Chap 3 3a to 3d

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Chap 3 3a to 3d

  1. 1. Introduction • This chapter teaches you how to deal with forces acting on an object • You will learn how to use several formulae (inspired by Isaac Newton) • You will learn how to model situations involving friction, particles moving on slopes and when joined over pulleys • You will also learn laws of momentum and impulse
  2. 2. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… You need to understand all the forces at work in various situations… R The Normal Reaction The normal reaction acts perpendicular to the surface which an object is resting on Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma Force is measured in Newtons (N). A Newton is: “The force that will cause a mass of 1kg to accelerate at 1ms-2” mg (mass x gravity) It is equal and opposite to the force exerted on the surface by the object, which is determined largely by gravity and the mass of the object The table matches the force from the brick, which is why the brick remains still on the table (there of course would be a maximum possible weight the table could take, but we will not worry about this for now! 3A
  3. 3. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma Force is measured in Newtons (N). A Newton is: “The force that will cause a mass of 1kg to accelerate at 1ms-2” You need to understand all the forces at work in various situations… Frictional Force Direction of motion Frictional Force The frictional force opposes motion between two ‘rough’ surfaces Although it is a force, friction does not cause movement in its own direction. It just reduces the effect of another force Surfaces will have a maximum level of friction where it is unable to completely prevent movement 3A
  4. 4. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration You need to understand all the forces at work in various situations… Tension in string Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma Force is measured in Newtons (N). A Newton is: Tension If an object is being pulled along (for example by a string), then the force acting on the object is called the Tension Tension = PULLING force “The force that will cause a mass of 1kg to accelerate at 1ms-2” 3A
  5. 5. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma Force is measured in Newtons (N). A Newton is: You need to understand all the forces at work in various situations… Thrust Thrust If an object is being pushed along (for example by a rod), then the force acting on the object is called the Thrust (or sometimes compression) Tension = PUSHING force “The force that will cause a mass of 1kg to accelerate at 1ms-2” 3A
  6. 6. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma Force is measured in Newtons (N). A Newton is: “The force that will cause a mass of 1kg to accelerate at 1ms-2” You need to understand all the forces at work in various situations… Resistance Any object moving through air, fluid or a solid will experience resistance caused by the particles in the way Gravity Gravity is the force between any object and the earth. The Force caused by gravity acting on an object is its weight Remember Newton’s formula…  The Force is called the weight  Mass is just mass!  The acceleration due to gravity is 9.8ms 2 (or can be left as ‘g’ 3A
  7. 7. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this chapter… Newton’s second law of motion “The force needed to accelerate a particle is equal to the product of the mass of the object and the acceleration required” F = ma You need to understand all the forces at work in various situations… Resolving When there are multiple forces acting on an object, we ‘resolve’ these forces in different directions One direction will usually be the direction of acceleration The other will be perpendicular to this Force is measured in Newtons (N). A Newton is: “The force that will cause a mass of 1kg to accelerate at 1ms-2” 3A
  8. 8. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Normal R Reaction Tension mg (mass x gravity) Frictional Force Direction of motion Thrust 3A
  9. 9. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Find the weight in Newtons, of a particle of mass 12kg The mass is already in kg, and use acceleration due to gravity Calculate As the acceleration was given to 2sf, you should give you answer to the same accuracy  Ensure you use the exact amount in any subsequent calculations though! 3A
  10. 10. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Find the acceleration when a particle of mass 1.5kg is acted on by a force of 6N Sub in F and m Divide by 1.5 3A
  11. 11. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Find the values of the missing forces acting on the object in the diagram below In this example you need to consider the horizontal forces and vertical forces separately (This is called resolving) Resolving Horizontally Take the direction of acceleration as the positive one Sub in values. You must subtract any forces acting in the opposite direction! 2ms-2 Y Calculate Add 4 4N 2kg X Resolving Vertically Take the direction of the force Y as positive 2g N Sub in values. Acceleration is 0 as there is none in the vertical direction Calculate Add 2g 3A
  12. 12. Dynamics of a Particle moving in a Straight Line You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration Find the values of the missing forces acting on the object in the diagram below In this example you need to consider the horizontal forces and vertical forces separately (This is called resolving) Resolving Horizontally Take the direction of acceleration as the positive one Sub in values. You must subtract any forces acting in the opposite direction! 2ms-2 Y Calculate 20N 80N 4kg Add X and Subtract 8 X Resolving Vertically Take the direction of the force Y as positive 4g N Sub in values. Acceleration is 0 as there is none in the vertical direction Calculate Add 20, add 4g 3A
  13. 13. Dynamics of a Particle moving in a Straight Line a ms-2 You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary A particle of mass 5kg is pulled along a rough horizontal table by a force of 20N, with a frictional force of 4N acting against it. Given that the particle is initially at rest, find: a)The acceleration of the particle b)The distance travelled by the particle in the first 4 seconds c)The magnitude of the normal reaction between the particle and the table Start by drawing a diagram R 4N 5kg 20N 5g N a) Resolve horizontally and sub in values. Take the direction of acceleration as positive Calculate a 3B
  14. 14. Dynamics of a Particle moving in a Straight Line 3.2ms-2 You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary A particle of mass 5kg is pulled along a rough horizontal table by a force of 20N, with a frictional force of 4N acting against it. Given that the particle is initially at rest, find: a)The acceleration of the particle – 3.2ms b)The distance travelled by the particle in the first 4 seconds c)The magnitude of the normal reaction between the particle and the table -2 Start by drawing a diagram R 4N 5kg 5g N 20N Use SUVAT b) Sub in values Calculate 3B
  15. 15. Dynamics of a Particle moving in a Straight Line 3.2ms-2 You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary Start by drawing a diagram R A particle of mass 5kg is pulled along a rough horizontal table by a force of 20N, with a frictional force of 4N acting against it. Given that the particle is initially at rest, find: 4N a)The acceleration of the particle – 3.2ms-2 b)The distance travelled by the particle in the first 4 seconds – 25.6m c)The magnitude of the normal reaction between the particle and the table c) 5kg 20N 5g N Resolve vertically, taking R as the positive direction Calculate 3B
  16. 16. Dynamics of a Particle moving in a Straight Line You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary T 2ms-2 A small pebble of mass 500g is attached to the lower end of a light string. Find the tension in the string when the pebble is: a)Moving upwards with an acceleration of 2ms-2 b)Moving downwards with a deceleration of 4ms-2 Start by drawing a diagram 0.5kg 0.5g N a) Resolve vertically, taking the direction of the acceleration as positive Calculate T 3B
  17. 17. Dynamics of a Particle moving in a Straight Line You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary T -4ms-2 A small pebble of mass 500g is attached to the lower end of a light string. Find the tension in the string when the pebble is: a)Moving upwards with an acceleration of 2ms-2 – 5.9N b)Moving downwards with a deceleration of 4ms-2 0.5kg 0.5g N b) Start by drawing a diagram In this case, the pebble is moving downwards at a decreasing rate, so you can put the acceleration on as negative Resolve vertically, taking the direction of movement as positive Calculate T Even though the pebble is moving downwards, there is more tension in the string as the pebble is decelerating – the string is working against gravity! 3B
  18. 18. Dynamics of a Particle moving in a Straight Line You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary A particle of mass 3kg is projected at an initial speed of 10ms-1 in the horizontal direction. As it travels, it meets a constant resistance of magnitude 6N. Calculate the deceleration of the particle and the distance travelled by the time it comes to rest. a ms-2 R 6N 3kg 3g N Start by drawing a diagram It is important to note that the initial projection speed is NOT a force, there are actually no forces acting in the positive direction Deceleration Take the direction of movement as positive – remember to include 0 as the positive force! Calculate a So the deceleration is 2ms-2 3B
  19. 19. Dynamics of a Particle moving in a Straight Line You can solve problems involving forces by drawing a diagram including all relevant forces, and then resolving in multiple directions if necessary A particle of mass 3kg is projected at an initial speed of 10ms-1 in the horizontal direction. As it travels, it meets a constant resistance of magnitude 6N. Calculate the deceleration of the particle and the distance travelled by the time it comes to rest. a ms-2 R 6N 3kg 3g N Start by drawing a diagram It is important to note that the initial projection speed is NOT a force, there are actually no forces acting in the positive direction Distance travelled Deceleration = 2ms-2 Sub in values Work through to calculate s 3B
  20. 20. Dynamics of a Particle moving in a Straight Line However, a force at an If a force at applied at an angle to the direction of motion you can resolve it to find the component of the force acting in the direction of motion A horizontal force has no effect on the object in the vertical direction angle will have some effect in BOTH the horizontal and vertical directions! S O H C A H T O A Opp = Sinθ x Hyp Cosθ x Hyp Adj = 10N Opp = Sin20Adj = Cos20 x 10 x 10 Hyp 10N Opp 10sin20 20° Adj 10N A vertical force has no effect on the object in the horizontal direction 10cos20 So a force can be split into its horizontal and vertical components using Trigonometry! 3C
  21. 21. Dynamics of a Particle moving in a Straight Line y If a force at applied at an angle to the direction of motion you can resolve it to find the component of the force acting in the direction of motion 9N 9Sin40 40° Find the component of each force in the x and y-directions 9Cos40 Force in the x-direction x Force in the y-direction = 9Cos40 = 9Sin40 = 6.89N = 5.79N 3C
  22. 22. Dynamics of a Particle moving in a Straight Line y If a force at applied at an angle to the direction of motion you can resolve it to find the component of the force acting in the direction of motion 12N 12Sin23 Find the component of each force in the x and y-directions 23° 12Cos23 Force in the x-direction x Force in the y-direction = 12Cos23 = 12Sin23 = 11.05N = 4.69N = -11.05N (This will be negative as it is the opposite direction to x!) 3C
  23. 23. Dynamics of a Particle moving in a Straight Line You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. It is dependent on two things: 1) The normal reaction between the two surfaces 2) The coefficient of friction between the two surfaces The maximum frictional force is calculated as follows: If a surface is described as ‘smooth’, the implication is that the coefficient of friction is 0. 3D
  24. 24. Dynamics of a Particle moving in a Straight Line a ms-2 You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. Draw a diagram R F 5kg 10N 5g N A block of mass 5kg is lying at rest on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A rough horizontal force, P, is applied to the block. Find the magnitude of the frictional force acting on the block and its acceleration when: a)P = 10N b)P = 19.6N c)P = 30N We need to find the maximum possible frictional force  To do this we need R, the normal reaction Resolve vertically Calculate R Now we can calculate the maximum possible frictional force Sub in values Calculate FMAX 3D
  25. 25. Dynamics of a Particle moving in a Straight Line a ms-2 You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. Draw a diagram R 10N F 5kg 10N 5g N A block of mass 5kg is lying at rest on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A rough horizontal force, P, is applied to the block. Find the magnitude of the frictional force acting on the block and its acceleration when: The maximum frictional force is 19.6 N Any force will be opposed by friction up to this value For part a), the force is only 10N Therefore, the frictional force will match this at 10N, preventing movement Hence, there is also no acceleration a)P = 10N b)P = 19.6N c)P = 30N 3D
  26. 26. Dynamics of a Particle moving in a Straight Line a ms-2 You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. Draw a diagram R 19.6N F 5kg 19.6N 5g N A block of mass 5kg is lying at rest on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A rough horizontal force, P, is applied to the block. Find the magnitude of the frictional force acting on the block and its acceleration when: a)P = 10N b)P = 19.6N c)P = 30N The maximum frictional force is 19.6 N Any force will be opposed by friction up to this value For part b), the force is only 19.6N Therefore, the frictional force will match this at 19.6N, preventing movement Hence, there is also no acceleration This situation is called ‘limiting equilibrium’, as the object is on the point of movement 3D
  27. 27. Dynamics of a Particle moving in a Straight Line a ms-2 You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. Draw a diagram R 19.6N F 5kg 30N 5g N A block of mass 5kg is lying at rest on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A rough horizontal force, P, is applied to the block. Find the magnitude of the frictional force acting on the block and its acceleration when: a)P = 10N b)P = 19.6N c)P = 30N For part c), the force is 30N The frictional force will oppose 19.6N of this, but no more. Hence, the object will accelerate… Resolve horizontally! Sub in values and resolve horizontally Calculate Divide by 5 So the acceleration will be 2.08ms-2 3D
  28. 28. Dynamics of a Particle moving in a Straight Line R You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. F Draw a diagram 5kg P 5g N Resolve vertically to find the normal reaction A 5g box lies at rest on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5. A force P is applied to the box. Calculate the value of P required to cause the box to accelerate if: a)P is applied horizontally b)P is applied at an angle of θ above the horizontal, where tanθ = 3/4 Sub in values and resolve vertically Calculate Now find the maximum frictional force Sub in values Calculate So P will have to exceed 24.5N to make the object move! 3D
  29. 29. Dynamics of a Particle moving in a Straight Line Draw a diagram R You can calculate the magnitude of a frictional force using the coefficient of friction F P 5kg Friction is a force which opposes movement between two ‘rough’ surfaces. 0.6P Psinθ θ Pcosθ 0.8P 5g N We need to find the values of Cosθ and Sinθ. The ratio for Tanθ can be used to find these! A 5g box lies at rest on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5. A force P is applied to the box. Calculate the value of P required to cause the box to accelerate if: a)P is applied horizontally – 24.5N b)P is applied at an angle of θ above the horizontal, where tanθ = 3/4 S O We can find the hypotenuse using Pythagoras’ Theorem! H C A H T Hyp 5 O Tanθ = 3/4 A So Opp = 3 And Adj = 4 3 Opp θ 4 Adj Sinθ = 3/5 Cosθ = 4/5 Sinθ = 0.6 Cosθ = 0.8 3D
  30. 30. Dynamics of a Particle moving in a Straight Line Draw a diagram R You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. F 5kg P 0.6P θ 0.8P 5g N Resolve vertically to find the normal reaction A 5g box lies at rest on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5. A force P is applied to the box. Calculate the value of P required to cause the box to accelerate if: a)P is applied horizontally – 24.5N b)P is applied at an angle of θ above the horizontal, where tanθ = 3/4 Sub in values and resolve vertically We find the normal reaction in terms of P Now find the maximum frictional force Sub in values Find Fmax in terms of P So, 0.8P will have to exceed this if the box is to move… 3D
  31. 31. Dynamics of a Particle moving in a Draw a diagram Straight Line R You can calculate the magnitude of a frictional force using the coefficient of friction Friction is a force which opposes movement between two ‘rough’ surfaces. A 5g box lies at rest on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5. A force P is applied to the box. Calculate the value of P required to cause the box to accelerate if: a)P is applied horizontally – 24.5N b)P is applied at an angle of θ above the horizontal, where tanθ = 3/4 24.5 – 0.3P F 5kg P θ 0.6P 0.8P 5g N We need to find the value for P for which the box is in ‘limiting equilibrium’ – that is, so the horizontal forces cancel each other out… Resolve horizontally… Sub in values and resolve horizontally Careful with the bracket! Rearrange and solve P must exceed 22N, which is less than when P was horizontal  The reason is because some of the force is upwards, this alleviates some of the friction between the surfaces… 3D

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