Cancelling couples, hung bodies and equivalent couples
الأسبوع 12 · إلغاء ازدواج، واتزان جسم معلق يؤثر فيه ازدواج، وتكافؤ الازدواجات وتمارين الدرس · 3 دروس
1
Cancelling a couple: a string tension and another force must form an opposite couple, and finding where the second force acts
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 2
GoalFind the second force that cancels the turning effect of a string pull and an existing couple, and where it must act.
New words
cancel a couple · add forces so that the total force is zero and the total moment is zeropivot (pin) · a support that can give a force through its centre but no turning effect about itself
Explanation
A light rod AB is pivoted at its midpoint O. Two forces at its ends, equal and perpendicular to the rod (one up, one down), form a couple, and it tries to turn the rod. A string also pulls at a point C. We want one more force that makes the rod balance. Reasoning: (1) the end forces already have resultant zero, so the string tension T and the extra force F must also have resultant zero: F is equal to T, opposite and parallel, acting at a different point D of the rod. (2) T and F then form a couple, and its moment must be equal and opposite to the moment of the end couple. (3) The arm of the new couple is the perpendicular distance between the lines of T and F. If CD is the distance along the rod and θ the angle between T and the rod, the arm is d = CD sin θ. (4) Put T × CD × sin θ equal to the moment of the end couple and solve for CD. The pivot gives no force and no turning effect here. Check that D lies on the rod. Sign convention: anticlockwise moments are positive, clockwise moments negative.
Worked examples
A light rod AB, 30 cm long, is pivoted at its midpoint O. Forces of 4 N, perpendicular to the rod, act at A (upward) and at B (downward). A string pulls with T = 12 N at the point C, 5 cm from A, at 30° to the rod, downward and towards B. Find the extra force that makes the rod balance and where it acts (anticlockwise positive). Solution: the end couple has up at the left and down at the right: clockwise, M₁ = −4 × 30 = −120 N·cm. So T and the extra force F must form a couple of +120 N·cm. F = 12 N, parallel to T and opposite, acting at D. Arm = CD sin 30° = 0.5 CD, so 12 × 0.5 CD = 120 and CD = 20 cm. D is 20 cm from C on the side of B (to give anticlockwise), that is 25 cm from A: on the rod. F acts upward and towards A at 30° to the rod.
A rod AB, 50 cm long, pivoted at its midpoint O, has perpendicular forces of 6 N at its ends (a couple of 300 N·cm clockwise). A string pulls with 15 N at C, and the extra force acts at D with CD = 40 cm. Find the angle θ between the string and the rod. Solution: T and F form a couple of +300 N·cm. Its arm is 40 sin θ, so 15 × 40 sin θ = 300, sin θ = 0.5 and θ = 30°. (Check: this is the same as 600 sin θ = 300.) In the figure C is 20 cm from the middle on the A side and D is 20 cm on the B side, so CD = 40 cm.
Class activity
Put a ruler flat on a table with a pencil under its middle as a pivot. Press down on one end and up under the other end with two fingers (a couple) and notice that it turns. Now cancel it with another pair of fingers placed closer to the middle, pressing in opposite directions. Where must the second pair of fingers be, and how hard must they press, compared with the first pair?
Practice
1
A light rod of length 40 cm is pivoted at its midpoint. Perpendicular forces of 5 N at its ends (up at A on the left, down at B on the right) form a couple. A string pulls down with 10 N at C, perpendicular to the rod. An extra force acts at D to balance the rod (anticlockwise positive). Find CD and say on which side of C the point D lies.
The end couple: up on the left, down on the right, clockwise: M = −5 × 40 = −200 N·cm. T and F must give +200: 10 × CD = 200, so CD = 20 cm. Down at the left and up at the right is anticlockwise, so D is on the side of B, with F = 10 N upward.
2
The rod is 30 cm long with 6 N forces at the ends (couple of magnitude 180 N·cm). A string pulls with 9 N at 30° to the rod. Find CD. Can the extra force act on the rod?
9 × CD × sin 30° = 180, so CD = 180 / 4.5 = 40 cm. The rod is only 30 cm long, so C and D cannot both lie on it: not possible with these numbers.
3
For the same rod (30 cm, couple of magnitude 180 N·cm), what is the smallest tension T of a string, at any angle, that can be cancelled by an extra force on the rod?
The arm d = CD sin θ is at most 30 cm (CD at most 30, sin θ at most 1). So T × 30 ≥ 180 and T ≥ 6 N. The smallest is 6 N, with the string perpendicular to the rod and C, D at the two ends.
4
Why must the extra force be equal and opposite to the string tension T?
The end forces already form a couple (resultant zero). For equilibrium the total force must be zero, so T and the extra force must add to zero, which means equal and opposite. They must not be on one line, otherwise they would have no moment to cancel the end couple.
5
A rod 20 cm long, pivoted at its midpoint, has 3 N perpendicular forces at its ends (a clockwise couple). A string pulls perpendicularly with 5 N at C. How far from C (along the rod) must the extra force act?
The end couple is 3 × 20 = 60 N·cm. 5 × CD = 60, so CD = 12 cm.
