Moments of any coplanar system, machines and mixed problems
الأسبوع 4 · تعميم النتيجة على أي نظام من القوى المستوية، وتطبيقات العزوم في الرافعات والقضبان المثبتة، ومسائل متنوعة · 3 دروس
2
Moments in machines: a crane arm, a hinged rod, a spanner and a forearm, finding the arm with trigonometry
Applied Mathematics Sec 3 Term 1, Unit 1, Lesson 1, Part 3
GoalFind the moment of forces in machines: a hinged arm held by a tie, a pedal crank and a hinged rod, by arms from trigonometry or by components.
New words
hinge · a fixed pivot about which a rod or arm can turn; its own reaction has no moment about the hingetie, tension T · a rope or cable holding a part in place; its force acts along the ropecrank · an arm that turns about an axle, such as a bicycle pedal arm
Explanation
Machines are made of arms that turn about a hinge or an axle, and to see what a force does to one we take its moment about the hinge. Two habits make this easy. First, take moments about the hinge itself: the hinge reaction passes through it, has arm 0 and drops out. Second, find each arm with a right-angled triangle, or break the force into components along the arm and across it. A force that acts along a tie has its arm equal to the perpendicular from the hinge to the tie: for a rod of length AB hinged at A and a tie from B to a point C of the wall (AC vertical), the arm of the tension is AB × AC / BC, which is the height of the right-angled triangle on its hypotenuse. For a rod at angle θ to the horizontal and a vertical force at its end, the arm is the horizontal distance, AB cos θ. For a horizontal force at the same end, the arm is the vertical height, AB sin θ. Always say which way each force turns the arm (anticlockwise positive), then add.
Worked examples
A sign arm AB, 1.2 m long, is hinged to a wall at A and held horizontal by a rope from B to the point C of the wall 0.9 m above A. The rope has tension 60 N and a sign of weight 45 N hangs from B. BC = √(1.2² + 0.9²) = 1.5 m. The arm of the tension about A is 1.2 × 0.9 / 1.5 = 0.72 m, and the tension pulls B up and towards the wall: anticlockwise, +60 × 0.72 = +43.2 N·m. The weight acts down at B with arm 1.2 m: −45 × 1.2 = −54 N·m. The sum of the moments about A is −10.8 N·m: the weight wins, so the arm would turn clockwise. (Check by components: the tension is (−48, 36) N and 1.2 × 36 = 43.2.)
A bicycle pedal crank is 17.5 cm long and points forward and up at 40° above the horizontal. The rider pushes vertically down on the pedal with 200 N. Take the axle as O. The pedal is 0.134 m across from O (0.175 cos 40°), so the arm of the vertical force is 0.134 m. The downward push on the right of O turns clockwise: M = −200 × 0.134 ≈ −26.81 N·m. In coordinates: the pedal is at (0.134, 0.112), the force is (0, −200), and M = x Fy − y Fx = 0.134 × (−200) − 0.
A light rod AB, 80 cm long, is hinged at A and rises at 30° above the horizontal. At B a force of 25 N pulls vertically down and a force of 15 N pulls horizontally towards the hinge side (to the left). B is at (0.6928, 0.4) m from A. The vertical force: arm 0.6928 m, clockwise: −17.32 N·m. The horizontal force to the left, at height 0.4 m above A, turns anticlockwise: +15 × 0.4 = +6 N·m. Total: −11.32 N·m, clockwise.
Class activity
Lean a ruler on a book as a hinged arm: hold one end at the edge of the desk as the hinge and hang a small bag (under 0.3 kg) from the other end with a loop of thread, then hold the ruler up with a second thread tied from the bag end to a point above the hinge, taken to your finger. Notice that a steeper thread needs less pull. Never hang weights over anyone’s head.
Practice
1
A rod hinged at A is 1.5 m long and makes 60° with the horizontal. A vertical force of 40 N acts downwards at its upper end. Find the moment about A (anticlockwise positive).
The horizontal distance of the end is 1.5 cos 60° = 0.75 m, and the downward force is to the right of A: M = −40 × 0.75 = −30 N·m.
2
A spanner 0.4 m long has a force of 50 N at its end making 30° with the spanner. Find the size of the moment about the nut.
|M| = 50 × 0.4 × sin 30° = 10 N·m.
3
A forearm is tilted 20° above the horizontal, with the elbow at A and the hand at B 0.28 m away, and it is turned about A. The muscle pulls up at right angles to the forearm with 250 N at a point 0.05 m from A, and a ball of weight 15 N is held in the hand. Find the sum of the moments about A (anticlockwise positive).
Muscle: the force is at right angles to the forearm, so +250 × 0.05 = +12.5 N·m. Ball: weight down, arm 0.28 cos 20° ≈ 0.263 m, clockwise: −15 × 0.263 ≈ −3.95 N·m. Sum ≈ +8.55 N·m.
4
In the sign-arm example, what tension would make the moments about A balance exactly (sum 0)? (AB = 1.2 m, AC = 0.9 m, weight 45 N.)
T × 0.72 = 45 × 1.2 = 54, so T = 75 N.
5
Why does the hinge reaction not appear when we take moments about the hinge?
Its line of action passes through the hinge, so its arm is 0 and its moment is 0.