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M1: Chapter 5 Moments

M1: Chapter 5 Moments. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Learning Objectives: Understand what is meant by the moment of a force. Use moments to solve problems involving unknown lengths, forces and masses. Last modified: 2 nd April 2014. Intro. This is a door.

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M1: Chapter 5 Moments

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  1. M1: Chapter 5Moments Dr J Frost (jfrost@tiffin.kingston.sch.uk) • Learning Objectives: • Understand what is meant by the moment of a force. • Use moments to solve problems involving unknown lengths, forces and masses. Last modified: 2nd April 2014

  2. Intro This is a door If I double the distance of my finger from the hinge, what happens to the force required to keep the door open? As the distance doubles, the force required halves (we’ll see why). Why do you think the handle is put on the other side of the door from the hinge? Increasing the distance of the force applied from the point of rotation increases the ‘turning effect’ of the force. ? ?

  3. Moments m = 70kg 10m Moment ? ! The ‘moment’ of a force… … measures the turning effect of the force on the body on which it is acting. ? perpendicular distance about a point

  4. Moments - Examples 1 2 3 We see what force is acting perpendicularly. Moment of force F about point P clockwise Moment of force F about point P There is no ‘turning’ effect. Moment of force F about point P clockwise ? ? ? 4 5 Moment: Moment: ? ?

  5. Summing moments We can also find the overall moment by summing them – just treat one of the directions (clockwise or anticlockwise) as negative. 2 1 Sum of moments (about ): ? Sum of moments (about ) clockwise 3 ? (Rod is light) Sum of moments (about ): ? 4 Sum of moments (about ): ?

  6. This whole chapter in a nutshell… m = 70kg 2m 10m ! If a rigid body is in equilibrium then: The resultant force in any direction is 0. The sum of the moments about any point is 0. i.e. Forces up = forces down a b i.e. Clockwise moments = anticlockwise moments

  7. Example Lewis and Tom are on a uniform seesaw of mass 20kg, where the pivot is not central. Lewis weighs 70kg and is 10m from the pivot. Tom is 2m from the pivot. The seesaw is horizontal. Determine the reaction force at the pivot of the seesaw. Determine how heavy Tom is. m = 70kg m = ? Lewis Tom 2m 10m Bro Tip: First draw on forces. Click to Brosketch Hint: Choose a suitable point to take moments about. Method 1: Equating moments about pivot: Equating forces in vertical direction, Method 2: Equating moments about Tom: Equating forces in vertical direction: ? ?

  8. Moments about which point? Key Point: You can choose different points to take your moments about, depending on which variable we wish to ‘ignore’. m = 70kg m = ? Lewis Tom 2m 10m If we take moments about the pivot:We don’t have to worry about since it doesn’t have any moment about the pivot. ? If we take moments about Tom:We don’t have to worry about Tom’s mass because Tom’s weight doesn’t contribute to the moment about Tom. ?

  9. Test Your Understanding Q A uniform rod of mass 20kg is supported at each end so that it is horizontal. The bar is 4m long and a mass of 10kg is placed 1m from one end. Find the magnitude of the reactions at each support. ? Diagram ? Working Moments about second support: Similarly taking moments about first:

  10. Finding centre of mass Q A non-uniform rod of length 4m and mass 5kg is in equilibrium in horizontal position resting on two supports at C and D where and . The magnitude of the reaction at is twice the magnitude of the reaction at . Find the distance of the centre of mass of the rod from . ? Diagram ? Working Let centre of mass be metres from . Forces up = forces down: Moments about :

  11. Exam Questions On provided hand-out. (Solutions on following slides)

  12. Answers ?

  13. Answers Hint: When the beam is on the verge of tilting about P, there is no reaction force at Q. ? ? ?

  14. Answers ? ?

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  18. Summary C LO: Understand what is meant by the moment of a force. • Moment of a force • What are the two crucial things to consider for systems in equilibrium? • Forces up = Forces down • Moments clockwise = Moment anticlockwise • Where tends to be best to take moments about? • When there is an unknown distance:Sometimes about the point where the distance is measured from (but if in doubt, take from the end). • When there are two unknown forces:About the point where one of the forces is acting, so that we don’t consider this force in our moments equation and thus have less unknowns. • How do we deal with rods with unknown centres of mass? • Create a distance say for the centre of mass. Usually best to find moments about the end of the rod. • If a rod is on two supports P and Q, and is on the verge of tilting about P? • There is no reaction force at Q. ? ? C LO: Solve problems involving moments. ? ? ? ?

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