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# Newton's Laws

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A set of slides created to teach Newton's Laws to learners at Bishops Diocesan College in Cape Town.

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### Transcript of "Newton's Laws"

1. 1. PHYSICAL SCIENCE Newton 1 & 2 GRADE 12 K WARNE
2. 2. Newton’s First Law <ul><li>” A body will................................ of ...... or ................. .................... in a straight line, unless acted on by a ......................................&quot; </li></ul><ul><li>This law describes - ................ </li></ul><ul><li>IMPORTANT POINTS: (A moving object…) </li></ul><ul><li>Continues in its state </li></ul><ul><li>Rest or uniform motion in a straight line </li></ul><ul><li>Unless acted upon by (external) forces </li></ul>
3. 3. Newton’s First Law <ul><li>” A body continues in its state of rest or of uniform motion in a straight line , unless acted on by a resultant force .&quot; </li></ul><ul><li>This law describes - Inertia. </li></ul><ul><li>IMPORTANT POINTS: (A moving object…) </li></ul><ul><li>Continues in its state </li></ul><ul><li>Rest or uniform motion in a straight line </li></ul><ul><li>Unless acted upon by (external) forces </li></ul>
4. 4. Inertia <ul><li>A body has a natural tendency to resist any changes to its state of motion. </li></ul><ul><li>This resistance is known as Inertia . </li></ul><ul><li>Examples: </li></ul><ul><ul><li>What will happen if the card in the picture is flicked ? Explain why. </li></ul></ul><ul><ul><li>The moon moves around the sun in a circular orbit . Explain why. </li></ul></ul>Every object in a state of uniform motion tends to remain in that state of motion unless an external (unbalanced)force is applied to it.
5. 5. Inertia <ul><li>A body has a ............................................ any changes to its state of motion. </li></ul><ul><li>This resistance is known as ....................... . </li></ul><ul><li>If the card in the picture is flicked ................. ....................................... . Inertia keeps the peg stationary when the card is moved quickly . </li></ul><ul><li>The peg’s Inertia is overcomes .................. ................ which try to keep it’s position on the card. </li></ul><ul><li>The moon was moving past the earth in a straight line but became ................. by the ............................ </li></ul><ul><li>Gravity does not act against the direction of motion (90 o ) so the motion continues because ............................................... to .................... . </li></ul><ul><li>(The question is who threw it in the first place!) </li></ul>Every object in a state of uniform motion tends to remain in that state of motion unless an external (unbalanced)force is applied to it. Gravity Motion Motion Friction Inertia Earth 1. 2.
6. 6. Inertia <ul><li>A body has a natural tendency to resist any changes to its state of motion. </li></ul><ul><li>This resistance is known as Inertia . </li></ul><ul><li>If the card in the picture is flicked the peg should fall into the glass . Inertia keeps the peg stationary when the card is moved quickly . </li></ul><ul><li>The peg’s Inertia is overcomes the friction forces which try to keep it’s position on the card. </li></ul><ul><li>The moon was moving past the earth in a straight line but became trapped by the earth’s gravity. </li></ul><ul><li>Gravity does not act against the direction of motion (90 o ) so the motion continues because there is no unbalanced force to stop it . </li></ul><ul><li>(The question is who threw it in the first place!) </li></ul>Every object in a state of uniform motion tends to remain in that state of motion unless an external (unbalanced)force is applied to it. Gravity Motion Motion Friction Inertia Earth 1. 2.
7. 7. Exam Question <ul><li>The diagram shows a truck with a crate immediately behind the cab in position A. </li></ul><ul><li>The truck suddenly accelerates forward and the crate slides towards the back of the truck and comes to rest in position (B) </li></ul><ul><li>Draw a force diagram, indication and labeling the horizontal force(s) acting on the crate while the truck accelerates. (2) </li></ul><ul><li>Briefly explain why the crate slides towards the back of the truck. (4) </li></ul><ul><li>Name and state the law or principle that you have applied in order to reach an answer in (b). (4) </li></ul><ul><li>WCED Nov 2001 TOTAL [10] </li></ul>A B
8. 8. Exam Question <ul><li>The diagram shows a truck with a crate immediately behind the cab in position A.The truck suddenly accelerates forward and the crate slides towards the back of the truck and comes to rest in position (B) </li></ul><ul><li>Draw a force diagram, indication and labeling the horizontal force(s) acting on the crate while the truck accelerates. </li></ul><ul><li>(2) </li></ul>Friction between the crate and truck A B
9. 9. Exam Question <ul><li>Briefly explain why the crate slides towards the back of the truck. </li></ul><ul><li>According to Newtons 1st law the crate will tend to maintain its state of rest . Due to the friction between the crate and truck the crate does experience a forward acceleration while the truck is accelerating. When the truck reaches a constant velocity, the friction will bring the crate to rest (relative to the truck). (4) </li></ul><ul><li>Name and state the law or principle that you have applied in order to reach an answer in (b). (4) </li></ul><ul><li>WCED Nov 2001 TOTAL [10] </li></ul>Friction between the crate and truck A B
10. 10. Newton’s Laws Second Law <ul><li>A resultant force on an object causes it to accelerate in the direction of that force. </li></ul>F res = F + f and F res = m x a These are the two most basic scenarios - any of the variables can be calculated if all the others are known. F m a F m T W F res = W + T = ma And W = mg
11. 11. Newton’s - Connected objects <ul><li>In both these cases the tension gets larger as more blocks are connected I.e. </li></ul><ul><li>T 1 < T 2 < T 3 </li></ul><ul><li>The systems can be treated as single units: </li></ul><ul><li>F res = m total a = (m 1 +m 2 +m 3 )a </li></ul><ul><li>Where </li></ul><ul><li>F res = T 3 + f or F res = T 3 + F g </li></ul><ul><li>And </li></ul><ul><li>T 3 + f( or w ) = (m 1 +m 2 +m 3 )a </li></ul>To find the tension between blocks you must consider their individual masses I.e. T 2 + w = (m 1 +m 2 )a and T 1 + w = m 1 a T 1 T 2 T 3 m 1 m 2 m 3 f m 3 m 2 m 1 T 3 T 2 T 1 w
12. 12. Newton’s Laws - Pully systems The accelerations of these systems can be easily found by considering them as a unit: F res = m total a where m total = m 1 +m 2 … And F res = w 1 - w 2 (always opposed) To find tension; T + w 1 = m 1 a or T + w 2 = m 2 a Individual masses or tensions can be found by considering masses individually: T 1 + w 1 = m 1 a or T 2 - w 2 = m 2 a Tension in any one string is the same throughout! w 1 w 2 m 2 m 1 T T m 1 m 2 m 3 T 1 T 2 w 1 w 2
13. 13. Examples Newton 1 & 2 <ul><li>1. A 25kg container is acted upon by a horizontal force of 65N. The frictional force between the container and the floor is 15N. Calculate the acceleration of the container. </li></ul>a = 2m.s -2
14. 14. Examples Newton 1 & 2 <ul><li>A ball with a mass of 18kg is suspended by a string. </li></ul><ul><ul><li>Calculate the tension in the string if the ball is </li></ul></ul><ul><ul><li>stationery, </li></ul></ul><ul><ul><li>accelerates upwards at 3m.s -2 ; </li></ul></ul><ul><ul><li>accelerates downwards at 3m.s -2 . </li></ul></ul>T = 180N upwards T = 234 upwards T = 126N upwards
15. 15. Examples Newton 1 & 2 <ul><li>Two blocks rest on a frictionless horizontal surface connected by a string. A force of 80N is exerted on the 7kg block. Calculate: </li></ul><ul><ul><li>the acceleration of the blocks </li></ul></ul><ul><ul><li>the force exerted by the string on the 13 kg block. </li></ul></ul>
16. 16. Examples Newton 1 & 2 <ul><li>4. A constant force of 120N is used to lift two masses of 4 kg and 6kg which are attached to each other with string. Ignore any mass of the string and calculate the acceleration of the system and the tension in the string. </li></ul>4kg 6kg 120 N
17. 17. Newton’s 3rd Law <ul><li>&quot;For every action there is an equal and opposite reaction .” </li></ul>........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ..............................................................................................................................................................................................................................................................................................................................
18. 18. Newton’s 3rd Law <ul><li>&quot;For every action there is an equal and opposite reaction .” </li></ul>The gasses experience a force backwards out of the rocket, this has an equal but opposite reaction force which drives the rocket forward! The force exerted by the hand on the head is the same magnitude as the force exerted by the head on the hand! The head hits the hand just as hard as the hand hits the head!
19. 19. Newton’s 3rd BOB (who isn't very bright) is tired of pushing his van to the petrol station. He gets an idea -- &quot;Hey! I go much faster when I am roller-blading than when I am walking. Why don't I wear my roller-blades!&quot; So BOB gets his roller blades from the back of his van, and starts pushing. Will this work? Explain using Newton's laws what will happen. (Calculate the acceleration and displacement each will experience. BOB weighs 50 kg, and the van weighs 2,000 kg the force is exerted for 1second.)
20. 20. Newton’s 3rd ANS : Since every action has an equal and opposite reaction, when BOB pushes the van, the van pushes BOB with the same size force but opposite direction. BOB weighs 50 kg, and the van weighs 2,000 kg. Taking right as positive: a bob = F/m =(-100/50)= -2m/s 2 s bob = ut + 1 / 2 at 2 =0(1)+1/2(-2)(1)= -1m a van = F/m =(100/2000)=0.05m/s 2 s van = ut+ 1 / 2 at 2 =0(1)+1/2(0.05)(1)= +0.025m Questions >> Questions>>
21. 21. Spot the odd one(s) out.
22. 22. Spot the odd ones out. “ Shooting - Inertia - 1st Newton - 1st. Falling - Gravity! “ Flying” -Inertia - 1st Orbit - Inertia - 1st Falling - Gravity!
23. 23. <ul><li>Instructions </li></ul><ul><li>Explanations </li></ul>
24. 24. Weight pull <ul><li>Rest or constant velocity - tension cannot overcome friction + inertia. </li></ul><ul><li>Acceleration - tension is greater than friction + inertia. </li></ul><ul><li>Tension is greatest just before acceleration. </li></ul>
25. 25. Suspended Weight <ul><li>Pulling slowly the top string breaks - </li></ul><ul><ul><li>The weight begins to move as the string stretches </li></ul></ul><ul><ul><li>Inertia + tention at the botom is greater than the tention.at the top. </li></ul></ul><ul><li>When the string is jerked it breaks below the weight </li></ul><ul><li>The wieght wants to remain stationary inertia acts with the top string and is greater than tention at the bottom. </li></ul>
26. 26. Accelerometer - air <ul><li>Accelerating </li></ul><ul><ul><li>- the ball “moves” in the opposite direction to the acceleration. </li></ul></ul><ul><ul><li>The ball has a greater inertia than the air in the flask which tends to remain its in its position of rest until the unbalanced force is applied through the string or side of flask. </li></ul></ul><ul><li>Constant velocity or at rest - the ball hangs vertically. </li></ul><ul><ul><li>The ball continues in its state of rest or constant velocity as it does not experience any unbalanced force. </li></ul></ul>
27. 27. Accelerometer - water <ul><li>Accelerating </li></ul><ul><li>The ball moves in the SAME direction as the motion - the water has a greater inertia and tends to remain in its position of rest (back of flask) forcing the ball to the opposite side of the container. </li></ul><ul><li>Rest or constant velocity </li></ul><ul><li>The water and ball remain in the neutral position as they do not experience any unbalanced force. </li></ul>water
28. 28. Swinging object
29. 29. Gravitation and Falling Bodies W = mg V = (-) a = -10m/s 2 <ul><li>The only force on the object is its weight. </li></ul><ul><li>The object continually accelerates downwards . </li></ul><ul><li>Take the upwards as positive , acceleration downwards negative . </li></ul><ul><li>Velocity momentarily zero at apex. </li></ul>W = mg V = (+) a = -10m/s 2 V=(0), a = -10m/s 2 An object is thrown vertically up into the air and allowed to fall back down.
30. 30. Gravitation and Falling Bodies W = mg V = (-) a = -10m/s 2 W = mg V = (+) a = -10m/s 2 V=(0), a = -10m/s 2 An object is thrown vertically up into the air and allowed to fall back down. t distance . . . . . . . t s . . . .
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