Cancelling couples, hung bodies and equivalent couples
Answers are for parents and teachers.
1
A rod AB, 36 cm long, is pivoted at its midpoint and has perpendicular forces of 7 N at its ends (a couple of magnitude 252 N·cm). A string pulls with 14 N. (a) If it is perpendicular to the rod, find CD. (b) If it makes 30° with the rod, find CD. What does the answer tell you about the positions of C and D?
(a) 14 × CD = 252, CD = 18 cm. (b) 14 × CD × 0.5 = 252, CD = 36 cm: the whole length of the rod, so C must be at one end and D at the other end.
2
A uniform rod AB, 50 cm long and weighing 12 N, is hinged at A and hangs in a vertical plane under an anticlockwise couple of 90 N·cm. Find the angle between AB and the downward vertical (nearest 0.1°) and the other angle that satisfies the equation.
AG = 25 cm, so 12 × 25 × sin θ = 90, sin θ = 0.3, θ ≈ 17.5°. The other angle is 180° − 17.5° = 162.5° (unstable).
3
A uniform square lamina of side 20 cm and weight 40 N hangs from a smooth pin at its corner A, under an anticlockwise couple of 400 N·cm in its plane. Find the angle between AC and the downward vertical.
AG = 10√2 cm. 40 × 10√2 × sin θ = 400, so sin θ = 1/√2 and θ = 45° (the stable one).
4
A couple has forces of 30 N and arm 0.6 m, anticlockwise. Find the forces of an equivalent couple with arm 0.4 m and of one with arm 0.9 m.
M = +18 N·m. F = 18 / 0.4 = 45 N, and F = 18 / 0.9 = 20 N.
5
A couple is F = (−8, 6) N at A(3, −2) and −F at B(−1, 2) (m). Find its moment (anticlockwise positive), its arm, and the forces of an equivalent couple with arm 2 m (same sense).
The vector from B to A is (4, −4): M = 4 × 6 − (−4)(−8) = 24 − 32 = −8 N·m (clockwise). |F| = 10 N, so the arm is 8 / 10 = 0.8 m. With arm 2 m: F = 8 / 2 = 4 N, clockwise.