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© Boardworks Ltd 2003. A slide contains teacher’s notes wherever this icon is displayed - To access these notes go to ‘Notes Page View’ (PowerPoint 97)

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Presentation on theme: "© Boardworks Ltd 2003. A slide contains teacher’s notes wherever this icon is displayed - To access these notes go to ‘Notes Page View’ (PowerPoint 97)"— Presentation transcript:

1 © Boardworks Ltd 2003

2 A slide contains teacher’s notes wherever this icon is displayed - To access these notes go to ‘Notes Page View’ (PowerPoint 97) or ‘Normal View’ (PowerPoint 2000). Normal ViewNotes Page View Teacher’s Notes Flash Files A flash file has been embedded into the PowerPoint slide wherever this icon is displayed – These files are not editable.

3 © Boardworks Ltd 2003 Moments

4 © Boardworks Ltd 2003 Force and rotation 5N A force acting on an object can cause it to turn about a pivot. What would happen to the see-saw above ? Would it turn? If so, clockwise or anti-clockwise? pivot

5 © Boardworks Ltd 2003 Force and rotation The left goes down - an anticlockwise turn. pivot A turning force is called a moment.

6 © Boardworks Ltd 2003 Moments Suppose you were trying to unscrew a nut using a spanner. The spanner exerts a moment or turning force on the nut. If the moment is big enough it will unscrew the nut. If not there are 2 ways of increasing the moment. Distance from force to pivot Pivot Force

7 © Boardworks Ltd 2003 Increasing the moment 1. Increase the distance from the force to the pivot - apply the force at the end or use a longer spanner. Distance from force to pivot Pivot Force

8 © Boardworks Ltd Increase the force applied - push/pull harder or get someone stronger to do it! Distance from force to pivot pivot Force Increasing the moment

9 © Boardworks Ltd 2003 Moment Moment = Force (N) x Distance (cm or m). The moment of a force is given by the relationship: Moments are measured in Newton centimetre (Ncm) or Newton metre (Nm). moment Fxd

10 © Boardworks Ltd 2003 pivot 500 N 0.5 m Gina weighs 500 N and stands on one end of a seesaw. She is 0.5 m from the pivot. What moment does she exert? moment = 500 x 0.5 = 250 Nm Click for solution Moments calculation

11 © Boardworks Ltd 2003 Principle of moments The green girl exerts an anti-clockwise moment equal to... her weight x distance from pivot. The yellow girl exerts a clockwise moment equal to... her weight x distance from pivot. pivot

12 © Boardworks Ltd 2003 If the two moments are equal then the seesaw is balanced. This is known as the principle of moments. When balanced Total clockwise moment = total anti-clockwise moment “c.m.” = “a-c.m.” Principle of moments pivot

13 © Boardworks Ltd 2003 The principle of moments can be investigated using 10g masses with this balance. moment (left) = 10 x 7 = 70 gcm moment (right) = (10 x 3) + (10 x 4) = 70 gcm Both moments are equal and so the seesaw is balanced. Principle of moments

14 © Boardworks Ltd 2003 Why don’t cranes fall over? Tower cranes are essential at any major construction site. load arm trolley loading platform tower Concrete counterweights are fitted to the crane’s short arm. Why are these needed for lifting heavy loads? counterweight

15 © Boardworks Ltd 2003 Why don’t cranes fall over? Using the principle of moments, when is the crane balanced? moment of = moment of load counterweight If a N counterweight is 3 metres from the tower, what weight can be lifted when the loading platform is 6 metres from the tower? 6 m6 m 3 m3 m N ?

16 © Boardworks Ltd 2003 Why don’t cranes fall over? moment of counterweight distance of counterweight from tower = = x 3 = Nm counterweight x moment of load = = ? x 6 load x distance of load from tower moment of load = moment of counterweight ? x 6 = ? = ? = 5000 N

17 © Boardworks Ltd 2003 Crane operator activity Where should the loading platform be on the loading arm to carry each load safely?

18 © Boardworks Ltd 2003 Moments activity

19 © Boardworks Ltd 2003 Using moments in calculations 1. Two girls are on a seesaw. One weighs 200N and is 1.5m from the pivot. Where must her 150N friend sit if the seesaw is to balance ?. Click for solution At balance total “c.m.” = total “a-c.m.” 200 x 1.5 = 150 x distance 200 x 1.5 = distance 150 distance = 2 m “c.m.” = clockwise moment “a-c.m.” = anti-clockwise moment

20 © Boardworks Ltd 2003 Pressure and Hydraulics

21 © Boardworks Ltd 2003 Pressure Pressure is exerted whenever a force is applied over an area. Which one exerts the biggest pressure, 1 or 2? 1. 2.

22 © Boardworks Ltd 2003 Case 1. The arm applies a force onto a board via a finger tip. The force applied produces a high pressure because the force acts over a small area. 1.

23 © Boardworks Ltd Case 2. The arm applies the same force onto the board. The force is now acting over a larger area - the area of the palm is greater than the finger tip. Thus, a lower pressure is produced.

24 © Boardworks Ltd 2003 Pressure is measured in: Newtons per metre squared (N/m 2 ) which is called a PASCAL (Pa) Pressure can also be measured in: Newtons per millimetre squared (N/mm2); Newtons per centimetre squared (N/cm 2 ). Pressure = Area Force P x A F Pressure is the force per unit area so is calculated using the expression shown below: Pressure

25 © Boardworks Ltd 2003 The same force spread over a big area means low pressure. Which shoes would you choose for walking over a muddy field?

26 © Boardworks Ltd 2003 The boots on the right spread the weight over a larger area. Therefore, the pressure exerted on the ground is low. In contrast, fashion shoes have a smaller area and exert a higher pressure. These shoes are likely to sink into soft ground.

27 © Boardworks Ltd 2003 A force spread over a large area means low pressure, e.g. skis and snowboards. Application of pressure The large surface area of the board means the boy exerts very little pressure on the snow. He therefore slides over the top and does not sink in.

28 © Boardworks Ltd 2003 A force concentrated on a small area means high pressure, e.g. razor blades, needles, high heeled shoes, spurs, ice skates, sharp knives. Application of pressure On the cutting edge of a knife a very high pressure is exerted - this makes it easier to cut. The high pressure on the cutting edge of an ice-skate melts the ice and helps the skater slide across the surface.

29 © Boardworks Ltd 2003 Pressure in liquids In a liquid: Pressure acts in all directions and pressure increases with depth.

30 © Boardworks Ltd 2003 High pressure low pressure The relationship between pressure and depth is shown by a water bottle with holes along its length. Pressure (N/m 2 ) = 10 N/Kg x depth (m) x density (Kg/m 3 ) The pull of gravity The deeper you go, the higher the pressure The denser the liquid, the heavier it is!

31 © Boardworks Ltd 2003 Hydraulics Hydraulic systems use the principle that pressure is transmitted throughout a liquid. They are used to transfer movement from one part of a machine to another without linking them mechanically. All hydraulic systems use two pistons linked via a pipe carrying a special oil called hydraulic fluid. Force applied here Force transferred here Pressure inside all parts of the hydraulic system is the same

32 © Boardworks Ltd 2003 Hydraulics brakes All hydraulic brake systems (e.g. in a car) use a small master piston and a bigger slave piston. The master piston is used to apply a force. This puts the liquid under pressure. The pressure is transmitted to the pistons on all four wheels. Pressure = Force applied Area master piston

33 © Boardworks Ltd 2003 The slave piston always has a much larger area than the master piston. The force exerted can be calculated from the same equation: So, a greater force is exerted by the brakes than the driver exerted on the pedal. Pressure = Force exerted Area slave piston Force exerted = Pressure x Area slave piston Much larger than master piston

34 © Boardworks Ltd 2003 The hydraulic brake Foot pedal Master piston Slave pistons drum Friction shoes Hydraulic fluid

35 © Boardworks Ltd 2003 The hydraulic brake - example The car master piston has an area of 5cm 2. If a force of 10N is applied to it, calculate the pressure created in the brake pipes. If the slave piston has an area of 50 cm 2, calculate the force exerted on the brake disc. At the master piston, P=F/A= 10/5 = 2 N/cm 2 At the slave piston, F= PxA =2x50 = 100 N (10 times the original force applied to the master piston). Click for solution

36 © Boardworks Ltd 2003 Hydraulics activity


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