☰ Year plan · Applied Mathematics

Moments of any coplanar system, machines and mixed problems

Week 4 Β· Moments: the general statement for a system of coplanar forces, moments in machines and mixed problems with unknowns Β· 3 lessons
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 Β· several forces in one plane, each with its own point of actionsum of moments Ξ£M_A Β· the algebraic sum of the moments of all the forces about the point Aequivalent Β· 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 Ξ£M_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 Ξ£M_A = Ξ£M_B, the line of action of R is parallel to AB. (2) If Ξ£M_A = βˆ’Ξ£M_B, it passes through the midpoint of AB. (3) If Ξ£M_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 βˆ’ y Rx = Ξ£M_O, for example, where it crosses the x-axis (y = 0) we have x Ry = Ξ£M_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.)

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Worked examples
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Three forces act on a plate: F₁ = (3, 1) N at P₁(0, 2), Fβ‚‚ = (βˆ’1, 2) N at Pβ‚‚(4, 0), F₃ = (1, βˆ’4) N at P₃(2, βˆ’3). Their resultant is R = (3, βˆ’1) N. About B(1, 1): P₁ βˆ’ B = (βˆ’1, 1): (βˆ’1)(1) βˆ’ 1 Γ— 3 = βˆ’4; Pβ‚‚ βˆ’ B = (3, βˆ’1): 3 Γ— 2 βˆ’ (βˆ’1)(βˆ’1) = 5; P₃ βˆ’ B = (1, βˆ’4): 1 Γ— (βˆ’4) βˆ’ (βˆ’4)(1) = 0; the sum is 1 NΒ·m. About C(7, βˆ’1): P₁ βˆ’ C = (βˆ’7, 3): (βˆ’7)(1) βˆ’ 3 Γ— 3 = βˆ’16; Pβ‚‚ βˆ’ C = (βˆ’3, 1): (βˆ’3)(2) βˆ’ 1 Γ— (βˆ’1) = βˆ’5; P₃ βˆ’ C = (βˆ’5, βˆ’2): (βˆ’5)(βˆ’4) βˆ’ (βˆ’2)(1) = 22; the sum is 1 NΒ·m. The two sums are equal, so the line of action of R is parallel to BC. Check: BC = (6, βˆ’2) is parallel to R = (3, βˆ’1).
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The same plate and R = (3, βˆ’1) N. About B(1, 3): the moments are (P₁ βˆ’ B = (βˆ’1, βˆ’1)) (βˆ’1)(1) βˆ’ (βˆ’1)(3) = 2; (Pβ‚‚ βˆ’ B = (3, βˆ’3)) 3 Γ— 2 βˆ’ (βˆ’3)(βˆ’1) = 3; (P₃ βˆ’ B = (1, βˆ’6)) 1 Γ— (βˆ’4) βˆ’ (βˆ’6)(1) = 2; the sum is 7 NΒ·m. About C(5, βˆ’3): (P₁ βˆ’ C = (βˆ’5, 5)) (βˆ’5)(1) βˆ’ 5 Γ— 3 = βˆ’20; (Pβ‚‚ βˆ’ C = (βˆ’1, 3)) (βˆ’1)(2) βˆ’ 3 Γ— (βˆ’1) = 1; (P₃ βˆ’ C = (βˆ’3, 0)) (βˆ’3)(βˆ’4) βˆ’ 0 = 12; the sum is βˆ’7 NΒ·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, βˆ’1) cut the x-axis? The sum of moments about O is βˆ’3 NΒ·m (from the same plate: (0, 2): 0 Γ— 1 βˆ’ 2 Γ— 3 = βˆ’6; (4, 0): 4 Γ— 2 βˆ’ 0 = 8; (2, βˆ’3): 2 Γ— (βˆ’4) βˆ’ (βˆ’3)(1) = βˆ’5; total βˆ’3). At the point (x, 0) of the line, x Ry βˆ’ 0 Γ— Rx = x Γ— (βˆ’1) = βˆ’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, Pβ‚‚, 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 βˆ’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 Fβ‚‚ = (βˆ’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 (βˆ’2, 4) N and the sum of moments about C(1, 3) is also +1 NΒ·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 NΒ·m (anticlockwise positive). Where does the line of action of R cross the x-axis?
4
A system has R β‰  0 and the sums of moments about A(0, 0) and B(4, 0) are +8 NΒ·m and βˆ’8 NΒ·m. Through which point does the line of action of R pass?
5
Why can the facts of this lesson fail if R = 0?