Theorem of moments, forces along rectangle sides, and two-point conclusions
Answers are for parents and teachers.
1
F₁ = (4, −2) N, F₂ = (−1, 5) N and F₃ = (2, 2) N act at A(−1, 3). Find the moment of their resultant about B(2, −1) in two ways: the sum of the three moments, and the moment of R (anticlockwise positive).
A from B is (−3, 4). M₁ = (−3)(−2) − 4 × 4 = −10; M₂ = (−3)(5) − 4 × (−1) = −11; M₃ = (−3)(2) − 4 × 2 = −14; sum −35 N·m. R = (5, 5): (−3)(5) − 4 × 5 = −35 N·m. They agree.
2
Rectangle ABCD has AB = 10 cm and BC = 6 cm. Forces 3 N, 2 N, 1 N and 1 N act along AB, BC, DC and AD (A to B, B to C, D to C, A to D). Show that the resultant passes through E on BC with BE = 4 cm, and find its size.
Moments about E (anticlockwise positive): AB is 4 cm below E, E on its left: +3 × 4 = +12. BC through E: 0. DC is 2 cm above E, E on its right: −1 × 2 = −2. AD is 10 cm to the left of E and points up: −1 × 10 = −10. Total 0, so R passes through E. R = (3 + 1, 2 + 1) = (4, 3), so R = 5 N.
3
Square ABCD has side 10 cm. Forces 6 N along AB (A to B), 7 N along DC (D to C) and F along BD (from B to D) have a resultant through E on BC with BE = 7 cm. Find F (anticlockwise positive).
Moments about E: AB is 7 cm below E: +6 × 7 = +42. DC is 3 cm above E, E on its right: −7 × 3 = −21. F acts at B, which is on BC, directly below E at 7 cm, with components (−k, k) where F = k√2: moment = 0 × k − (−7)(−k) = −7k. So 42 − 21 − 7k = 0, k = 3 and F = 3√2 ≈ 4.24 N.
4
F = (2, −4) N acts at A(3, 1). Find its moments about B(0, 0), C(1, −2) and D(6, 2) (anticlockwise positive). What do the three results tell you about the line of action?
About B: 3 × (−4) − 1 × 2 = −14. About C: A from C is (2, 3), 2 × (−4) − 3 × 2 = −14. About D: A from D is (−3, −1), (−3)(−4) − (−1)(2) = 14. B and C: equal, so the line is parallel to BC (BC = (1, −2) is parallel to F). B and D: opposite, so the line passes through the midpoint of BD, which is (3, 1) = A.
5
A force F acts at A(3, −1). Its moment about O is +7 N·m and its moment about B(1, 2) is 0. Find F.
Let F = (a, b). About O: 3b − (−1)a = 3b + a = 7. About B: A from B is (2, −3), so 2b − (−3)a = 2b + 3a = 0, a = −2b/3. Then 3b − 2b/3 = 7, 7b/3 = 7, b = 3 and a = −2. F = (−2, 3) N (its line passes through B, as a zero moment says).