☰ 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
1

The conditions of equilibrium of a rigid body under coplanar forces: ΣX = 0, ΣY = 0 and ΣM = 0, and why both the force condition and the moment condition are needed

Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 1
GoalState the three conditions ΣX = 0, ΣY = 0, ΣM = 0 for a rigid body in equilibrium and explain why the force condition and the moment condition are both needed.
New words
rigid body · an object whose shape does not change, like a rod or a plankreaction · the force a support or a wall exerts on the bodyequilibrium · the body stays at rest: it neither moves nor turns
Explanation

A rigid body under coplanar forces is in equilibrium when two conditions hold. (1) The resultant force is zero. Splitting every force into a horizontal part (direction i) and a vertical part (direction j), this means ΣX = 0 and ΣY = 0, where X and Y are the sums of the horizontal and vertical parts. For vertical forces only, ΣY = 0 is the only one that matters. (2) The sum of the moments about one point is zero, ΣM = 0 (anticlockwise positive, clockwise negative). Why do we need both? Two equal and opposite forces on different lines have ΣF = 0, yet they make a couple and the body turns (see the table). A single force through a point O has ΣM = 0 about O, yet it pushes the body along. Only when both conditions are true does nothing happen. A useful habit: take moments about the point where most of the unknown forces act, because a force through the chosen point has no moment and drops out of the equation.

Forces on the rodΣFΣM about OWhat happens
two equal opposite forces, different lines0not 0the rod turns (couple)
one force through Onot 00the rod moves along
both sums are 000equilibrium
Worked examples
20 cm40 cm20 cmRA30 N50 NRBAPQB
A light rod AB, 80 cm long, rests horizontally on supports at A and B. It carries 30 N at 20 cm from A and 50 N at 60 cm from A. Find the reactions RA and RB. Up and anticlockwise are positive. Take moments about A, where RA drops out: −30 × 20 − 50 × 60 + 80 RB = 0, so RB = 3600 / 80 = 45 N. Vertical forces: RA + RB − 30 − 50 = 0, so RA = 35 N. Check with moments about B: 35 × 80 − 30 × 60 − 50 × 20 = 2800 − 1800 − 1000 = 0.
6 cm10 N10 NAB
Is the following in equilibrium? A rod has 10 N up at A and 10 N down at B, 6 cm apart (picture). Up and anticlockwise are positive. ΣY = 10 − 10 = 0, so the first condition is true. Moments about A: 6 × (−10) = −60 N·cm, not zero. So the rod is not in equilibrium: it turns clockwise. Second case: two weights of 10 N hang 0.4 m to the left and 0.4 m to the right of a point O of a rod with nothing supporting it. About O: (−0.4)(−10) + (0.4)(−10) = 0, so the moment condition is true, but ΣY = −20 N, so the rod falls. A support giving 20 N upward at O is needed.
Class activity

Rest a ruler on two paper cups, one under each end, and put one coin 5 cm from the left end. Push down gently on the left cup with a fingertip, then on the right: which cup feels the extra load more? Move the coin to the middle and compare. (Use a coin only; never put anything heavy on a cup you hold.)

Practice
1
Up and anticlockwise are positive. A light rod AB, 60 cm long, rests on supports at A and B and carries 24 N at 15 cm from A. Find RA and RB.
2
Up and anticlockwise are positive. A light rod has 15 N up at A, 25 N up at B (10 cm from A) and a downward force of 40 N at C, AC = x. For which x is the rod in equilibrium?
3
A rod has 30 N up at A and 30 N down at a point 4 cm from A. The forces add up to zero. Is the rod in equilibrium? Find ΣM about A.
4
Why is it a good idea to take moments about a point where an unknown force acts?
5
Take g = 10 N per kg. A uniform rod AB, 1.2 m long and of mass 3 kg, rests horizontally on supports at A and B. Find RA and RB.