☰ خطة العام · الرياضيات التطبيقية (بالإنجليزية)

Moments of any coplanar system, machines and mixed problems

الأسبوع 4 · تعميم النتيجة على أي نظام من القوى المستوية، وتطبيقات العزوم في الرافعات والقضبان المثبتة، ومسائل متنوعة · 3 دروس
3

Mixed problems: unknown constants or an unknown force from given moments, and the least force that turns a nut or a nail

Applied Mathematics Sec 3 Term 1, Unit 1, Lesson 1, Part 3
GoalSolve for unknown constants or an unknown force from given moments, and find the least force that produces a required moment.
New words
unknown constants · letters such as a and b in a force, found from equations given by momentsleast force · the smallest force that gives a required moment: it acts at right angles to the arm
Explanation

Every moment is an equation. If a force contains an unknown constant, the condition on its moment (zero, or a given value) gives an equation, and two conditions give two equations to solve. If the unknown is a force and we know where it acts, a moment about a point gives one equation in its two components, and a second point gives another. A useful trick: when a moment about a point is zero, the force passes through that point, so measure positions from it. Least force. For a spanner of length OA the moment of a force F making angle θ with it is F × OA × sin θ. To produce a required moment M with the smallest force we want sin θ as large as possible, which is 1 at θ = 90°. So the least force is M / OA, applied at right angles. At any other angle the force must be bigger, F = M / (OA sin θ).

FθLOA
Worked examples
F₁ = (p, 2) N acts at (2, 1) and F₂ = (3, q) N acts at (−1, 2). The sum of the moments about O is zero and so is the sum about B(1, 0). Find p and q (anticlockwise positive). About O: 2 × 2 − 1 × p = 4 − p, and (−1) × q − 2 × 3 = −q − 6; sum = −2 − p − q = 0, so p + q = −2. About B: (2, 1) − B = (1, 1): 1 × 2 − 1 × p = 2 − p; (−1, 2) − B = (−2, 2): (−2) q − 2 × 3 = −2q − 6; sum = −4 − p − 2q = 0, so p + 2q = −4. Subtract: q = −2, and then p = 0. So F₁ = (0, 2) N and F₂ = (3, −2) N.
xy−11234567−1123450FOAB
A force F acts in the plane of the triangle OAB, with O(0, 0), A(0, 3) and B(5, 0) (metres). Its moment about O is +45 N·m, about A it is −15 N·m, and about B it is 0. Find F. The zero moment about B means the line of action passes through B, so take B as the point of action. About O: x Fy − y Fx = 5 Fy = 45, so Fy = 9. About A: B − A = (5, −3), so 5 × 9 − (−3) Fx = 45 + 3Fx = −15, so Fx = −20. F = (−20, 9) N, of size √481 ≈ 21.93 N.
F50°0.30 mLOA
A rusty bolt needs a moment of 36 N·m to start turning, and the spanner is 0.30 m long. The least force is at right angles: F = 36 / 0.30 = 120 N. If there is a wall in the way and the force can only make 50° with the spanner, F = 36 / (0.30 sin 50°) ≈ 156.6 N, about 30% more. A longer spanner reduces the force: with 0.45 m the least force is 36 / 0.45 = 80 N.
Class activity

Take a closed jar lid that is loose and turn it with two fingers resting on its rim: first push at right angles to the radius, then push at a slant. Which direction feels easiest? Do the same with a loose tap handle if an adult is present. This is the least-force idea: right angles to the arm.

Practice
1
A force (a, 5) N acts at (4, 2) and its moment about O is zero. Find a.
2
Pulling a nail needs a moment of 24 N·m about the hammer head and the handle is 0.30 m. Find the least force and the force when it makes 60° with the handle.
3
A force F acts at A(2, 1) and has moment +7 N·m about O and +1 N·m about C(3, 0). Find F (anticlockwise positive).
4
A door handle is 0.8 m from the hinge and a moment of 12 N·m is needed to open it. What is the least force, and what is its direction relative to the door?
5
Why does a force making a very small angle with the spanner need to be so large?