The moment of a force: turning effect, coordinates and Varignon
Answers are for parents and teachers.
1
A spanner 30 cm long is pushed at its end with 45 N, at 40° to the spanner. Find the arm and the size of the moment about the nut. Give the moment in N·m.
L = 0.3 sin 40° ≈ 0.1928 m; |M| = 45 × 0.1928 ≈ 8.68 N·m.
2
F = (−5, 12) N acts at A(−2, 3). Find its moment about O and about B(1, −1), and the arm about O (anticlockwise positive).
About O: M = (−2) × 12 − 3 × (−5) = −24 + 15 = −9 N·m. About B: A − B = (−3, 4), M = (−3) × 12 − 4 × (−5) = −36 + 20 = −16 N·m. F = 13 N, so the arm about O is 9/13 ≈ 0.692 m.
3
A force (k, 4) N acts at (3, 2) m and its moment about O is +7 N·m. Find k (anticlockwise positive).
M = 3 × 4 − 2k = 12 − 2k = 7, so k = 2.5.
4
A force of 120 N, 25° below the horizontal and pointing right, acts at P(0.6, 0.9) m. Find its moment about O (anticlockwise positive) in two ways: by components, and by the arm.
Components: Fx ≈ 108.76, Fy ≈ −50.71; M = 0.6 × (−50.71) − 0.9 × 108.76 ≈ −128.3 N·m. Arm: L = 0.6 sin 25° + 0.9 cos 25° ≈ 1.0692 m, and 120 × 1.0692 ≈ 128.3 N·m, clockwise.
5
A force of 10 N acts at (4, 3) and points from (4, 3) towards O, so F = (−8, −6). Show with the formula (anticlockwise positive) that its moment about O is zero, and explain in words.
M = 4 × (−6) − 3 × (−8) = −24 + 24 = 0. The line of action runs straight through O, so the arm is 0.