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
2
Rods and boards with loads: unknown forces and positions from figures, then the exercises on a motorcycle, a wedge with a string, strings with a maximum tension and two carriers
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 2
GoalFind unknown forces and positions on loaded rods and boards in equilibrium, including a wheeled load, a rod on one support with a string, and strings with a greatest tension.
New words
greatest tension · the largest pull a string can take before it breaksslack string · a string whose tension has just become zero
Explanation
Every problem of this lesson is solved by the same three moves. (1) Draw the rod, mark every force with its direction (up arrows for supports and strings, down arrows for weights) and the distances. (2) Write ΣY = 0. (3) Take moments about the point where the most unknown forces act, with anticlockwise positive, and solve. Some cases to know. A wheeled load (a bicycle, a trolley) is a plank on two supports, the wheels. A rod resting on one support and held level by a vertical string: the string is a second support that pulls up. The string is about to go slack when its tension is zero. For a greatest tension, write the tension as a formula in the unknown position and use the condition ‘tension at most the greatest value’, which gives a range of positions.
Situation
Treat it as
Take moments about
bicycle or trolley
a plank on two wheel supports
one wheel
rod on a support with a vertical string
two upward forces: reaction and tension
the support
string about to go slack
the tension is 0
the other support
Worked examples
A light rod rests horizontally on two supports. K acts upwards at A, 20 N downwards 6 cm from A, F upwards 24 cm from A, and 30 N downwards at B, 28 cm from A. Find F and K. Up and anticlockwise are positive. About A: 24F − 6 × 20 − 28 × 30 = 0, so 24F = 960 and F = 40 N. Vertical forces: K + 40 − 20 − 30 = 0, so K = 10 N. Check about B: −28 × 10 + 22 × 20 − 4 × 40 = −280 + 440 − 160 = 0.
A bicycle’s wheels touch the ground 100 cm apart. The bicycle weighs 150 N, acting midway between the wheels, and the rider weighs 600 N, acting 60 cm behind the front wheel. Find the ground reaction on each wheel. Up and anticlockwise are positive. About the front wheel: 100 Rrear − 150 × 50 − 600 × 60 = 0, so Rrear = 43500 / 100 = 435 N and Rfront = 750 − 435 = 315 N. The rear wheel carries more because the rider sits nearer to it.
Class activity
Balance a ruler on a pencil at its middle. Now slide the pencil 5 cm towards one end and hold the ruler level with a fingertip pushing up (or down) at the far end. Does the finger push up or down, and is the push big or small compared with the ruler’s weight? Predict first by moments about the pencil.
Practice
1
A uniform rod AB, 90 cm long and weighing 36 N, rests on a support 30 cm from A and is held horizontal by a light vertical string at B. Up and anticlockwise are positive. Find the tension and the reaction of the support. Then find the weight that, hung at A, makes the string slack.
About the support S: the weight is 15 cm to the right, B is 60 cm to the right: 60T − 15 × 36 = 0 gives T = 9 N and RS = 36 − 9 = 27 N. String slack means T = 0: 30w = 15 × 36 gives w = 18 N (and then RS = 54 N).
2
A uniform rod AB, 80 cm long and weighing 24 N, hangs level from vertical strings at A and at C, where AC = x cm. A load of 36 N hangs 20 cm from A. Each string can bear at most 40 N. Find the range of x and the least and greatest tension in each string. Up and anticlockwise are positive.
About A: xT_C = 24 × 40 + 36 × 20 = 1680, so TC = 1680 / x and TA = 60 − 1680 / x. TC ≤ 40 needs x ≥ 42; TA ≤ 40 is true when x ≤ 84, always true on the rod (x ≤ 80). So 42 ≤ x ≤ 80. TC goes from 40 N (x = 42) down to 21 N (x = 80). TA goes from 20 N up to 39 N.
3
Two workers carry a load of 480 N hung from a light pole. They stand 2 m apart, and the load hangs x m from worker A. Worker B can carry at most 300 N. Up and anticlockwise are positive. Find the greatest x, and A’s share then.
About A: 2 RB = 480x, so RB = 240x. 240x ≤ 300 gives x ≤ 1.25 m. Then RA = 480 − 300 = 180 N.
4
A bicycle of weight 100 N has wheels 100 cm apart and its weight acts midway. A rider of weight 500 N sits so that his weight acts 60 cm behind the front wheel. Find the reaction on each wheel.
About the front wheel: 100 Rrear = 100 × 50 + 500 × 60 = 35000, so Rrear = 350 N and Rfront = 600 − 350 = 250 N.
5
A light rod on two supports has 20 N down at 8 cm and 20 N down at 28 cm from A. F acts up at 20 cm from A and K up at A. Find F and K. Up and anticlockwise are positive.
About A: 20F = 20 × 8 + 20 × 28 = 720, so F = 36 N. Then K = 40 − 36 = 4 N.
