الأسبوع 11 · الازدواج وعزمه، وكون عزم الازدواج متجهًا ثابتًا، واتزان جسم تحت تأثير عدة ازدواجات · 3 دروس
3
Equilibrium of a body under several couples: the sum of their algebraic moments is zero
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 1
GoalShow that a body acted on by several coplanar couples is in equilibrium when the signed sum of their moments is zero, and use it to find an unknown force or arm.
New words
sum of moments · ΣM = M₁ + M₂ + … with anticlockwise positive and clockwise negativebalance of couples · the body stays at rest when ΣM = 0
Explanation
Every couple has the resultant zero, so any number of couples gives a zero total force: the only thing left that could turn the body is the moments, and they simply add (a couple’s moment is the same about every point, so they can all be taken about one point). A rigid body acted on by couples in one plane is therefore in equilibrium when M₁ + M₂ + … + Mₙ = 0, with signs. Method: (1) give each couple a sign from its turning, anticlockwise +, clockwise −; (2) write F × L for each; (3) put the sum equal to zero and solve for the unknown force or arm. Notice that couples can only be balanced by couples: one single force on top of couples would leave a resultant, and the body would move.
Worked examples
Couple
F
L
Sense
M
1
12 N
0.5 m
clockwise
−6 N·m
2
20 N
0.2 m
anticlockwise
+4 N·m
3
P
0.4 m
anticlockwise
+0.4 P
A light plate has three couples in its plane (anticlockwise positive). Couple 1: forces of 12 N, arm 0.5 m, clockwise. Couple 2: forces of 20 N, arm 0.2 m, anticlockwise. Couple 3: forces of P newtons, arm 0.4 m, anticlockwise. Find P for equilibrium. Solution: M₁ = −12 × 0.5 = −6, M₂ = +20 × 0.2 = +4, M₃ = +0.4P (all in N·m). Equilibrium: −6 + 4 + 0.4P = 0, so 0.4P = 2 and P = 5 N.
A rectangle ABCD has AB = 40 cm and BC = 30 cm. Forces of 10 N act along AB (from A to B) and along CD (from C to D). Forces of Q newtons act along CB (from C to B) and along AD (from A to D). Find Q for equilibrium (anticlockwise positive). Solution: the forces along AB and CD push the bottom side to the right and the top side to the left: anticlockwise, arm BC = 30 cm, M₁ = +10 × 30 = +300 N·cm. The forces along CB and AD push the right side down and the left side up: clockwise, arm AB = 40 cm, M₂ = −40Q. Equilibrium: 300 − 40Q = 0, so Q = 7.5 N.
Class activity
Lay a ruler flat on a table with a coin at each end. Push down gently on the two coins, one with each of two fingers on opposite sides of the middle of the ruler, so that you try to turn it: this is a couple of coins. Add a second couple that tries to turn the ruler the other way (your other two fingers, close together) and find by trial how big the second couple must be so that the ruler does not turn.
Practice
1
A plate has two couples: 30 N with arm 0.2 m anticlockwise, and 20 N with arm 0.15 m clockwise. Find the total moment (anticlockwise positive). Is the plate in equilibrium?
ΣM = 30 × 0.2 − 20 × 0.15 = 6 − 3 = +3 N·m. Not zero, so the plate is not in equilibrium: it turns anticlockwise.
2
A third couple with forces of 5 N is added to the plate of question 1 to bring it to equilibrium. Find its arm and its sense.
It must have moment −3 N·m: clockwise, with 5 × L = 3, so L = 0.6 m.
3
Three couples act on a plate (anticlockwise positive): 8 N with arm 1.5 m anticlockwise, F with arm 0.25 m clockwise, and 6 N with arm 1 m clockwise. Find F for equilibrium.
12 − 0.25F − 6 = 0, so 0.25F = 6 and F = 24 N.
4
Can a single force of 10 N balance a couple? Explain.
No. A couple has resultant zero, so with the extra force the resultant would be 10 N, not zero. Only another couple can balance a couple.
5
In rectangle ABCD, AB = 50 cm and BC = 20 cm. Forces of 6 N act from A to B and from C to D (anticlockwise couple). Forces of P newtons act from C to B and from A to D. Find P for equilibrium (anticlockwise positive).
