الرياضيات التطبيقية (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 7: If one of two like forces moves parallel to itself by x, the resultant moves by F₂x / (F₁ + F₂): the proof by moments

Moving a force, forces as vectors and practice

الأسبوع 7 · أثر انتقال إحدى القوتين في المحصلة، والقوى المتوازية بالمتجهين، وتمارين الدرس · 3 دروس

If one of two like forces moves parallel to itself by x, the resultant moves by F₂x / (F₁ + F₂): the proof by moments

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

GoalShow by moments that when one of two like forces moves parallel to itself by x, the resultant moves by F₂x / (F₁ + F₂), and use the result.

New words

parallel to itself · the force slides along the rod and keeps its direction and its sizeshift of the resultant · the distance CC′ between the old and the new line of action of R

Explanation

Two like forces F₁ at A and F₂ at B have the resultant R = F₁ + F₂ acting at C. Take moments about A. F₁ has no moment about A, so R × AC = F₂ × AB (1). Now slide F₂ along the rod to B′, a distance x further from A, so AB′ = AB + x. The two forces are the same, so R is the same size, and it now acts at C′: R × AC′ = F₂ × AB′ (2). Subtract (1) from (2): R × (AC′ − AC) = F₂ × (AB′ − AB), that is R × CC′ = F₂ × x. Therefore CC′ = F₂x / (F₁ + F₂). In words: the resultant moves the same way as the moving force, by the fraction F₂/R of the distance moved. If F₂ moves towards A, the same formula holds and R moves towards A. If the two forces are equal, R moves half as far as the force.

Worked examples

Two like forces, 12 N at A and 20 N at B, act on a light rod with AB = 40 cm. Find R and AC. Then the 20 N force slides 8 cm towards A. Find the new position of R. Solution: R = 12 + 20 = 32 N and 32 × AC = 20 × 40, so AC = 25 cm. The force that moves is F₂ = 20 N and x = 8 cm towards A, so R moves 20 × 8 / 32 = 5 cm towards A and AC′ = 25 − 5 = 20 cm. Check with moments about A: 32 × AC′ = 20 × 32 gives AC′ = 20. In the picture, the moved force and the new resultant are the ones marked with the arrows nearer to A.

Two like forces, 8 N at A and 12 N at B, act on a rod with AB = 30 cm. How far, and in which direction, must the 12 N force slide so that the resultant moves 3 cm away from A? Solution: R = 20 N. We need 12 × x / 20 = 3, so x = 5 cm and the force must slide 5 cm away from A. Check: before, 20 × AC = 12 × 30 gives AC = 18; after, 20 × AC′ = 12 × 35 gives AC′ = 21, which is 3 cm further.

Class activity

Balance a ruler on a pencil with one coin at the 6 cm mark and two coins stacked at the 24 cm mark; the pencil must be at the 18 cm mark. Now slide the stack of two coins 6 cm towards the end of the ruler, to the 30 cm mark, and find the new balance point. The formula predicts 2 × 6 / 3 = 4 cm further along, so about the 22 cm mark. Is that what you find?

Practice

Two equal like forces of 6 N act at A and B. The force at B slides 10 cm along the rod away from A. How far does the resultant move?

A force of 10 N acts at A and 15 N at B, AB = 20 cm. (a) Find AC. (b) The 15 N force slides 5 cm towards A. By how much and which way does R move, and what is the new AC?

Forces of 4 N and 12 N act at A and B, 16 cm apart. By how much does R move if (a) the 12 N force slides 8 cm along the rod, (b) the 4 N force slides 8 cm? Which move is bigger, and why?

Forces of 5 N at A and 7 N at B act on a rod. The 5 N force slides 6 cm towards B. How does R move?

A force of 9 N acts at A and a force F at B, both pointing the same way. When F slides 6 cm along the rod the resultant moves 4.5 cm. Find F.