180 Moments.p65
Transcription
180 Moments.p65
Physics Factsheet www.curriculum-press.co.uk Number 180 Moments A force acting on a body will cause that body to accelerate in a straight line. This is what we tend to confine ourselves to at GCSE level. The unit of the moment of a force is the newton-metre (Nm). Generally the force will be given in newtons. However it may be necessary to convert a mass (in kg) into a weight (in N) using w = mg. a The distance involved may be given in cm (or mm, etc). It may or may not be necessary to change this to metres. Quite often it is possible to work in Ncm throughout the calculation. F However a force acting on a body can also cause a body to rotate about an axis, causing a rotational acceleration. Notice that a moment is a vector. It has a magnitude and direction (clockwise or anticlockwise). α Example 1 O F Moment = (2.5 × 9.81) × 30 × sin120 = 640Ncm or 6.4Nm clockwise (to 2sf). O 30 cm Another factsheet will deal with rotational mechanics (or rotational dynamics). 2.5 kg 1200 In this Factsheet we are going to confine ourselves to the moments caused by a turning force or torque. W Exam Hint:- Sometimes it is useful to redraw the setup, showing just forces and distances. There are two ways that moments can be drawn with the perpendicular distances shown. Both give the same results, so you can use either. There is no need for confusion. Turning force The turning effect of a force about an axis depends on two factors: (a) the magnitude of the force (b) the perpendicular distance between the line of action of the force and the axis. Examples: O d Small torque (F small, d small) F O F sinθ F F Large torque (F large, d large) d (a) Redrawn showing the perpendicular component of the distance, d O (b) Redrawn showing the perpendicular component of the force, F In both cases, the moment equals Fd sinθ. In general you can just use the moment equation without any need for further drawing. F Small torque (F and d not perpendicular) θ d F θ d O d sinθ d Example 2 O Calculate the moment of the force about axis O: Definition of moment of a force: F Moment of force = F × d × sin θ d O 20 “The moment of a force equals the product of the force and the perpendicular distance from the line of action of the force to the axis.” θ 0 75cm Example: Moment = Fd sinθ = 15 × 0.75 × sin70 = 11 Nm anticlockwise. Notice that the angle supplied (200) is not angle θ. Exam Hint:- Take care that the angle you use in your calculation is the angle between the line of action of the force and the line to the axis. 6N 2m F = 15N O Moment = 2 × 6 = 12Nm 1 Physics Factsheet 180. Moments Example 3 Typical Exam Question 1 A cyclist of mass 55kg uses his weight to push down on the pedal at points A, B, and C. F A B C 20 cm 20 cm 450 The long bar shown has a fixed axis at X. A pair of forces is applied as shown. Find the resultant moment about X. 210 cm X 12N F 200 12N 270 cm F Answer: Moment = (12 × 270) – (12 × 210) = 720 Ncm or 7.2 Nm clockwise 20 cm (a) Find the moment applied about the axle in each position. (b) Explain why the greatest turning effect occurs at point B. *Notice that this is exactly the same couple as exerted on the steering wheel, and results in exactly the same turning force about X (as predicted in the keypoint above). The fact that the two forces in the couple work against each other does not affect the resultant moment. Answer: (a) A: moment = (55×9.81)×0.20×sin45 = 76Nm clockwise B: moment = (55×9.81)×0.20 = 110Nm clockwise C: moment = (55×9.81)×0.20×sin70 = 100Nm clockwise Moment is a vector. In calculations it is often convenient to make clockwise moments positive and anticlockwise moments negative, as in this example. *Notice that θ=700 (not 200) at C. (b) Greatest turning effect at B, as W and d unchanged, and sinθ = 1 (perpendicular situation). Principle of Moments Turning forces are often applied in pairs, where the two forces are equal and opposite. The two forces are antiparallel and often act on opposite sides of the axis, so their effects add together. F Often a body is acted on by several forces, each of which can exert a turning force about an axis. If the body is to have no rotational acceleration, these turning effects must balance each other. The statement of the Principle of Moments says: “For a body to be in rotational equilibrium, the total clockwise moment about any axis must equal the total anticlockwise moment about the same axis.” O F Both forces contribute to the clockwise moment about the axis in this situation. Example 4 F1 = 25 N 8.0 cm A simple example is the use of our hands on the steering wheel of a car when making a right turn. O left hand pushes up 6.0 cm F2 Find the value of F2 for equilibrium about axis O. Answer: Clockwise moments = anticlockwise moments 25 × 8.0 = F2 × 6.0 F2 = 33N right hand pulls down We can put some numbers on these forces and distances. For this steering wheel: And in a real-life situation: Example 5 12 N 30 cm A A uniform ladder of mass 20kg and length 8.0m leans against a shiny wall (no friction) at an angle of 250 to the vertical. The bottom of the ladder is fixed to the floor (it cannot slide). Find the force of reaction of the wall against the ladder. 30 cm B C 12 N Moment = (12 × 0.30) + (12 × 0.30) = 7.2 Nm clockwise. R Answer: The weight of the uniform ladder is at its centre. If we forget about this being a steering wheel, and choose point A as a fixed axis: 8.0 m Moment about A = 12 × 0.60 = 7.2Nm clockwise again (the upward force through A has no turning effect about A). Clockwise moment about O = (20 × 9.81) × (4.0 × cos65) = 332Nm. 650 Or if we choose C to be a fixed axis, we will find that the moment about C has the same value once again. Check this yourself. O The total moment of a couple is always the same, no matter what fixed axis you use. 2 W Anticlockwise moment about O = R × (8.0 × sin65) = 7.3R Nm 7.3R = 332, R = 45N Physics Factsheet 180. Moments Practice Questions 3. An accurate balance is used to calculate the mass of an object by balancing moments. A small mass slides along a scale to achieve equilibrium. 1. In which situation, A or B, does the force F exert a greater moment about the axis? A B 10 cm O 13 cm 65 1.50 mm 220 mm 0 2.50g F F M 2. A screwdriver is used to pry off the lid of a paint tin. The balance position is shown. Calculate the unknown mass in kg. (Ignore the mass of the scale itself.) 16.0 cm 4. A van of mass 520 kg has its weight equally shared on each wheel. It is stationary on a small uniform bridge as shown. (The mass of the bridge is 200kg.) 20 N (a) In which direction is the 20N force best applied, assuming the screwdriver is horizontal? (b) What acts as the axis of rotation? (c) Calculate the moment exerted about this axis by the 20N force? (d) If the length of the tip of the screwdriver to the left of the pivot is 0.5cm, what is the upward force on the edge of the lid just before it lifts? A B 1.5 m 2.5 m 3.0 m Find the upward force of the ground on the bridge at A and at B. Notice: (a) The forces on each end of the bridge are different. (b) FA + FB = 2990 + 4080 = 7070 N Weight of bridge + van = 7060 N So the vertical forces on the bridge are in balance (as expected). Take moments about point B (and ignore force acting through B). 7.0FB = (260 × g × 3.0) + (200 × g × 3.5) + (260 × g ×5.5) = 28500 FB = 4080 N 4. Take moments about point A (and ignore force acting through A). 7.0FB = (260 × g × 1.5) + (200 × g × 3.5) + (260 × g × 4.0) = 20,900 FB = 2990 N 3. Principle of Moments 1.50 × M × g = 2.5×10-3 × g ×220 M = 0.37 kg (Notice that g cancels out of the calculation. We don’t need to use it.) The moment is greater in B. B: moment = 13F × sin 65 = 12 Ncm clockwise 1. A: moment = 10F Ncm clockwise Answers 3 2. (a) (b) (c) (d) Vertically downward (perpendicular). The point on the screwdriver touching the top of the wall of the tin. moment = 20 × 16.0 = 320 Ncm clockwise Upward force on lid = Downward force on tip of screwdriver Use Principle of Moments. 20 × 16.0 = F × 0.5, F = 640 N Acknowledgements: This Physics Factsheet was researched and written by Paul Freeman The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136