☰ خطة العام · الرياضيات التطبيقية (بالإنجليزية)

Moving a force, forces as vectors and practice

الأسبوع 7 · أثر انتقال إحدى القوتين في المحصلة، والقوى المتوازية بالمتجهين، وتمارين الدرس · 3 دروس
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Practice on resultants and balance points: R, AC and AB from tables of data, and the point where a light rod loaded with weights balances

Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 3
GoalPractise finding R, AC and AB from tables of data, and find the point where a light rod loaded with weights balances.
New words
balance point · the point where a single support holds the loaded rod level; it is where the resultant of the weights actsoutside the rod · the line of the resultant of unlike forces, beyond the larger force
Explanation

For two parallel forces F₁ at A and F₂ at B, choose the case first. Like forces: R = F₁ + F₂; C is between A and B; F₁ × AC = F₂ × CB; AB = AC + CB. Unlike forces with F₂ the larger: R = F₂ − F₁ in the direction of F₂; C is outside AB beyond B; F₁ × CA = F₂ × CB; AB = CA − CB. Each row of a table of data gives enough numbers for these three equations, so you can find the missing ones. For a light rod carrying weights w₁, w₂, … at the positions x₁, x₂, …, the weights are like forces, and the single support that holds the rod level must be at the position of their resultant: x = (w₁x₁ + w₂x₂ + …) / (w₁ + w₂ + …). The weight of the rod is ignored when the rod is light.

CaseRWhere is C?Moment equation
like forcesF₁ + F₂between A and BF₁·AC = F₂·CB
unlike forcesF₂ − F₁ (F₂ larger)beyond B, the larger forceF₁·CA = F₂·CB
Worked examples
A table of data for pairs of parallel forces (F₁ at A, F₂ at B). Row 1: like forces, F₁ = 14 N, F₂ = 22 N, AB = 36 cm: find R and AC. Row 2: like forces, F₁ = 8 N, R = 20 N, AC = 15 cm: find F₂ and AB. Row 1: R = 36 N; 14 × AC = 22 × (36 − AC) gives 36 × AC = 792 and AC = 22 cm. Row 2: F₂ = 20 − 8 = 12 N; 8 × 15 = 12 × CB gives CB = 10 cm, so AB = 25 cm.
10 cm20 cm15 cm4 N6 N2 N8 NABCD
Weights of 4 N, 6 N, 2 N and 8 N hang from the marks 0, 10, 30 and 45 cm of a light rod. Where must a single string be tied so that the rod hangs level? The weights are like forces, R = 4 + 6 + 2 + 8 = 20 N. x = (4 × 0 + 6 × 10 + 2 × 30 + 8 × 45) / 20 = 480 / 20 = 24 cm. The string pulls up with 20 N at the 24 cm mark.
20 cm30 cm9 N15 NR = 6 NABC
Unlike forces: F₁ = 9 N up at A and F₂ = 15 N down at B, AB = 20 cm. Find R and its line. R = 15 − 9 = 6 N downwards, the direction of F₂. C is beyond B: 9 × CA = 15 × CB with CA = CB + 20, so 9CB + 180 = 15CB and CB = 30 cm, CA = 50 cm.
Class activity

Make a see-saw from a ruler on a pencil. Put coins in two piles so that one pile has three times as many coins as the other. Predict, using F₁ × distance = F₂ × distance, where the pencil must be if the piles are 20 cm apart (the pencil is 5 cm from the big pile), then test. Then move both piles 5 cm in the same direction and see how the balance point moves.

Practice
1
Two like forces of 9 N at A and 21 N at B act on a rod with AB = 40 cm. Find R and AC.
2
Unlike forces: 5 N at A upwards and 8 N at B downwards, AB = 12 cm. Find the resultant and the distance of its line from B.
3
Two like forces: F₁ = 6 N at A, F₂ at B, the resultant is 15 N and acts 12 cm from A. Find F₂ and AB.
4
Weights of 3 N, 5 N and 2 N hang from the 0 cm, 10 cm and 30 cm marks of a light rod. Where is the balance point?
5
A light plank 2 m long has a child of weight 300 N at one end and a child of weight 200 N at the other end. How far from the heavier child must the pivot be so that the plank balances?