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Moving a force, forces as vectors and practice

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2

Parallel forces given as i and j vectors: the resultant, and where its line crosses AB by the section formula for like and unlike forces

Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 3
GoalFind the resultant of two parallel forces given as vectors, and the point where its line cuts AB, using the section formula.
New words
force as a vector · a force written as a pair (a, b) = a i + b j, with i and j the unit vectors along the axessection formula · the point C on AB with AC : CB = m : n is (nA + mB) / (m + n)
Explanation

A force can be given by its components: F = (a, b) means a i + b j. Two forces are parallel exactly when one is a multiple of the other, F₂ = k F₁. If k > 0 they are like forces, if k < 0 they are unlike. Their resultant is R = F₁ + F₂ = (1 + k) F₁, which is another multiple of F₁ (if k = −1 it is zero: a couple). Where does R act? Let F₁ act at the point A and F₂ at the point B, and let the line of R cut AB at C. The forces are parallel, so the distances of C from their lines are in the same ratio as CA and CB, and the moment balance F₁ × (distance) = F₂ × (distance) gives CA : CB = |k| : 1. For like forces C is inside AB, nearer the larger force. For unlike forces C is outside AB, beyond the larger force. In both cases one formula works if k keeps its sign: C = (A + kB) / (1 + k), the average of the positions of A and B in which A counts 1 time and B counts k times. Coordinates are added one at a time: x with x, y with y.

xy−2246810246810120F₁F₂RABC
Worked examples
In the figure, F₁ = (1, 2) acts at A(−2, 1) and F₂ = (3, 6) at B(6, 5) (forces in newtons). Find the resultant and the point C where its line cuts AB. Since (3, 6) = 3 × (1, 2), the forces are parallel and like, with k = 3. R = (1 + 3, 2 + 6) = (4, 8), of size √80 ≈ 8.94 N. Then C = (A + 3B) / 4 = ((−2 + 18) / 4, (1 + 15) / 4) = (4, 4). Check: AC = (6, 3) is three quarters of AB = (8, 4), so AC : CB = 3 : 1, C is nearer the larger force. Moments about C: (−6)(2) − (−3)(1) = −9 for F₁ and (2)(6) − (1)(3) = 9 for F₂, sum 0. (The moment about C of a force (a, b) at the point (x, y) relative to C is x b − y a.)
xy−1123456789−11234567891011120F₁F₂RABC
F₁ = (2, −1) acts at A(1, 2) and F₂ = (−4, 2) at B(4, 6). Find the resultant and the point where its line cuts the line AB. (−4, 2) = −2 × (2, −1), so k = −2: the forces are parallel and unlike. R = (2 − 4, −1 + 2) = (−2, 1). C = (A + kB) / (1 + k) = ((1, 2) − 2(4, 6)) / (−1) = (7, 10). C lies beyond B (B is the midpoint of AC), on the side of the larger force, with CA : CB = 2 : 1. Moments about C: (−6)(−1) − (−8)(2) = 22 for F₁ and (−3)(2) − (−4)(−4) = −22 for F₂: the sum is 0.
Class activity

On squared paper mark A(0, 0) and B(6, 0). Draw a force of 2 squares upwards at A and a force of 4 squares upwards at B (scale: 1 square for 1 N). Predict where the resultant crosses AB using the formula with k = 2, C = (A + 2B) / 3, then check by cutting a strip of card, balancing it on a pencil with 2 coins at one end and 4 at the other (in the same ratio).

Practice
1
F₁ = (1, −3) acts at A(1, 2) and F₂ = (2, −6) at B(4, 5). Show they are parallel, and find R and the point C where its line cuts AB.
2
F₁ = (4, 2) acts at A(−1, 3) and F₂ = (−2, −1) at B(5, 6). Find R and where its line cuts the line AB.
3
Are F₁ = (2, 5) and F₂ = (6, 10) parallel? Explain.
4
F₁ = (3, −2) acts at A(−2, 4) and F₂ = (−9, 6) at B(2, 0). Find R and the point where its line cuts AB.
5
What happens to the formula C = (A + kB) / (1 + k) when k = −1, that is when the two forces are equal and opposite?