Parallel forces: like, unlike and the theorem of moments
Week 5 · Parallel coplanar forces: the resultant of two like forces, of two unlike forces, and the theorem of moments for parallel forces · 3 lessons
2
Parallel forces in opposite directions: R = F₁ − F₂ along the larger force, acting outside the segment on the larger force’s side, and equal opposite forces
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 1
GoalFind the resultant of two parallel forces in opposite directions, R = F₁ − F₂, and show that it acts outside AB on the side of the larger force.
New words
unlike forces · parallel forces that point opposite waysoutside the segment · on the line AB but beyond A or beyond B, not between them
Explanation
Now let F₁ act upwards at A and a smaller F₂ act downwards at B (F₁ > F₂). The resultant is R = F₁ − F₂, in the direction of the larger force. Where does it act? Let C be a point on the line AB at which R acts: the turning effects of F₁ and F₂ about C must cancel, because R has arm 0. Between A and B the two forces turn the rod in the same sense about any point (up at the left and down at the right: both clockwise about a point between them), so they cannot cancel there. The point must lie outside AB, on the side of the larger force: beyond A, where the larger force has the shorter arm. Then F₁·AC = F₂·BC with BC = AC + AB, which gives AC = F₂ × AB / (F₁ − F₂). If the forces are nearly equal, R is small and C is very far away. If F₁ = F₂ there is no single resultant at all: equal and opposite parallel forces form a couple, which turns a body but does not push it anywhere (Unit 3). Say which direction is positive: here upwards; moments anticlockwise positive.
Worked examples
A force of 8 N acts upwards at A and a force of 5 N acts downwards at B, with AB = 30 cm (upwards positive). R = 8 − 5 = 3 N upwards. AC = 5 × 30 / 3 = 50.0 cm, beyond A (on the side of the 8 N force), so BC = 80.0 cm. Check: 8 × 50.0 = 400.0 and 5 × 80.0 = 400.0.
Let F₁ = 10.2 N upwards at A and F₂ = 10 N downwards at B, AB = 0.4 m. R = 0.2 N upwards and AC = 10 × 0.4 / 0.2 = 20 m beyond A. A tiny difference in the forces throws the resultant 20 m away. With F₁ = F₂ = 10 N the formula would divide by zero: there is no single resultant, and the pair of forces is a couple. A child turning a tap with two fingers pushing in opposite directions on either side of the tap makes such a pair.
F₁ = 10 N acts upwards at A and a smaller force F₂ acts downwards at B. The resultant R = 4 N upwards acts 0.9 m beyond A. Find F₂ and AB. F₂ = 10 − 4 = 6 N. BC = F₁ × AC / F₂ = 10 × 0.9 / 6 = 1.5 m, so AB = BC − AC = 0.6 m.
Class activity
Put a ruler on the desk and lay a pencil under it near the middle. Press down with a finger at one end and press up from below with a finger a little way along from the other side of the pencil. Notice that two pushes in opposite directions at different points turn the ruler. Try pushing with equal strength on both sides of the pencil. Keep every push light.
Practice
1
A force of 12 N acts upwards at A and 4 N acts downwards at B, with AB = 0.6 m. Find R and where it acts.
R = 8 N upwards. AC = 4 × 0.6 / 8 = 0.3 m beyond A (BC = 0.9 m; check 12 × 0.3 = 3.6 = 4 × 0.9).
2
A force of 7 N acts downwards at A and 9 N acts upwards at B, with AB = 0.2 m. Find R and where it acts.
R = 2 N upwards, on the side of the larger force: BC = 7 × 0.2 / 2 = 0.7 m beyond B (AC = 0.9 m; check 7 × 0.9 = 6.3 = 9 × 0.7).
3
Two forces of 15 N act in opposite directions along parallel lines 0.4 m apart. What is their resultant?
There is none: they are equal and opposite and not on one line, so they form a couple.
4
A force of 20 N acts upwards at A and an unknown F acts downwards at B, 0.3 m from A. R acts 0.1 m beyond A. Find F and R.
BC = 0.4 m. 20 × 0.1 = F × 0.4, so F = 5 N and R = 20 − 5 = 15 N upwards.
5
Why can the resultant of two unlike forces not act between A and B?
About a point between A and B both forces turn the rod the same way, so their moments cannot cancel, but the moments about the point where R acts must cancel.
