Non-uniform boards, loaded rods and a bar on a wedge
Week 9 · Equilibrium of coplanar forces: a non-uniform board, rods and loads with unknown forces, and a bar on rough ground and a smooth wedge · 3 lessons
1
A non-uniform board on two supports: its weight and the place its weight acts from the two positions at which a man makes it just tip
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 2
GoalFind the weight and the line of action of the weight of a non-uniform board from the two positions at which a person makes it just tip.
New words
non-uniform · its weight does not act at the midpoint, so we call the distance x and find itabout to tip · the board is just about to turn over a support; the other support’s reaction has just become zero
Explanation
A board AB rests on two supports C and D. A uniform board has its weight at the midpoint, but a non-uniform one has two unknowns: its weight w and the distance x of its line of action from one end. The trick is the idea ‘about to tip’. If a person walks beyond support D towards B, the board is about to turn about D at the moment the reaction at C becomes zero. Then only three forces act, the weight, the person and the reaction at D, and the moments about D must cancel. In the same way, when the person walks beyond C towards A, the board is about to turn about C and the reaction at D is zero. Each position gives one equation with w and x; two positions give two equations. Dividing, or adding the two moment equations, gives w. The sign convention in the sums below: moments are written as magnitudes, the person’s moment equal to the board’s moment about the same support.
Worked examples
A non-uniform board AB is 6 m long and rests on supports at C and D, with AC = 1 m and AD = 4 m. A man of weight 400 N walks along it. He can go as far as 4.9 m from A before the board tips about D, and as far as 0.4 m from A (going the other way) before it tips about C. Find the weight w of the board and the distance x of its line of action from A. About to tip about D: RC = 0, and moments about D give 400 × (4.9 − 4) = w × (4 − x), so 360 = w(4 − x) (1). About to tip about C: RD = 0 and 400 × (1 − 0.4) = w × (x − 1), so 240 = w(x − 1) (2). Adding: 600 = 3w, so w = 200 N. From (2): x − 1 = 1.2, so x = 2.2 m from A. Check (1): 200 × 1.8 = 360.
For the same board (w = 200 N acting 2.2 m from A) can the 400 N man stand safely at the end A? Up and anticlockwise are positive. Take moments about C, with the man 1 m to the left of C and the weight 1.2 m to the right. Suppose the board does not tip: 3RD + (−1)(−400) + (1.2)(−200) = 0, that is 3RD + 400 − 240 = 0 and RD = −53.3 N. A support cannot pull down, so RD = −53.3 N is impossible: the board tips about C. The man must stay at least 0.4 m from A.
Class activity
Rest a ruler on two erasers placed 10 cm apart on a desk, with a little of the ruler hanging over each eraser. Place a coin on the overhanging end on one side and slide it out until the ruler just starts to tip: note the distance beyond the eraser. Do the same on the other side. If the two distances are different, where do you think the ruler’s weight acts? (Coins only; keep your fingers clear.)
Practice
1
A non-uniform plank AB, 5 m long, has weight 150 N acting 2 m from A. It rests on supports at C and D with AC = 1 m and AD = 4 m. What is the greatest weight P that can be put at the end B before the plank tips about D? Moments are taken as magnitudes about D.
About D: P × 1 = 150 × (4 − 2) = 300, so P = 300 N (RC is then zero).
2
For the same plank, can a child of weight 300 N stand at the end A? Up and anticlockwise are positive; take moments about C.
The child is 1 m left of C and the weight is 1 m right of C: 3RD + 300 − 150 = 0 gives RD = −50 N, impossible. The plank tips about C, so the child cannot stand at A.
3
A non-uniform board rests on supports C and D, 2 m apart. A 300 N man can stand 0.6 m beyond D (before it tips about D) and 0.4 m beyond C on the other side (before it tips about C). Find the board’s weight and the distance of its line of action from C.
With y the distance from C: 300 × 0.6 = w(2 − y) and 300 × 0.4 = w y. Adding: 300 = 2w, so w = 150 N, and y = 120 / 150 = 0.8 m from C.
4
At the moment a board is about to tip about D, what is the reaction at the other support C, and why?
It is zero: the board is just lifting off C, so C no longer pushes on it.
5
A board of weight 200 N has a man of weight 400 N standing where it is about to tip about D. Find RD.
RC = 0, so RD = 200 + 400 = 600 N by the vertical balance.
