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
3
Parallel forces up and down on one line: the magnitude, direction and point of action of the resultant, with a positive direction chosen first
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 2
GoalFind the size, direction and line of action of the resultant of up and down parallel forces, and use a given resultant to find unknown forces.
New words
positive direction · the direction you choose to count as plus; forces the other way get a minus signcouple · two equal and opposite parallel forces whose lines are apart: no resultant force, but a turning effect
Explanation
Step 1: choose a positive direction (here up) and a reference point A on the rod, with positions measured to the right of A. Step 2: add the forces with their signs: R = F₁ + F₂ + …. The sign of R tells its direction. Step 3: add the moments about A: Σ M = x₁F₁ + x₂F₂ + … (ΣM is the sum of moments). Step 4: xR = (Σ M) / R. A positive xR means the line of action of R is to the right of A. This position can lie outside the rod: when up and down forces almost cancel, R is small and acts far away. Check the result by taking moments about a second point. The table shows the three cases that can happen.
What you find
What it means
R ≠ 0
one resultant force; xR = (sum of moments) / R
R = 0 and the sum of moments is not 0
no resultant: the forces make a couple
R = 0 and the sum of moments is 0
the forces are in equilibrium
Worked examples
Up is positive. A light rod carries 40 N down at A, 25 N up at 3 cm from A, 55 N up at 7 cm from A and 20 N down at 10 cm from A. Find the resultant. R = −40 + 25 + 55 − 20 = 20 N upwards. Moments about A: 0 + 3 × 25 + 7 × 55 + 10 × (−20) = 75 + 385 − 200 = 260 N·cm. So xR = 260 / 20 = 13 cm from A: the line of R lies 3 cm beyond the last force. Check about the point at 10 cm: (−10)(−40) + (−7)(25) + (−3)(55) = 400 − 175 − 165 = 60 and 20 × 3 = 60.
Up is positive, all lengths in cm. A light rod AB is 12 cm long. It carries 30 N down at A, an upward force K at C with AC = 3, 30 N down at D with AD = 8 and an upward force F at B. The resultant is 60 N upwards and acts 5 cm from A. Find K and F. Forces: K + F − 30 − 30 = 60, so K + F = 120. Moments about A: 3K + 12F − 8 × 30 = 60 × 5, so 3K + 12F = 540 and K + 4F = 180. Subtract: 3F = 60, so F = 20 N and K = 100 N.
Class activity
Lay a ruler on a table with the 0 cm end on your left. Put a coin on the 5 cm mark and another on the 25 cm mark, and imagine the coins pull DOWN. Now imagine a third force pulling UP at the 15 cm mark with the same strength as one coin. Without doing any weighing, decide the direction of the resultant of the three forces (use the rule R = sum of the signed forces) and say where its line of action is.
Practice
1
Up is positive, anticlockwise is positive. Forces on a light rod AB, 20 cm long: 10 N up at A, 30 N down at 5 cm from A, 40 N up at 12 cm from A, and 20 N down at B. What is the resultant?
R = 10 − 30 + 40 − 20 = 0. The sum of the moments about A is 5 × (−30) + 12 × 40 + 20 × (−20) = −150 + 480 − 400 = −70 N·cm. There is no resultant force: the forces make a couple of 70 N·cm that turns the rod clockwise.
2
Up is positive. Forces on a light rod: 8 N down at A, 15 N up at 2 cm from A, 5 N up at 6 cm from A and 9 N down at 9 cm from A. Find the resultant and its line of action.
R = −8 + 15 + 5 − 9 = 3 N upwards. Moments about A: 0 + 2 × 15 + 6 × 5 + 9 × (−9) = 30 + 30 − 81 = −21. So xR = −21 / 3 = −7 cm: R acts 7 cm to the left of A, outside the rod.
3
Up is positive. A light rod carries 20 N down at A, 35 N up at 4 cm from A and 25 N down at 10 cm from A. Find the size, direction and line of action of the resultant.
R = −20 + 35 − 25 = −10 N: 10 N downwards. Moments about A: 4 × 35 + 10 × (−25) = 140 − 250 = −110. xR = −110 / −10 = 11 cm from A, 1 cm beyond the 25 N force.
4
Up is positive, lengths in cm. A light rod AB is 30 cm long. It carries 20 N up at A, an unknown force F down at 12 cm from A and 50 N up at B. Their resultant is 40 N upwards. Find F and the distance of the line of R from A.
20 − F + 50 = 40 gives F = 30 N. Moments about A: 12 × (−30) + 30 × 50 = −360 + 1500 = 1140, so xR = 1140 / 40 = 28.5 cm from A.
5
True or false: for up and down forces on a rod, the resultant always acts between the outer forces. Give an example.
False. In the first example of this lesson the resultant acts 3 cm beyond the last force, and in practice item 2 it acts 7 cm to the left of A. Only when all the forces point the same way must R lie between the outer forces.
