The theorem of moments for parallel forces: the sum of their moments about any point equals the moment of the resultant
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 1
GoalState the theorem of moments for parallel forces and use it to find the resultant of three or more parallel forces and where it acts.
New words
theorem of moments (parallel forces) · the sum of the moments of parallel forces about any point equals the moment of their resultant about that pointsigned distance x · position along the rod measured from A, positive to the right
Explanation
Take a horizontal rod with A as the origin of positions x (positive to the right) and vertical forces Fi (upwards positive) at positions xi. The moment of a vertical force about A is xi Fi (anticlockwise positive), the formula you know with y = 0. The theorem says: the sum of the moments of the forces about any point equals the moment of the resultant R about that point. About A: Σ Fi xi = R × xR, with R = Σ Fi and xR the position where R acts. Why is it true? About the point C where R acts, we saw in the last two lessons that the moments of the forces cancel, so their sum is 0, the same as the moment of R about C (its arm is 0). For two like forces the equation reads F₁x₁ + F₂x₂ = (F₁ + F₂) xR, which is exactly xR = (F₁x₁ + F₂x₂) / (F₁ + F₂), the point dividing AB in the inverse ratio of the forces. The theorem makes many forces easy: add up the forces for R, add up the products F × x, divide to get xR. A negative xR means R acts to the left of A. If R = 0 the forces form a couple.
Worked examples
Three weights hang from a light rod at 0.1 m, 0.6 m and 1.2 m from A: 20 N, 30 N and 50 N (downwards). R = 100 N downwards. About A (downwards force at a positive x: clockwise): the moments are −20 × 0.1 = −2, −30 × 0.6 = −18, −50 × 1.2 = −60, total −80 N·m. The moment of R is −100 × xR. So xR = 80 / 100 = 0.8 m from A. A single 100 N support at 0.8 m would hold the rod level.
A rod is acted on by 30 N upwards at A (x = 0), 20 N downwards at x = 0.4 m and 20 N downwards at x = 0.8 m (upwards positive). R = 30 − 20 − 20 = −10 N, that is 10 N downwards. Moments about A: 30 × 0 + (−20)(0.4) + (−20)(0.8) = −8 − 16 = −24 N·m. R xR = (−10) xR = −24, so xR = 2.4 m. The resultant of 10 N downwards acts 2.4 m from A, beyond the end of the rod at 0.8 m: it is just a line of action, and could be reached by an extension of the rod.
Three downward forces of 10 N, F and 30 N act at x = 0, 1 m and 2 m. Their resultant acts at x = 1.25 m. Find F. R = 40 + F. About A: F × 1 + 30 × 2 = (40 + F) × 1.25, that is F + 60 = 50 + 1.25F, so 10 = 0.25F and F = 40 N. Then R = 80 N.
Class activity
Tape three small coins at different places along a ruler and find the point where the ruler balances on a pencil. Predict it first with the theorem: measure the positions of the coins from one end, multiply each coin’s weight (use 1 unit per identical coin) by its position, add, and divide by the number of coins. Compare with the balance point you find.
Practice
1
Downward forces of 10 N, 20 N and 30 N act at 0, 0.6 m and 1.0 m from A. Find R and where it acts.
R = 60 N downwards. Σ F x = 0 + 12 + 30 = 42, so xR = 42 / 60 = 0.7 m from A.
2
Like forces of 8 N at x = 0.2 m and 12 N at x = 0.7 m. Use the theorem about A to find where the resultant acts.
R = 20 N. 8 × 0.2 + 12 × 0.7 = 1.6 + 8.4 = 10, so xR = 10 / 20 = 0.5 m.
3
16 N acts upwards at x = 0 and 6 N downwards at x = 0.5 m (upwards positive). Find R and xR.
R = 10 N upwards. Moments about A: 16 × 0 + (−6)(0.5) = −3 = 10 xR, so xR = −0.3 m: R acts 0.3 m to the left of A, outside the segment, on the side of the larger force.
4
Downward forces 6 N at x = 0 and F at x = 1.2 m have a resultant acting at x = 0.8 m. Find F.
F × 1.2 = (6 + F) × 0.8, so 1.2F = 4.8 + 0.8F, 0.4F = 4.8 and F = 12 N.
5
Why is the sum of the moments of the forces about the point where R acts equal to zero?
By the theorem it equals the moment of R about that point, and R passes through the point, so its arm and moment are 0.
