle 12.5,\ 12.5\sqrt{3} \rangle \approx \langle 12.5, 21.65 \rangle \end{aligned}

The air velocity \mathbf{a} = \langle p, q \rangle is unknown, with p^2 + q^2 = 150^2. The sum \mathbf{a} + \mathbf{w} must point due east: its north component is zero and its east component positive. So \displaystyle q = -12.5\sqrt{3} \approx -21.65, and

p^2 = 22{,}500 - 468.75 = 22{,}031.25

The negative root would make p + 12.5 negative, a westward track, so \displaystyle p = \sqrt{22{,}031.25} \approx 148.43. The ground speed is p + 12.5 \approx 160.9 miles per hour.

For the heading, \mathbf{a} \approx \langle 148.43, -21.65 \rangle points into quadrant IV, so \displaystyle \alpha = \arctan\frac{-21.65}{148.43} \approx -8.3^\circ and \theta = 90^\circ - (-8.3^\circ) = 98.3^\circ. She flies heading 098.3^\circ, nose 8.3^\circ south of east, against the wind’s northward push. Her ground speed beats her airspeed because the wind also has an eastward part.

Boundaries

Only like quantities add. Velocities add to velocities, and forces add to forces acting on the same object. A force and a velocity never add.

One direction convention. Bearings, direction angles and “from” winds must all be turned into one convention before resolving. A wind’s bearing names where it comes from; its vector points the other way.

Steady wind and current. These models treat the wind or current as the same everywhere along the trip and constant in time. When it is not (a river usually runs faster in the middle), the answers are estimates.

Straight across needs the faster boat. \displaystyle \sin\varphi = \frac{\|\mathbf{c}\|}{\|\mathbf{b}\|} has a solution only when \|\mathbf{c}\| \le \|\mathbf{b}\|, and an actual crossing only when \|\mathbf{c}\| < \|\mathbf{b}\|.

Two equations, two unknowns. Equilibrium in the plane gives two component equations, so it settles at most two unknown magnitudes. A cable under a hanging weight can never be exactly horizontal: it would have no upward component to hold any weight.

This is the plane. Forces in space have three components and give three equations; see vectors in space.

Where this goes next

How much of a force acts along a given direction (along a ramp, along a road) is the scalar projection, computed with the dot product. The same tools measure the work a force does along a displacement. The angle juggling of the river and the wind correction returns whenever a motion must be aimed at a moving target.

Exercises

Planes and wind

Give speeds and bearings to the nearest tenth.

A plane has heading 000^\circ and airspeed 200 miles per hour. A west wind (from 270^\circ) blows at 40 miles per hour. Find the ground speed and true course.
A plane has heading 135^\circ and airspeed 240 miles per hour, with a 35 mile per hour wind from 270^\circ. Find the ground speed and true course.
Aigul flies heading 300^\circ at airspeed 120 miles per hour, with a 20 mile per hour wind from 045^\circ. Find the ground speed and true course.
A plane flies heading 080^\circ at airspeed 210 miles per hour, but its ground velocity is 195 miles per hour on course 072^\circ. Find the wind’s speed and the bearing it blows from.

Crossing a river

Baurzhan points a boat that moves 3 meters per second through still water straight across a river 90 meters wide, flowing at 1.2 meters per second. Find the crossing time, how far downstream he lands, and his actual speed.
For the boat and river in Exercise 5, find the angle upstream of straight across that lands him directly opposite, and the crossing time.
A boat moves 5 meters per second through still water on a river 200 meters wide flowing at 4 meters per second. Find the heading, measured upstream from straight across, that crosses directly, the across speed, and the time.

Resultants and equilibrants

Forces of 60 N at direction angle 0^\circ and 80 N at 90^\circ act on one point. Find the resultant’s magnitude and direction angle.
Forces of 300 N at 15^\circ and 250 N at 75^\circ act on one point. Find the resultant by components, and check its magnitude with the law of cosines.
Forces of 40 N at 0^\circ, 50 N at 120^\circ and 30 N at 250^\circ act on one ring. Find the resultant and the equilibrant, each as a magnitude and direction angle.
Find the equilibrant of \langle 35, -12 \rangle N and \langle -20, 48 \rangle N, in components and as a magnitude and direction angle.

Held at rest

A gong of weight 120 N hangs from two cables, each at 30^\circ to the horizontal on opposite sides. Find the tension in each.
A weight of 200 N hangs from a knot held by two cables, the left at 40^\circ to the horizontal and the right at 60^\circ. Find both tensions, to the nearest tenth.
A rope parallel to a smooth ramp tilted 18^\circ holds a cart of weight 600 N at rest. Find the tension and the ramp’s push, to the nearest tenth.
A rope parallel to a smooth ramp holds a cart of weight 400 N at rest with a tension of 150 N. Find the ramp’s tilt and the ramp’s push, to the nearest tenth.

Reasoning

Explain why a boat slower than the current cannot reach the point directly opposite, whatever its heading.
Explain why, when the boat points straight across, the crossing time does not depend on the current.
A weight W hangs from two cables, each at angle \beta above the horizontal on opposite sides. Show that each tension is \displaystyle \frac{W}{2\sin\beta}, and explain what happens as \beta shrinks toward 0^\circ.
Use the triangle inequality to explain why the ground speed always lies between \|\mathbf{a}\| - \|\mathbf{w}\| and \|\mathbf{a}\| + \|\mathbf{w}\| when the airspeed exceeds the wind speed, and say when each end is reached.
Explain why the equilibrant of a set of forces has the same magnitude as their resultant and the opposite direction, and why no other single force brings them into equilibrium.

Answers

204.0 miles per hour on 011.3^\circ. \mathbf{g} = \langle 0, 200 \rangle + \langle 40, 0 \rangle = \langle 40, 200 \rangle; \displaystyle \sqrt{41{,}600} \approx 203.96; \alpha = \arctan 5 \approx 78.7^\circ, so \theta \approx 11.3^\circ.
265.9 miles per hour on 129.7^\circ. The wind blows toward 090^\circ: \mathbf{g} \approx \langle 169.71, -169.71 \rangle + \langle 35, 0 \rangle, which is \langle 204.71, -169.71 \rangle; \alpha \approx -39.66^\circ, so \theta \approx 129.7^\circ.
126.7 miles per hour on 291.2^\circ. The wind blows toward 225^\circ: \mathbf{g} \approx \langle -103.92, 60 \rangle + \langle -14.14, -14.14 \rangle, which is \langle -118.07, 45.86 \rangle; \alpha \approx 158.8^\circ, so \theta \approx -68.8^\circ + 360^\circ.
About 32.0 miles per hour, from 138.1^\circ. \mathbf{w} = \mathbf{g} - \mathbf{a}, and \langle 185.46, 60.26 \rangle - \langle 206.81, 36.47 \rangle is \langle -21.35, 23.79 \rangle, which points toward 318.1^\circ, so it comes from 318.1^\circ - 180^\circ.
30 seconds, 36 meters downstream, about 3.23 meters per second. \displaystyle \frac{90}{3} = 30; 1.2 \cdot 30 = 36; \displaystyle \sqrt{9 + 1.44} = \sqrt{10.44}.
About 23.6^\circ upstream, about 32.7 seconds. \displaystyle \sin\varphi = \frac{1.2}{3} = 0.4; across speed \displaystyle 3\cos\varphi = \sqrt{9 - 1.44} \approx 2.750; \displaystyle \frac{90}{2.750}.
About 53.1^\circ upstream, 3 meters per second, about 66.7 seconds. \sin\varphi = 0.8, so \cos\varphi = 0.6 and 5(0.6) = 3; \displaystyle \frac{200}{3}.
100 N at about 53.1^\circ. \langle 60, 80 \rangle; \displaystyle \sqrt{3600 + 6400} = 100; \displaystyle \arctan\frac{4}{3}.
About 477.0 N at 42.0^\circ. \langle 289.78, 77.65 \rangle + \langle 64.70, 241.48 \rangle is \langle 354.48, 319.13 \rangle. The forces are 60^\circ apart, so the triangle’s angle opposite \mathbf{R} is 120^\circ, and 300^2 + 250^2 - 2(300)(250)\cos 120^\circ is 227{,}500; and \displaystyle \sqrt{227{,}500} \approx 476.97.
Resultant about 15.8 N at 72.6^\circ; equilibrant about 15.8 N at 252.6^\circ. \langle 40, 0 \rangle + \langle -25, 43.30 \rangle + \langle -10.26, -28.19 \rangle is \langle 4.74, 15.11 \rangle.
\langle -15, -36 \rangle N, which is 39 N at about 247.4^\circ. The resultant is \langle 15, 36 \rangle; \displaystyle \sqrt{225 + 1296} = \sqrt{1521} = 39; \displaystyle \arctan\frac{36}{15} + 180^\circ \approx 67.4^\circ + 180^\circ.
120 N each. By Exercise 18, \displaystyle T = \frac{120}{2\sin 30^\circ} = \frac{120}{1}.
About 101.5 N (left) and 155.6 N (right). x: T_1\cos 40^\circ = T_2\cos 60^\circ, so T_2 \approx 1.532\,T_1; y: T_1\sin 40^\circ + T_2\sin 60^\circ = 200 gives T_1(0.6428 + 1.3268) = 200.
Tension about 185.4 N, push about 570.6 N. As in Example 5, T = 600\sin 18^\circ and P = 600\cos 18^\circ.
Tilt about 22.0^\circ, push about 370.8 N. T = 400\sin\beta gives \sin\beta = 0.375; \displaystyle P = 400\cos\beta = \sqrt{400^2 - 150^2} = \sqrt{137{,}500}.
Pointing at angle \varphi upstream, the downstream component of the boat’s motion is \|\mathbf{c}\| - \|\mathbf{b}\|\sin\varphi. Since \sin\varphi \le 1, this is at least \|\mathbf{c}\| - \|\mathbf{b}\| > 0, so it is never zero: every heading drifts downstream.
Pointed straight across, \mathbf{b} is all across-the-river and \mathbf{c} has no across component, so the across component of \mathbf{b} + \mathbf{c} is \|\mathbf{b}\|. The time is the width divided by \|\mathbf{b}\|; the current only changes where he lands.
The x-equation is T_2\cos\beta - T_1\cos\beta = 0, and \cos\beta \ne 0, so T_1 = T_2 = T. The y-equation is 2T\sin\beta = W, so \displaystyle T = \frac{W}{2\sin\beta}. As \beta shrinks, \sin\beta shrinks toward 0 and the tension grows without bound; at \beta = 0^\circ there is no solution at all.
\mathbf{g} = \mathbf{a} + \mathbf{w}, and the triangle inequality gives \|\mathbf{a}\| - \|\mathbf{w}\| \le \|\mathbf{a} + \mathbf{w}\| and \|\mathbf{a} + \mathbf{w}\| \le \|\mathbf{a}\| + \|\mathbf{w}\|. The top is reached with a tailwind (wind pointing the same way as the heading), the bottom with a headwind (pointing exactly opposite).
\mathbf{R} + (-\mathbf{R}) = \mathbf{0}, so -\mathbf{R} brings the forces into equilibrium; \|-\mathbf{R}\| = \|\mathbf{R}\| and -\mathbf{R} points the opposite way (a multiple by -1). If \mathbf{R} + \mathbf{X} = \mathbf{0} for some force \mathbf{X}, adding -\mathbf{R} to both sides gives \mathbf{X} = -\mathbf{R}, so it is the only one.