A fielder on the boundary at Eden Gardens throws the ball to the wicket-keeper at \(20\,\text{m/s}\), at \(37°\) above the ground. How long is the ball in the air, how high does it rise, and how far does it go? The ball moves forward and upward at the same time. Gravity changes only the upward motion. The forward motion carries on unchanged. To use this, the one velocity must be split into two.
Two parts at right angles
Draw a vector \(\vec A\) from the origin of the \(x\) and \(y\) axes, at an angle \(\theta\) above the \(x\) axis. Drop a perpendicular from its tip to each axis. The feet of the perpendiculars mark two vectors along the axes, and by the triangle law they add to \(\vec A\). Their signed lengths are the components of \(\vec A\). In the right triangle with hypotenuse \(A\), the side next to the angle is \(A\cos\theta\) and the side opposite it is \(A\sin\theta\).
The directions of the axes are written with unit vectors. \(\hat i\) is the unit vector along \(+x\), \(\hat j\) along \(+y\) and \(\hat k\) along \(+z\). Each has magnitude 1 and no unit. A vector is then its components times these directions: \(\vec A = A_x\hat i + A_y\hat j\) in a plane, and \(\vec A = A_x\hat i + A_y\hat j + A_z\hat k\) in space. For example, \(3\hat i + 4\hat j\) metres is \(3\,\text{m}\) along \(x\) and \(4\,\text{m}\) along \(y\).
A component is a signed number, not a magnitude. It is negative when the vector points towards the negative side of that axis. At \(150°\) the vector points up and to the left, so \(A_x\) is negative and \(A_y\) positive. The formula gives these signs by itself when \(\theta\) is measured anticlockwise from \(+x\), since \(\cos 150° = -0.866\).
Which component takes the cosine
The rule "\(A_x = A\cos\theta\)" holds only when the angle is measured from the \(x\) axis. Physics often measures from elsewhere. A string may be at \(60°\) to the vertical, or a slope at \(30°\) to the horizontal. The safe rule is simple. The component along the side that touches the angle takes the cosine, and the other component takes the sine. A string pulling with \(40\,\text{N}\) at \(60°\) to the vertical has a vertical component \(40\cos 60° = 20\,\text{N}\) and a horizontal component \(40\sin 60° = 34.6\,\text{N}\).
A string pulls a lamp with a tension of \(40\,\text{N}\), at \(60°\) to the vertical. What is the vertical component of the tension?
Back from components to magnitude and direction
The components and the magnitude belong to one right triangle, so the theorem of Pythagoras returns the magnitude. The derivation below shows it in three steps. The same idea works in space, where \(A^2 = A_x^2 + A_y^2 + A_z^2\). For \(\vec A = 2\hat i + 3\hat j + 6\hat k\), the magnitude is \(\sqrt{4 + 9 + 36} = 7\).
\[ A_x^2 + A_y^2 = A^2\cos^2\theta + A^2\sin^2\theta \]
Why. square the two components and add them.
\[ A_x^2 + A_y^2 = A^2(\cos^2\theta + \sin^2\theta) = A^2 \]
Why. take out the common factor \(A^2\); the bracket is 1 for every angle.
\[ A = \sqrt{A_x^2 + A_y^2}, \quad \tan\theta = \dfrac{A_y}{A_x} \]
Read it. a magnitude is never negative, so take the positive root. Dividing the two components cancels \(A\) and leaves the tangent of the angle.
The tangent needs one care. A calculator gives \(\tan^{-1}\) only between \(-90°\) and \(90°\). Take \(\vec A = -3\hat i + 3\hat j\). The calculator gives \(\tan^{-1}(3/(-3)) = -45°\). But the vector points up and to the left, into the second quarter, so its angle is \(180° - 45° = 135°\). Always read the quarter from the signs of the components.
Dividing a vector by its magnitude gives its unit vector. For \(\vec v = 3\hat i + 4\hat j\) m/s, \(v = 5\,\text{m/s}\) and \(\hat v = 0.6\hat i + 0.8\hat j\), which has magnitude 1 and no unit. Two vectors are equal only when every component is equal. So \(\vec A = \vec B\) in a plane means two equations, \(A_x = B_x\) and \(A_y = B_y\).
A displacement has components \(d_x = -3\,\text{m}\) and \(d_y = 3\,\text{m}\). What is its direction, measured anticlockwise from the \(+x\) axis?
Adding by components
Drawing is slow for three or more vectors, and it is never exact. Components turn addition into arithmetic. Resolve every vector, add all the \(x\) components, and add all the \(y\) components. The two sums are the components of the resultant. Then find its magnitude and direction as above.
Three forces act on a ring: \(6\,\text{N}\) along \(+x\), \(10\,\text{N}\) at \(53°\) above \(+x\), and \(3\,\text{N}\) along \(-y\). Take \(\cos 53° = 0.6\) and \(\sin 53° = 0.8\). Then \(R_x = 6 + 6 + 0 = 12\,\text{N}\) and \(R_y = 0 + 8 - 3 = 5\,\text{N}\). The resultant is \(\sqrt{144 + 25} = 13\,\text{N}\), at \(\tan^{-1}(5/12) = 23°\) above \(+x\).
The axes need not be horizontal and vertical. On an incline, choose one axis along the slope and one at right angles to it. The normal force and a string along the slope then lie on the axes, and only the weight needs resolving. The angle between the weight and the perpendicular to the slope equals the angle \(\theta\) of the slope. So the weight splits into \(W\sin\theta\) down the slope and \(W\cos\theta\) into the slope.
A block of mass \(2\,\text{kg}\) is held at rest on a smooth incline of \(30°\) by a string parallel to the slope. Take \(g = 10\,\text{m/s}^2\). Find the tension in the string, in N.
This is the worked example "A throw from the boundary" of Motion in Two Dimensions, lesson 4, done with the tool. A fielder throws the ball at \(20\,\text{m/s}\) at \(37°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find the time of flight, the maximum height and the range.
The angle is measured from the horizontal, so the horizontal component takes the cosine: \(u_x = 20 \times 0.8 = 16\,\text{m/s}\) and \(u_y = 20 \times 0.6 = 12\,\text{m/s}\).
Horizontally the velocity stays \(16\,\text{m/s}\). Vertically the ball is in free fall from \(12\,\text{m/s}\) upward, so \(T = 2u_y/g\), \(H = u_y^2/2g\) and \(R = u_xT\).
\(T = 24/10 = 2.4\,\text{s}\). \(H = 144/20 = 7.2\,\text{m}\). \(R = 16 \times 2.4 = 38.4\,\text{m}\).
The ball is in the air for \(2.4\,\text{s}\), rises \(7.2\,\text{m}\) and lands \(38.4\,\text{m}\) away, the same answers as the physics chapter.
A fielder runs from the point \((10, 5)\,\text{m}\) to the point \((22, 14)\,\text{m}\) of the ground. Find the magnitude of the displacement, in m.
Two errors are common. The first is the wrong axis. \(A_x = A\cos\theta\) holds only when \(\theta\) is measured from the \(x\) axis; measured from the \(y\) axis, the sine and the cosine swap. The second is the quarter of the plane. \(\tan^{-1}(A_y/A_x)\) cannot tell \(3\hat i + 3\hat j\) from \(-3\hat i - 3\hat j\), so read the direction from the signs. Remember also that "\(\sin 37° = 0.6\)" is a rounding. The exact angle of a 3, 4, 5 triangle is \(36.87°\).
Simon Stevin was an engineer in the Low Countries. In 1586 he published a book on the art of weighing. In it he draped a loop of fourteen equal balls over a triangular prism with one long slope and one short slope. If the loop could turn by itself, it would turn for ever, and he knew that no machine does that. So the balls on the long slope must balance the fewer balls on the short slope. From this he found the pull along a slope, which today we write as \(W\sin\theta\). He printed the loop on the title page of his book, with the words "A wonder, yet no wonder".
The throw from the boundary is now two simple motions. The forward component of \(16\,\text{m/s}\) never changes, because gravity has no forward component. The upward component of \(12\,\text{m/s}\) is a free fall that lasts \(2.4\,\text{s}\). One split of one vector turned a curved path into two straight lines.
Components are the most used tool of the book. Motion in Two Dimensions splits every projectile into a forward and an upward motion. Objects in Equilibrium, Accelerated Objects and both chapters on friction resolve weights and pulls along and across an incline. Circular motion splits an acceleration into tangential and radial parts, and a banked road splits the normal force. Momentum is conserved along each direction separately in collisions. Electrostatics adds the fields of charges by components. A charge in a magnetic field moves in a helix, because its velocity has parts along and across the field.
- Using \(A\cos\theta\) for the \(x\) component when the angle is measured from the vertical.The component along the line the angle is measured from takes the cosine. Draw the angle before you choose.
- Reading the direction from \(\tan^{-1}(A_y/A_x)\) without the signs.Place the vector in its quarter from the signs of \(A_x\) and \(A_y\), then fix the angle.
- Adding magnitudes of vectors instead of their components.Add the \(x\) components and the \(y\) components separately, then find one magnitude at the end.