Week 6 · Parallel coplanar forces: an unknown force from the resultant, sums of moments of parallel forces, and the resultant of several parallel forces · 3 lessons
2
The sum of the moments of parallel forces about a point, and the resultant of several like forces and where it acts
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 2
GoalAdd up the moments of parallel forces about a point, and find the size and line of action of the resultant of several forces that point the same way.
New words
sum of moments · all the moments added with their signs: anticlockwise positive, clockwise negativetheorem of moments · the sum of the moments of the forces about a point equals the moment of the resultant about that point
Explanation
Take forces that are all perpendicular to a straight rod, and a point O on the rod. Measure the position x of each force along the rod, positive to the right of O. Count a force F as positive when it points up. A force at position x then has the moment x × F about O: an upward force to the right of O turns the rod anticlockwise (positive), and so does a downward force to the left of O. The sum of the moments about O is the sum of all the products x × F. The theorem of moments says this sum equals the moment of the resultant, R × xR, where xR is the position of the line of action of R. Hence xR = (Σ x × F) / R whenever R ≠ 0 (the sign Σ means ‘the sum of’). If all the forces point the same way, R is just the sum of their sizes, and xR is an average of the positions in which each position counts as many times as the force there is large. It therefore lies between the outer forces and nearer the larger forces; with equal forces it is the plain average of the positions. The picture shows four forces: 6 N and 5 N up, 4 N and 8 N down.
Worked examples
For the four forces in the picture (6 N up at A, 4 N down at B, 8 N down at C, 5 N up at D) find the sum of the moments about B and about D, then the resultant. Up is positive, anticlockwise is positive. About B the positions are A: −3, B: 0, C: 5, D: 7, so the sum is (−3)(6) + 0 + 5(−8) + 7(5) = −18 − 40 + 35 = −23 N·cm (clockwise). About D the positions are −10, −7, −2, 0 and the sum is −60 + 28 + 16 + 0 = −16 N·cm. The resultant is R = 6 − 4 − 8 + 5 = −1 N, that is 1 N downwards. From R × xR = −23 we get xR = 23 cm to the right of B, which is 16 cm beyond D: the check about D gives (−1)(16) = −16.
A, B, C, D lie on a line with AB = 2 cm, BC = 3 cm, CD = 5 cm. Parallel forces of 15 N, 25 N, 10 N and 50 N, all pointing down, act at A, B, C, D. Find the resultant and where it acts. R is the sum: 15 + 25 + 10 + 50 = 100 N downwards. Take moments about A; the positions are 0, 2, 5, 10, so the sum of the moments is 0 + 25 × 2 + 10 × 5 + 50 × 10 = 600 in size. Then 100 × AO = 600 and AO = 6 cm. The line of R is between C and D, nearer the big 50 N force.
Class activity
On a 30 cm ruler lay coins in three places: 1 coin at the 0 cm end, 1 coin at the 10 cm mark and 2 coins at the 25 cm mark. Predict the balance point with the average rule, (0 + 10 + 2 × 25) / 4 = 15, then slide the ruler over a pencil to find it. Why is it nearer the 25 cm end than the middle of the ruler?
Practice
1
Three like forces of 12 N, 18 N and 30 N, all pointing up, act at A, B, C on a light rod with AB = 5 cm and BC = 10 cm. Find the resultant and its distance from A.
R = 12 + 18 + 30 = 60 N upwards. Moments about A: 18 × 5 + 30 × 15 = 540, so 60 × xR = 540 and xR = 9 cm from A.
2
Up is positive, anticlockwise is positive. A light rod AB is 40 cm long. It carries 20 N up at A, 25 N down at the point 15 cm from A, and 15 N up at B. Find the sum of the moments about A and about B, then the resultant.
About A: 0 + 15 × (−25) + 40 × 15 = −375 + 600 = 225 N·cm. About B (positions −40, −25, 0): (−40)(20) + (−25)(−25) + 0 = −800 + 625 = −175 N·cm. R = 20 − 25 + 15 = 10 N upwards, acting at 225 / 10 = 22.5 cm from A (check: 10 × (22.5 − 40) = −175).
3
Equal like forces act at A, C, D and B on a line, where AC : CD : DB = 2 : 3 : 5. Find the ratio AO : OB in which the resultant divides AB.
Positions from A in units: 0, 2, 5, 10. For equal forces the resultant acts at the average (0 + 2 + 5 + 10) / 4 = 4.25, so AO = 4.25, OB = 5.75 and AO : OB = 17 : 23.
4
Weights of 2 N, 3 N and 5 N hang from the 0 cm, 10 cm and 30 cm marks of a light rod. At which mark must a single string be tied so that the rod hangs level?
The string pulls up with the resultant of the weights, so it goes at the position of that resultant: (2 × 0 + 3 × 10 + 5 × 30) / 10 = 18 cm.
5
Why must R be different from zero before we write xR = (Σ x × F) / R? What do we have when R = 0 but the sum of the moments is not zero?
We cannot divide by zero, and there is no single force to put at a position. Forces with R = 0 and a moment that is not zero make a couple: they turn the rod without moving it as a whole (Unit 3).
