☰ خطة العام · الرياضيات التطبيقية (بالإنجليزية)

The two conditions of equilibrium: boards, rods and strings

الأسبوع 8 · اتزان مجموعة من القوى المستوية: شرطا الاتزان، ولوح أو قضيب على حاملين، وقضيب معلق بخيطين · 3 دروس
2

A board or rod on two supports: the two reactions from moments about one support, and what happens as the load moves

Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 1
GoalFind the two reactions of a board or rod resting on two supports by taking moments about one support, and see what happens as the load moves.
New words
uniform · the weight acts at the midpointweight and mass · weight in newtons = mass in kg × 10 (g = 10). Egyptian books also write it as kg.wt
Explanation

A plank AB rests on supports at A and B, and a load stands on it. The reactions RA and RB push up, and the weights pull down. Take g = 10 N per kg, so a mass of 15 kg weighs 150 N (the Egyptian books write ‘kg.wt’ for this: we always use newtons). A uniform plank has its weight at the midpoint. Method: (1) ΣY = 0 gives RA + RB = total weight. (2) Take moments about B, where RB drops out: RA × AB = Σ weight × its distance from B. (3) Subtract to get RB, and check by moments about A. The table shows a 5 m plank of weight 150 N with a child of weight 300 N standing x metres from A: the support nearer the child carries more, and as the child walks from A to B, RA falls and RB rises by the same amount, 60 N for each metre.

x, m012345
RA, N37531525519513575
RB, N75135195255315375
Worked examples
1.5 m1 m2.5 mRA300 N150 NRBACGB
A uniform board AB of mass 15 kg and length 5 m rests on supports at its ends. A child of mass 30 kg stands 1.5 m from A. Take g = 10. Up and anticlockwise are positive. Weights: board 15 × 10 = 150 N at 2.5 m from A, child 30 × 10 = 300 N at 1.5 m from A. Moments about B: 5 RA − 150 × 2.5 − 300 × 3.5 = 0, so RA = (375 + 1050) / 5 = 285 N. Then RB = 450 − 285 = 165 N. Check about A: 5 × 165 − 150 × 2.5 − 300 × 1.5 = 825 − 375 − 450 = 0. The child is nearer A, and RA is the bigger reaction.
x1 m2 m1 mRAbox 240 Nplank 240 NRCAGCB
A uniform plank AB, 6 m long and weighing 240 N, rests on a support at A and a support at C, where C is 1 m from B. A box of weight 240 N is placed on the plank at x metres from A. Where must it be put so that the two reactions are equal? Up and anticlockwise are positive. Equal reactions share the total: RA = RC = (240 + 240) / 2 = 240 N. Moments about C, where AC = 5 m. RA is up and 5 m to the left of C, which turns the plank clockwise (negative). The plank’s weight (2 m to the left of C) and the box (5 − x to the left) are down on the left, which turns it anticlockwise (positive). So −5 × 240 + 2 × 240 + (5 − x) × 240 = 0, so 5 − x = 3 and x = 2 m from A.
Class activity

Lay a ruler on two kitchen scales, one under each end, and zero both scales. Put a coin at different marks and read the two scales, if they show grams: the scale nearer the coin shows more. Loads under 0.5 kg only. Predict by moments how the readings share a 20 g coin placed 10 cm from one end of a 30 cm ruler of negligible weight. Then test.

Practice
1
Use the board of the example (5 m, weight 150 N, supports at the ends A and B) with the child of weight 300 N now 4 m from A. Up and anticlockwise are positive. Find RA and RB.
2
A light plank AB, 4 m long, rests on end supports and carries 200 N at 1 m from A and 100 N at 3 m from A. Up and anticlockwise are positive. Find RA and RB.
3
Take g = 10. A uniform plank AB of mass 20 kg and length 4 m rests on end supports. A box of mass 30 kg stands 1 m from B. Up and anticlockwise are positive. Find RA and RB in newtons.
4
A person walks along a plank resting on two end supports from A to B. What happens to RA and RB?
5
A uniform plank AB, 6 m long and weighing 180 N, rests on a support at A and a support at C, with AC = 4 m. Up and anticlockwise are positive. Find the two reactions.