الأسبوع 11 · الازدواج وعزمه، وكون عزم الازدواج متجهًا ثابتًا، واتزان جسم تحت تأثير عدة ازدواجات · 3 دروس
1
Why a zero resultant does not always mean no motion: the couple, its arm and its moment M = F × L, constant when the force and the arm change together
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 1
GoalRecognise a couple, find its arm and its moment M = F × L with a sign, and use the fact that the moment stays the same when the force and the arm change together.
New words
couple · two forces of equal size, opposite directions and different lines of actionarm of a couple · the perpendicular distance between the two lines of action
Explanation
Push the rim of a tap with one finger and it turns, but it also tries to slide sideways. Push with two fingers on opposite sides, with equal pushes in opposite directions, and it only turns. Two forces of equal size, opposite directions and different lines of action are called a couple. Their resultant is zero (F + (−F) = 0), so there is no push in any direction, yet the body is not in equilibrium: this is why Week 8 needed the moment condition as well as the force condition. The arm L of the couple is the perpendicular distance between the two lines of action. Its moment is M = F × L, written with the sign of the turning: anticlockwise positive, clockwise negative. Why F × L? Take moments about a point on the line of action of one force: that force has no moment, and the other force F, with arm L, gives F × L. (Next lesson shows that every point gives the same answer.) If the force doubles and the arm halves, M does not change, so the same turning effect can be made with a small force and a long arm. The unit is N·m (or N·cm).
Worked examples
Two vertical forces of 15 N act on a flat plate, in its plane: one upward on the right and one downward on the left, with their lines of action 0.4 m apart. Find the moment of the couple (anticlockwise positive). Solution: the resultant is 15 − 15 = 0, but the plate turns. The arm is L = 0.4 m. The right side is pushed up and the left side down, which turns the plate anticlockwise, so M = +15 × 0.4 = +6 N·m.
Arm L
0.3 m
0.4 m
0.5 m
0.6 m
Force F
40 N
30 N
24 N
20 N
M = F × L
12 N·m
12 N·m
12 N·m
12 N·m
A bolt needs a turning effect of 12 N·m (anticlockwise). Find the force of each of the two pushes of a couple when the arm is 0.3 m, 0.4 m, 0.5 m and 0.6 m. Solution: F = M / L, so 12 / 0.3 = 40 N, 12 / 0.4 = 30 N, 12 / 0.5 = 24 N, 12 / 0.6 = 20 N (table). A longer arm needs a smaller force. If only 15 N is available, the arm must be 12 / 15 = 0.8 m.
Class activity
Put a bottle cap on a table. Turn it with two fingertips on opposite sides of the rim, pushing in opposite directions (a couple). Then turn it with one fingertip on the rim only. What else does the single finger do to the cap? Now try the couple with the two fingertips close to the middle and then at the rim: with which arm is it easier to turn?
Practice
1
A couple has two forces of 24 N and their lines of action are 0.25 m apart. The forces turn the body anticlockwise. Find the moment (anticlockwise positive).
M = +24 × 0.25 = +6 N·m.
2
A couple of moment −90 N·cm (clockwise) has forces of 15 N. Find its arm.
L = |M| / F = 90 / 15 = 6 cm.
3
A couple of two 25 N forces with arm 8 cm is to be replaced by a couple of two 10 N forces with the same moment and the same sense. What arm is needed?
M = 25 × 8 = 200 N·cm, so L = 200 / 10 = 20 cm.
4
Two horizontal forces of 7 N act along two parallel lines 0.6 m apart: the upper force points left and the lower force points right. Find the moment (anticlockwise positive).
The upper part is pushed left and the lower part right, which is anticlockwise: M = +7 × 0.6 = +4.2 N·m.
5
Why is the resultant of a couple zero, and why does the body still turn?
The two forces are equal and opposite, so they add to zero and there is no push in any direction. But each force turns the body the same way about the middle of the arm, so the two moments add instead of cancelling.