☰ 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
1

The general statement for any coplanar system: equal moments about A and B, or opposite moments, and what a zero moment about a point tells you, checked with forces given as i and j components

Applied Mathematics Sec 3 Term 1, Unit 1, Lesson 1, Part 3
GoalExtend the two-point rules to any coplanar system of forces acting at different points, and use the zero-moment point to locate the line of action of the resultant.
New words
system of forces · several forces in one plane, each with its own point of actionsum of moments ΣM_A · the algebraic sum of the moments of all the forces about the point Aequivalent · having the same resultant force and the same moment about every point
Explanation

Now the forces act at different points of a body, each with its own position (x, y) and components (Fx, Fy), all in one plane. To get the total moment about a point P, find each moment with its position measured from P and add them: this is ΣM_P. If the forces add up to a resultant R that is not zero, the whole system is equivalent to R acting along one line of action: it has the same total moment about every point as R does. So the facts of last week hold for the totals. (1) If ΣM_A = ΣM_B, the line of action of R is parallel to AB. (2) If ΣM_A = −ΣM_B, it passes through the midpoint of AB. (3) If ΣM_C = 0, then C lies on the line of action of R (or R is zero). There is also a way to locate the line: any point (x, y) of it satisfies x Ry − y Rx = ΣM_O, for example, where it crosses the x-axis (y = 0) we have x Ry = ΣM_O. To work, write each force as i and j components, compute each moment with its own position from the chosen point, and add. Anticlockwise is positive. (If R is zero the system is balanced or is a couple, which Unit 3 studies.)

xy−112345678−5−4−3−2−1123450F₁F₂F₃P₁P₂P₃BC
Worked examples
xy−112345678−5−4−3−2−1123450F₁F₂F₃P₁P₂P₃BC
Three forces act on a plate: F₁ = (3, 1) N at P₁(0, 2), F₂ = (−1, 2) N at P₂(4, 0), F₃ = (1, −4) N at P₃(2, −3). Their resultant is R = (3, −1) N. About B(1, 1): P₁ − B = (−1, 1): (−1)(1) − 1 × 3 = −4; P₂ − B = (3, −1): 3 × 2 − (−1)(−1) = 5; P₃ − B = (1, −4): 1 × (−4) − (−4)(1) = 0; the sum is 1 N·m. About C(7, −1): P₁ − C = (−7, 3): (−7)(1) − 3 × 3 = −16; P₂ − C = (−3, 1): (−3)(2) − 1 × (−1) = −5; P₃ − C = (−5, −2): (−5)(−4) − (−2)(1) = 22; the sum is 1 N·m. The two sums are equal, so the line of action of R is parallel to BC. Check: BC = (6, −2) is parallel to R = (3, −1).
xy−112345678−5−4−3−2−1123450F₁F₂F₃BCM
The same plate and R = (3, −1) N. About B(1, 3): the moments are (P₁ − B = (−1, −1)) (−1)(1) − (−1)(3) = 2; (P₂ − B = (3, −3)) 3 × 2 − (−3)(−1) = 3; (P₃ − B = (1, −6)) 1 × (−4) − (−6)(1) = 2; the sum is 7 N·m. About C(5, −3): (P₁ − C = (−5, 5)) (−5)(1) − 5 × 3 = −20; (P₂ − C = (−1, 3)) (−1)(2) − 3 × (−1) = 1; (P₃ − C = (−3, 0)) (−3)(−4) − 0 = 12; the sum is −7 N·m. The sums are opposite, so the line of action of R passes through the midpoint of BC, which is (3, 0).
Where does the line of action of R = (3, −1) cut the x-axis? The sum of moments about O is −3 N·m (from the same plate: (0, 2): 0 × 1 − 2 × 3 = −6; (4, 0): 4 × 2 − 0 = 8; (2, −3): 2 × (−4) − (−3)(1) = −5; total −3). At the point (x, 0) of the line, x Ry − 0 × Rx = x × (−1) = −3, so x = 3. The line of action cuts the x-axis at (3, 0): it is the same point as the midpoint found in the last example, so the sum of moments about (3, 0) is zero.
Class activity

On squared paper mark the three points P₁, P₂, P₃ of the first example and draw each force as a small arrow (1 square for 1 N). Mark B and C and the point (3, 0), and draw the resultant arrow through (3, 0) with slope −1/3. Check by eye that B and C are the same distance from this line, on the same side.

Practice
1
F₁ = (2, 3) N at (1, 0) and F₂ = (−4, 1) N at (0, 2) act on a plate. Find the sum of their moments about B(2, 1) (anticlockwise positive).
2
For the same two forces the resultant is (−2, 4) N and the sum of moments about C(1, 3) is also +1 N·m. What does this say about the resultant?
3
A system has resultant R = (0, 6) N and the sum of moments about O is +30 N·m (anticlockwise positive). Where does the line of action of R cross the x-axis?
4
A system has R ≠ 0 and the sums of moments about A(0, 0) and B(4, 0) are +8 N·m and −8 N·m. Through which point does the line of action of R pass?
5
Why can the facts of this lesson fail if R = 0?