Two dimensional motion is motion in a plane. Specification of the motion variables requires a pair of numbers. The kind of algebra used to operate variables represented by a pair of numbers is called vector algebra.
Physical quantities are classified into vectors and scalars based on how they are represented. Scalars are physical quantities that can be specified by a number and a unit completely. For example when we say the length of an object i 5 m, the number 5 and the unit m specify the length of the object completely. Therefore, length is a scalar quantity. Other examples of scalar physical quantities are mass, time, temperature and volume. Vectors are physical quantities that require the specification of direction in addition to a number and a unit. For example velocity is a vector quantity because, in addition to how fast an object is going, we also need to know in what direction it is going. Other examples of vectors are displacement, acceleration, force and area.
In text books, a vector is symbolically represented by a letter in bold face while in exercise books, it is represented by a capital letter with an arrow on top. Graphically, a vector is represented by means of an arrow. The direction of the vector is represented by the direction of the arrow. The magnitude (numerical value) of the vector is represented by the length of the arrow. The arrow is drawn in such a way that its length is proportional to the magnitude of the vector. This is done by specifying a scale and drawing the length according to the scale.
The negative of a vector is the vector with the same magnitude but opposite direction. For example if A = 4 m east, then -A = 4 m west. A vector can be multiplied by a number. The effect of multiplying a vector by a number is to multiply the magnitude of the vector. If the number is positive the direction remains the same. If the number is negative the direction becomes opposite. For example if A = 2 m north, then 4A = 8 m north and -4A = 8 m south.
The sum of two or more vectors (sometimes called the resultant) is the single vector with the same effect. For example, suppose a particle is displaced 4 m east from point A to point B. And then again displaced 4 m north from point B to point C. This can be accomplished by a single vector displacing the particle from point A to point C directly in a north east direction. The later is the sum of the two vectors.
To add vectors graphically, first connect all the vectors tail to head. Then the sum vector is the vector whose tail is the tail of the first vector and whose head is the head of the last vector. The magnitude of the sum vector is obtained by measuring the length and multiplying by the scale. The direction is obtained by measuring the angle between the sum vector and the positive x-axis by means of a protractor.
Subtracting vector B from vector A means adding the negative of B to A. That is A - B = A + (-B). To subtract vector B from A graphically, first find the negative of vector B. Then add vector A and the negative of vector B using the rules of addition.
Example: Determine the magnitude and the direction of the following vectors.
300 m/s west
Solution: Its magnitude is 300 m/s and its direction as measured from positive x-axis (east) is 180°.
400 m 30° west of north
Solution: The magnitude is 400 m. Since the angle as measured from north to the west is 30° and and the angle for north with respect to east is 90°, The angle for this vector must be 120°
60 m 40° south of east.
Solution: The magnitude is 60 m. This vector makes an angle of 320° with the positive x-axis when measured in a counterclockwise direction or 40° when measured in a clockwise direction. Thus, this angle can be represented either as 320° or -40°.
7 m/s² at 200°
Solution: The magnitude is 7 m/s². The default reference line is east. So, if no reference line is specified, the reference line is east. Thus, the angle for this vector is 200°
The Cartesian coordinates of a vector are its projection on the x-axis (horizontal line) and its projection on the y-axis (vertical line). The projection on the x-axis is called the x-component or the horizontal component of the vector. The projection on the y-axis is called the y-component or the vertical component of the vector. Components can be positive or negative. A horizontal component is taken to be positive if it is to the right and negative f it is to the left. A vertical component is taken to be positive if its direction is up and negative if its direction is down. For example, for a displacement vector, 5 m 37° north of east, the projections on the x-axis and y-axis (as can be shown by dropping the perpendiculars on the x-axis and the y-axis) are 4m and 3 m respectively.
The magnitude of a vector will be represented by the symbol of the vector in italics. For example, the magnitude of the vector A will be represented by A. The direction of a vector will be represented by θ. The x-component (y-component) of a vector will be represented by the symbol of the vector with subscript x (y) in italics. For example, the x- and y- components of the vector A will be represented by Ax and Ay respectively.
For simplicity, consider a vector A on the first quadrant (whose angle is less than 90 °). The vector and its horizontal and vertical components form a right angled triangle. The hypotenuse of this triangle is equal to the magnitude of the vector, A. The horizontal component, Ax, is adjacent to the angle of the vector, θ. The vertical component, Ay, is opposite to the angle of the vector, θ.
The definitions of cosine and sine can be used to obtain the following expressions for the horizontal and vertical components of the vector in terms of the magnitude and direction of the vector:
Ax = A cos θ
Ay = A sin θ
Example: Find the horizontal and vertical components of the following vectors.
Solution: A = 100 m; θ = 53° Ax = ?; Ay = ?.
Ax = A cos θ = 100 m cos 53° = 60 m
Ay = A sin θ = 100 m sin 53° = 80 m
A = 10 m 30° south of west.
Solution: A = 10 m; θ = 180° + 30° = 210°; Ax = ?; Ay = ?
Ax = A cos θ = 10 m cos 210° = -5 m
Ay = A sin θ = 10 m sin 210° ≈ -8.7 m.
Applying Pythagorean theorem to the right angled triangle considered earlier, the following equation for the magnitude can be obtained in terms of the components of the vector.
A = √(Ax² + Ay²)
The angle θ can be related to the component with the help of the trigonometric function tangent: tan θ = Ay/Ax. And an expression for θ can be obtained by applying arctan to both sides of this equation.
θ = arctan (Ay/Ax)
The calculator will give only values between -90° and 90° because the period of tangent is 180° and not 360°. And thus, if Ax is negative 180° should be added to the angle obtained from the calculator. If Ax is zero, the calculator will give an error because division by zero is not allowed. If Ax is zero and Ay is positive, the value of θ is 90°. If Ax is zero and Ay is negative the value of θ is -90° or 270°.
Example: Calculate the magnitude and direction of the vector whose horizontal and vertical components (respectively) are.
3 m and 4 m
Solution: Ax = 3 m; Ay = 4 m; A = ?; θ = ?.
A = √(Ax² + Ay²) = √(3² + 4²) m = 5 m.
θ = arctan (Ay/Ax) = arctan (4/3) = 53°
-3 m and 4 m
Solution: Ax = -3 m; Ay = 4 m; A = ?; θ = ?.
A = √(Ax² + Ay²) = √{(-3)² + (-4)²} m = 5 m
θ = arctan (Ay/Ax) + 180° = arctan {(4)/(-3)} + 180° = -53° + 180° = 127°
180° is added because Ax is negative.
If
R = A + B
Then
Rx = Ax + Bx
Ry = Ay + By
The magnitude and direction of the sum vector can be obtained from these components.
R = √(Rx² + Ry²) = √{(Ax + Bx)² + (Ay + By)²}
θ = arctan (Ry/Rx) = arctan {(Ay + By)/(Ax + Bx)}
Example: Given the vectors A = 100 m 37° north of east and B = 100 m 53° south of west.
Calculate the horizontal and vertical components of their sum vector
Solution: A = 100 m; θA = 37°; B = 100 m; θB = 53° + 180° = 233° (180° is added to find the angle with respect to the positive x-axis); Rx = Ax + Bx = ?; Ry = Ay + By = ?
Ax = A cos θA = 100 m cos 37° = 80 m
Ay = A sin θA = 100 m sin 37° = 60 m
Bx = B cos θB = 100 m cos 233° = -60 m
By = B sin θB = 100 m sin 233° = -80 m
Rx = Ax + Bx = (80 - 60) m = 20 m
Ry = Ay + By = (60 - 80) m = -20 m
Calculate the magnitude and the direction of the sum vector.
Solution: Rx = 20 m; Ry = -20 m; R = ?; θ = ?
R = √(Rx² + Ry²) = √{20² + (-20)²} m ≈ 28 m
θ = arctan (Ry/Rx) = arctan (-20/20) = -45°