Cancelling couples, hung bodies and equivalent couples
Week 12 · Couples: cancelling a couple, a hanging body with a couple acting on it, equivalent couples and the exercises of Lesson 3-1 · 3 lessons
2
A body hung from a pin with a couple acting on it: the weight and the pin’s reaction form a couple that balances it, giving the hanging position
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 2
GoalFind the position in which a body hung from a smooth pin rests when a couple acts on it, using the couple formed by the weight and the pin reaction.
New words
hung from a pin · the body can turn freely about the pin A, which gives a reaction R through Aweight at G · W acts downward at the centre G of a uniform body (the middle of a rod, the centre of a square or a disc)
Explanation
A flat body hangs from a smooth pin at A, in a vertical plane, and a couple of moment M acts on it in its plane. It turns until it rests. In that position the forces on it are the weight W at G, the pin reaction R at A, and the couple. Total force zero gives R = W upward. Now R (up, line through A) and W (down, line through G) are equal and opposite, so they form a couple. Their lines are vertical lines AG sin θ apart, where θ is the angle between AG and the downward vertical. So the weight and the reaction form a couple of size W × AG × sin θ, and it must be equal and opposite to the applied couple: W × AG × sin θ = |M|. Therefore sin θ = |M| / (W × AG). Two angles θ and 180° − θ fit: G below A (stable, the usual hanging position) or G above A (balance, but it falls away at the slightest push). If |M| > W × AG no angle works and the body keeps turning. For an anticlockwise applied couple, the weight must turn clockwise, so G rests to the right of A.
Worked examples
A uniform rod AB, 60 cm long and of mass 0.8 kg, is hinged smoothly at A and hangs in a vertical plane. A couple of 120 N·cm (anticlockwise) acts on it. Take g = 10. Find the angle between AB and the downward vertical, and the hinge reaction. Solution: the weight is 0.8 × 10 = 8 N and AG = 30 cm. W × AG × sin θ = 120 gives 8 × 30 sin θ = 120, so sin θ = 0.5 and θ = 30° (the other solution, 150°, is the unstable position with the rod pointing up). The reaction is R = W = 8 N, upward. The rod rests with G to the right of A.
A uniform square lamina ABCD of side 40 cm and weight 50 N hangs from a smooth pin at the corner A. An anticlockwise couple of 1000 N·cm acts in its plane. Find θ, the angle between AC and the downward vertical, and the greatest couple it could balance. Solution: AG = half the diagonal = 20√2 cm. 50 × 20√2 × sin θ = 1000 gives sin θ = 1/√2, so θ = 45° (or 135°, unstable). The greatest couple is for sin θ = 1: 50 × 20√2 = 1000√2 ≈ 1414 N·cm.
Class activity
Cut a card square and make a small hole near one corner. Hang it on a pencil held horizontally (a loose fit). Where does the corner-to-corner line from the hole hang when nothing else acts? Now hold the card by its middle with a finger of the other hand and turn it a little: the card tries to swing back. How is this swing-back like the couple of the weight and the pin reaction?
Practice
1
A uniform rod 40 cm long and weighing 6 N is hinged at one end and hangs in a vertical plane with an anticlockwise couple of 72 N·cm. Find θ from the downward vertical (nearest 0.1°) and the hinge reaction.
AG = 20 cm, so 6 × 20 × sin θ = 72, sin θ = 0.6, θ ≈ 36.9°. The reaction is 6 N upward.
2
What is the greatest anticlockwise couple the rod of question 1 can balance, and what is then its position?
W × AG × sin θ is greatest for sin θ = 1: 6 × 20 = 120 N·cm, with the rod horizontal (θ = 90°).
3
A uniform circular disc of radius 20 cm and weight 15 N hangs from a smooth pin at a point A on its rim. A couple of 150 N·cm acts in its plane (anticlockwise). Find the angle between AG and the downward vertical.
G is the centre, AG = 20 cm. 15 × 20 × sin θ = 150, so sin θ = 0.5 and θ = 30° (the stable one).
4
True or false: a hung body can rest with G exactly below the pin only when no couple acts on it. Explain.
True. G below A means θ = 0, so sin θ = 0 and the couple W × AG × sin θ of the weight and the reaction is zero, so there can be no applied couple.
5
Why are there two angles θ and 180° − θ, and which one do you see in practice?
sin θ = sin (180° − θ). In the first G is below the pin and the body returns if pushed slightly (stable). In the second G is above the pin and the body falls away at the slightest push, so we only see the first.
