Term 1 review: moments, parallel forces, equilibrium and couples
Week 15 · Term 1 review: moments, parallel forces and equilibrium of rods and ladders, couples · 3 lessons
1
Review: moments of a force about a point, the theorem of moments, and equal or opposite moments about two points
Applied Mathematics Sec 3 Term 1, Units 1–3, Review
GoalReview the moment of a force about a point, the theorem of moments and what equal or opposite moments about two points tell us, with new problems.
New words
moment of a force · |M| = F × L, with L the perpendicular distance from the point to the line of action; M = x Fy − y Fx for F at (x, y); anticlockwise positivetheorem of moments · for forces at one point, the sum of their moments about any point equals the moment of the resultant
Explanation
Weeks 2 to 4 in one page. (1) The moment of a force about a point O is F × L, where L is the perpendicular distance from O to the line of action; anticlockwise is positive and clockwise negative. (2) With coordinates, a force (Fx, Fy) acting at (x, y) has moment x Fy − y Fx about the origin; for another point, measure x and y from that point. (3) Varignon: the moment of a force equals the sum of the moments of its components. (4) Theorem of moments: for forces acting at one point, the sum of their moments about any point equals the moment of their resultant R = ΣF. (5) The moment about a point on the line of action is zero. (6) If the moments of the same system about two points A and B are equal, the line of action of the resultant is parallel to AB; if they are equal in size and opposite in sign, it passes through the midpoint of AB. Units: N·m, N·cm.
Worked examples
A force F = (−6, 8) N acts at A(3, 2) (coordinates in m). (a) Find its moment about O (anticlockwise positive) and the arm. (b) Find its moment about C(0, 6) and explain the answer. Solution: (a) M = x Fy − y Fx = 3 × 8 − 2 × (−6) = 24 + 12 = +36 N·m, anticlockwise. |F| = √(36 + 64) = 10 N, so the arm is L = 36 / 10 = 3.6 m. (b) From C, the point A is at (3, −4): M = 3 × 8 − (−4)(−6) = 24 − 24 = 0. The moment is zero because C lies on the line of action: the line goes from A in the direction (−6, 8), that is (−3, 4), and A + (−3, 4) = (0, 6) = C.
Forces F₁ = (2, 3) N and F₂ = (4, −1) N act at A(1, 2) (m). Find the moments of F₁, F₂ and of the resultant R about B(0, 4) and about C(3, 5) (anticlockwise positive). What does the result tell you about the line of action of R? Solution: R = (6, 2). About B, the vector from B to A is (1, −2): F₁ gives 1 × 3 − (−2)(2) = 7, F₂ gives 1 × (−1) − (−2)(4) = 7, and R gives 1 × 2 − (−2)(6) = 14 = 7 + 7. About C, the vector from C to A is (−2, −3): F₁ gives (−2)(3) − (−3)(2) = 0, F₂ gives (−2)(−1) − (−3)(4) = 14, and R gives (−2)(2) − (−3)(6) = 14 = 0 + 14. The two moments about B and C are equal (14 N·m), so R is parallel to BC; indeed BC = (3, 1) and R = (6, 2) = 2 × (3, 1).
Class activity
Hold a ruler at one end and push its other end with one finger at right angles to it, then at a slant. Notice which push turns it more easily, and explain with F × L. Then push the ruler along its own length: why does it not turn?
Practice
1
A force of 40 N acts at the end of a spanner 0.25 m long at 30° to the spanner. Find the size of its moment about the nut.
L = 0.25 × sin 30° = 0.125 m, so M = 40 × 0.125 = 5 N·m.
2
A force with components 120 N to the right and 50 N upward acts at the point (2, 3) m. Use components to find its moment about the origin (anticlockwise positive) and its arm.
M = 2 × 50 − 3 × 120 = 100 − 360 = −260 N·m (clockwise). |F| = 130 N, so L = 260 / 130 = 2 m.
3
F = (3, −4) N acts at A(2, 1). Find its moments about B(−1, 2) and C(2, −2) (anticlockwise positive). What do you conclude?
About B: vector (3, −1): 3 × (−4) − (−1)(3) = −9. About C: vector (0, 3): 0 × (−4) − 3 × 3 = −9. Equal, so the line of action is parallel to BC; indeed BC = (3, −4) is parallel to F.
4
F = (1, 2) N acts at A(0, 0). Find its moments about B(3, 0) and C(−1, 4) (anticlockwise positive). What do you conclude?
About B: vector (−3, 0): (−3)(2) − 0 = −6. About C: vector (1, −4): 1 × 2 − (−4)(1) = +6. Equal in size, opposite in sign, so the line passes through the midpoint (1, 2) of BC (check: it lies on y = 2x).
5
A force has zero moment about a point P. What can you say about the force or its line of action?
Either the force is zero, or P lies on its line of action (the arm is zero).
