Moments of any coplanar system, machines and mixed problems
Answers are for parents and teachers.
1
Forces F₁ = (1, 2) N at (0, 1) and F₂ = (−3, 1) N at (2, 0) act on a plate. Find the sums of their moments about B(1, 1) and about C(−1, 4) (anticlockwise positive) and say what they show about the resultant.
About B: (0,1) − B = (−1, 0): (−1)(2) − 0 = −2; (2,0) − B = (1, −1): 1 × 1 − (−1)(−3) = −2; sum −4 N·m. About C: (0,1) − C = (1, −3): 1 × 2 − (−3)(1) = 5; (2,0) − C = (3, −4): 3 × 1 − (−4)(−3) = −9; sum −4 N·m. The sums are equal and R = (−2, 3) ≠ 0, so its line of action is parallel to BC = (−2, 3).
2
A boom AB, 1.2 m long, is hinged at A on a wall and held horizontal by a rope from B to a point C of the wall 0.5 m above A. The tension is 65 N and a load of 30 N hangs from B. Find the arm of the tension and the sum of the moments about A (anticlockwise positive).
BC = √(1.2² + 0.5²) = 1.3 m. Arm = 1.2 × 0.5 / 1.3 ≈ 0.4615 m. Tension: +65 × 0.4615 = +30 N·m. Load: −30 × 1.2 = −36 N·m. Sum = −6 N·m.
3
A forearm is tilted 25° above the horizontal, elbow at A, hand at B 0.30 m from A. The muscle pulls at right angles to the forearm with 300 N at a point 0.04 m from A, and the hand holds a ball of weight 12 N. Find the sum of the moments about A (anticlockwise positive) and say which way the forearm tends to turn.
Muscle: +300 × 0.04 = +12 N·m. Ball: −12 × 0.30 cos 25° ≈ −3.26 N·m. Sum ≈ +8.74 N·m, anticlockwise: the forearm tends to lift (the muscle wins).
4
F₁ = (a, −1) N acts at (1, 3) and F₂ = (2, b) N acts at (−1, 3). The sum of their moments about O is zero, and so is the sum about B(−1, 2). Find a and b, and describe the line of action of the resultant.
About O: 1 × (−1) − 3a + (−1) b − 3 × 2 = −7 − 3a − b = 0. About B: (1,3) − B = (2, 1): 2 × (−1) − 1 × a = −2 − a; (−1,3) − B = (0, 1): 0 × b − 1 × 2 = −2; sum −4 − a = 0, so a = −4. Then b = −7 + 12 = 5. F₁ = (−4, −1), F₂ = (2, 5), R = (−2, 4). The line of action passes through both O and B, so it is the line OB: direction (−1, 2), parallel to R.
5
A nail puller needs a moment of 18 N·m about its pivot and its handle is 0.30 m long. Find the least force and its direction, and the force needed if it can only make 60° with the handle.
Least force: 18 / 0.30 = 60 N, at right angles to the handle. At 60°: 18 / (0.30 sin 60°) ≈ 69.28 N.