Egypt Secondary 3 Applied Mathematics, Week 5: Parallel forces in the same direction: the balanced ruler, the resultant R = F₁ + F₂ and the point where it acts, F₁·AC = F₂·BC, and the case of equal forces

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

Parallel forces in the same direction: the balanced ruler, the resultant R = F₁ + F₂ and the point where it acts, F₁·AC = F₂·BC, and the case of equal forces

Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 1

GoalFind the resultant of two parallel forces in the same direction and the point where it acts, using R = F₁ + F₂ and F₁·AC = F₂·BC.

New words

like forces · parallel forces that point the same waypoint of action of the resultant · the point C on AB through which the line of action of R passeslight rod · a rod whose own weight is so small that we ignore it

Explanation

Hang weights F₁ and F₂ from the ends A and B of a light rod. Can one single force have the same effect? Yes: a support that holds the rod up must push up with F₁ + F₂, so the resultant of the two weights is R = F₁ + F₂, in the same direction as both. Where does it act? Support the rod at a point C and let it balance. The support force R has arm 0 about C, so the two weights must have equal and opposite turning effects about C: F₁ × AC (one sense) equals F₂ × BC (the other sense). So F₁·AC = F₂·BC, and AC : BC = F₂ : F₁. The bigger force has the smaller arm: C lies between A and B, nearer the larger force. If the forces are equal, C is the midpoint. Rule to use: R = F₁ + F₂, and distance of C from A = F₂ × AB / (F₁ + F₂). (Directions: say which way is positive, here downwards for weights; every moment about C is then of one of the two senses, anticlockwise or clockwise.)

Worked examples

Weights of 30 N and 50 N hang from the ends A and B of a light rod AB, 1.6 m long. Take downwards as positive. R = 30 + 50 = 80 N downwards. AC = F₂ × AB / R = 50 × 1.6 / 80 = 1 m from A, so BC = 0.6 m. Check: 30 × 1 = 30 and 50 × 0.6 = 30. A single support of 80 N upwards at C, 1.0 m from the 30 N end, would hold the rod level.

Two equal forces of 25 N act downwards at the ends of a light rod 0.6 m long. R = 50 N, and since F₁ = F₂ the arms are equal: C is the midpoint, 0.3 m from each end. This is why a load carried by two people with a light pole is shared equally when it hangs from the middle.

Two like forces, F₁ = 18 N at A and F₂ at B, with AB = 0.9 m, have a resultant R = 45 N. Then F₂ = 45 − 18 = 27 N. AC = F₂ × AB / R = 27 × 0.9 / 45 = 0.54 m, BC = 0.36 m. Check: 18 × 0.54 = 9.72 and 27 × 0.36 = 9.72.

Class activity

Lay a ruler across a pencil so that it balances at the middle. Put one coin at each end: it still balances. Now put two coins on one end and one at the other, and slide the ruler along the pencil until it balances again. Measure the two arms: the one for the two coins should be half as long. Keep the coins light and small.

Practice

Like forces of 9 N and 6 N act at the ends A and B of a light rod 30 cm long. Find R and the distance AC.

Two equal like forces of 18 N act 0.5 m apart. Find R and where it acts.

Weights of 2 N and 3 N hang from the ends of a light ruler 30 cm long. Where should it be supported to stay level?

Like forces F₁ = 24 N at A and F₂ at B, with AB = 1 m, have R = 60 N. Find F₂ and AC.

True or false: the resultant of two like parallel forces lies between them, nearer the smaller force.