VECTORS IN MECHANICS

Magnitude and Direction If a vector, a, is given in component form e.g. a = 4i + 3j, we can use Pythagoras’ theorem to find the magnitude, and basic trigonometry to find the direction. 4 3 a θ ~ a Note, when printed, a vector is written in bold print. When handwritten, vectors are underlined: The magnitude of a is written as either a or just: a It would be handwritten as: a ~ or a Here we have: = a = 5 3 4 The direction to the horizontal is θ, where tan θ = θ = 36.9º

Example 1: Given the vectors a = 3i + j, and b = 6i – 7j, find the magnitude and direction of i) 2a + b ii) a – b. Give the directions as three figure bearings. Note, the vector pi + qj can be written as a column vector a = b = 2 + = i) 2a + b = 12 5 θ 2a + b = = 13 5 12 tan θ = θ = 22.6º The bearing is 90 + 22.6 = 113º ( to the nearest degree).

(to the nearest degree). ii) a = b = i j θ 3 8 a – b = a – b = = 8.54 (3 s.f.) 8 3 tan θ = θ = 69.4º The bearing is 270 + 69.4 = 339º (to the nearest degree).

Finding the components of a vector: j 40º 25 Consider a vector of magnitude 25 which makes an angle of 40º to the horizontal. adjacent 25 The horizontal component can be found: cos 40 = adjacent = 25 cos 40 opposite 25 The vertical component can be found: sin 40 = opposite = 25 sin 40 = Hence the vector is: (or 19.2i + 16.1j).

Example 2: A vector of magnitude 17 is shown. Find the vector in component form. i j 17 23º 17 23º adjacent 17 cos 23 = adjacent = 17 cos 23 Note the negative for both of the components. opposite 17 sin 23 = opposite = 17 sin 23 = Hence the vector is:

Example 3: A vector of magnitude 35 is shown. Find the vector in component form. i j 51º 35 51º 35 opposite 35 sin 51 = opposite = 35 sin 51 adjacent 35 cos 51 = adjacent = 35 cos 51 = Hence the vector is:

If a vector is given in component form, Pythagoras’ theorem can Summary of key points: If a vector is given in component form, Pythagoras’ theorem can be used to find the magnitude, and basic trigonometry to find the direction. θ k h R The component of a vector opposite a given angle is found using sin. i.e. h = R sin θ The component of a vector adjacent to a given angle is found using cos. i.e. k = R cos θ This PowerPoint produced by R.Collins; ©ZigZag Education 2010