Review: parallel forces and their resultant, and equilibrium of boards, rods and ladders under coplanar forces
Applied Mathematics Sec 3 Term 1, Units 1–3, Review
GoalReview parallel forces and their resultant, and the equilibrium of boards, rods and ladders under coplanar forces, with new problems.
New words
equilibrium of a rigid body · ΣX = 0, ΣY = 0 and ΣM = 0 (moments about a convenient point, anticlockwise positive)about to tip or slip · tip: the reaction at the other support is zero; slip: friction F = μN
Explanation
Weeks 5 to 10 in one page. Like parallel forces F₁ at A and F₂ at B: R = F₁ + F₂ acts at C between them with F₁ × AC = F₂ × BC (C is nearer the larger force). Unlike forces (F₁ > F₂): R = F₁ − F₂ in the direction of the larger, acting at C outside AB on the side of the larger force, with F₁ × AC = F₂ × BC. For equilibrium of a rigid body under coplanar forces: ΣX = 0, ΣY = 0 and ΣM = 0, taking moments about the point where most unknown forces act. A board on two supports: moments about one support give the other reaction; the board is about to tip when the reaction at the other support is zero. A ladder on rough ground against a smooth wall: the wall reaction S is horizontal, N equals the total weight, friction F = S, and F ≤ μN (about to slip when F = μN); moments about the foot find S. Remember the units: forces in N, with g = 10 so a mass of m kg weighs 10m N (a kg.wt is 10 N).
Worked examples
A uniform plank 6 m long, of mass 12 kg, rests on supports at its two ends A and B, with g = 10. A child of mass 30 kg stands 4.5 m from A. Find the reactions. Solution: weights: plank 120 N at the middle (3 m from A), child 300 N at 4.5 m from A. Moments about A (anticlockwise positive; RB upward at 6 m): 6 RB − 120 × 3 − 300 × 4.5 = 0, so 6 RB = 360 + 1350 = 1710 and RB = 285 N. Vertical: RA + RB = 420, so RA = 135 N.
A uniform ladder of mass 20 kg and length 5 m leans against a smooth wall, its foot 3 m from the wall (top 4 m up) on rough ground with μ = 0.5. A painter of mass 40 kg climbs it (g = 10). How far along the ladder can he go? Solution: W = 200 N, P = 400 N. N = 600 N, so the greatest friction is 0.5 × 600 = 300 N, and S = F = 300 N. Moments about the foot A (anticlockwise positive): the wall reaction S pushes left at 4 m up: +4S. The weights push down at horizontal distances 1.5 m and 3f m, where f is the fraction of the length he has climbed: −200 × 1.5 − 400 × 3f. So 4 × 300 − 300 − 1200 f = 0, so 1200 f = 900 and f = 0.75. He can climb 0.75 × 5 = 3.75 m along the ladder.
Class activity
Place a ruler leaning on a stack of books against a wall of books (the book wall is the smooth wall). Put a small eraser near the top of the ruler and tilt it to see when it slips. Explain with F ≤ μN why the ruler slips sooner when the eraser is higher, and what changes if you put the foot of the ruler on a cloth.
Practice
1
Unlike parallel forces of 9 N (at A) and 15 N (at B), 12 cm apart, act on a light rod. Find the size and direction of R and where it acts (distance from the larger force).
R = 15 − 9 = 6 N in the direction of the 15 N force. Let C be x cm beyond B: 9 × (12 + x) = 15 × x gives 108 = 6x and x = 18 cm from the 15 N force, outside the segment on its side.
2
Weights of 40 N and 60 N hang downward from the ends of a light rod 90 cm long. Where must it be supported to balance?
40 × AC = 60 × BC with AC + BC = 90: AC = 54 cm from the 40 N end (36 cm from the 60 N end).
3
A uniform beam AB, 5 m long and weighing 240 N, rests on supports at A and at C, where AC = 4 m. A load of 160 N hangs down from B. Find the reactions (upward) at A and C.
Moments about A (anticlockwise positive): 4 RC − 240 × 2.5 − 160 × 5 = 0, so RC = 1400 / 4 = 350 N. RA = 400 − 350 = 50 N.
4
A uniform plank AB, 4 m long and weighing 200 N, rests on supports at A and at C, with AC = 3 m. How far from A can a person of weight 400 N stand, beyond C, before the plank is about to tip about C?
About to tip means RA = 0. Moments about C: the plank’s weight is 1 m on the A side of C, the person is d beyond C: 200 × 1 = 400 × d, so d = 0.5 m. The person can stand 3.5 m from A.
5
A uniform ladder 13 m long and weighing 240 N rests with its foot 5 m from a smooth wall (top 12 m up) on rough ground. With nobody on it, find the wall reaction (horizontal) and the least μ for it not to slip.
Moments about the foot: 12 S = 240 × 2.5, so S = 50 N. N = 240 N and F = S = 50 N, so μ ≥ 50 / 240 = 5/24 ≈ 0.21.
