Theorem of moments, forces along rectangle sides, and two-point conclusions
Answers are for parents and teachers.
1
Fβ = (4, β2) N, Fβ = (β1, 5) N and Fβ = (2, 2) N act at A(β1, 3). Find the moment of their resultant about B(2, β1) in two ways: the sum of the three moments, and the moment of R (anticlockwise positive).
A from B is (β3, 4). Mβ = (β3)(β2) β 4 Γ 4 = β10; Mβ = (β3)(5) β 4 Γ (β1) = β11; Mβ = (β3)(2) β 4 Γ 2 = β14; sum β35 NΒ·m. R = (5, 5): (β3)(5) β 4 Γ 5 = β35 NΒ·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 Γ 4 = +12. BC through E: 0. DC is 2 cm above E, E on its right: β1 Γ 2 = β2. AD is 10 cm to the left of E and points up: β1 Γ 10 = β10. 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 Γ 7 = +42. DC is 3 cm above E, E on its right: β7 Γ 3 = β21. F acts at B, which is on BC, directly below E at 7 cm, with components (βk, k) where F = kβ2: moment = 0 Γ k β (β7)(βk) = β7k. So 42 β 21 β 7k = 0, k = 3 and F = 3β2 β 4.24 N.
4
F = (2, β4) N acts at A(3, 1). Find its moments about B(0, 0), C(1, β2) and D(6, 2) (anticlockwise positive). What do the three results tell you about the line of action?
About B: 3 Γ (β4) β 1 Γ 2 = β14. About C: A from C is (2, 3), 2 Γ (β4) β 3 Γ 2 = β14. About D: A from D is (β3, β1), (β3)(β4) β (β1)(2) = 14. B and C: equal, so the line is parallel to BC (BC = (1, β2) 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, β1). Its moment about O is +7 NΒ·m and its moment about B(1, 2) is 0. Find F.
Let F = (a, b). About O: 3b β (β1)a = 3b + a = 7. About B: A from B is (2, β3), so 2b β (β3)a = 2b + 3a = 0, a = β2b/3. Then 3b β 2b/3 = 7, 7b/3 = 7, b = 3 and a = β2. F = (β2, 3) N (its line passes through B, as a zero moment says).