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Vectors and Scalars Define Vector and Scalar quantities and give examples of each.

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A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities: Length Area Volume Time Mass

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A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quantities: Displacement Velocity Acceleration Force

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Vector diagrams are shown using an arrow The length of the arrow represents its magnitude The direction of the arrow shows its direction

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When two vectors are joined tail to tail Complete the parallelogram The resultant is found by drawing the diagonal When two vectors are joined head to tail Draw the resultant vector by completing the triangle

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Solution: CComplete the parallelogram (rectangle) θ TThe diagonal of the parallelogram ac represents the resultant force Two forces are applied to a body, as shown. What is the magnitude and direction of the resultant force acting on the body? 5 N 12 N 5 12 a bc d The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc RResultant displacement is 13 N 67 º with the 5 N force 13 N

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When resolving a vector into components we are doing the opposite to finding the resultant We usually resolve a vector into components that are perpendicular to each other y v x HHere a vector v is resolved into an x component and a y component

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HHere we see a table being pulled by a force of 50 N at a 30 º angle to the horizontal WWhen resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 43.3 N 50 N y=25 N x=43.3 N 30 º We can see that it would be more efficient to pull the table with a horizontal force of 50 N

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60 º A force of 15 N acts on a box as shown. What are the horizontal and vertical components of the force? Vertical Component Horizontal Component Solution: 15 N 7.5 N N

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Moments: turning forces What is the equation for calculating moment of a force?

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Key Equation: Moment of a force = force x perpendicular distance moment = fd Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment of a force about a point (torque) (C) Most: apply principle of moments to simple balanced situations (B) Some: employ principle of centre of mass in calculations (A)

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Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment of a force about a point (torque) (C) Most: apply principle of moments to simple balanced situations (B) Some: employ principle of centre of mass in calculations (A) The Principle of moments: When an object is in equilibrium: Sum of clockwise moments about any point Sum of anti- clockwise moments about any point = Note definition clockwise moments – anticlockwise moments = zero (resultant force)

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Centre of gravity is important for balancing

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You have to keep the centre of gravity above the base if you want to balance.

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My head is big for my body compared to an adult’s head. That’s why I find it harder to balance.

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Poor balance produces an overall turning force. This requires a turning force in the opposite direction to counteract it.

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A uniform horizontal shelf of width 0.38 m is attached to a wall a shown in the diagram. The total weight of the shelf and books is 70 N. This weight acts from the middle of the shelf (0.19 m from the wall). 1.Calculate the turning effect of the weight about point P. 2.Calculate the tension in the support wire. Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment in single support situations (C) Most: calculate the moment in twin support situations (B) Some: determine the moment of a couple (torque) (A) T 40 o 70 N P

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1. Moment = 70 N 0.19 m = 13.3 N m 2. The component of T at 90 to the shelf must provide the moment to balance the moment of the weight. T sin 40° 0.38 m = 13.3 N m T sin 40° = 13.3 / 0.38 = 35 N T = 35 / sin 40° T = 54.5 N Note that we take moments about point P. This is because there is a third force which acts on the shelf; this is the contact force (or ‘reaction’) of the wall on the shelf. We do not know its magnitude or direction but, since it acts through point P, it has no turning effect about P. Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment in single support situations (C) Most: calculate the moment in twin support situations (B) Some: determine the moment of a couple (torque) (A) T 40 o 70 N P

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1) The plank is set up as shown and the balance zeroed. When the student lies on the plank the reading is 600 N. The balance is 2 m from the student’s feet and the centre of gravity of the student is 1.5 m from their feet. What is the student’s weight?

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Taking moments about the feet. 600 x 2 = W x 1.5 so W = 800 N

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Couples A couple is a pair of equal and opposite forces NOT acting along the same line Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment in single support situations (C) Most: calculate the moment in twin support situations (B) Some: determine the moment of a couple (torque) (A)

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Couples Show that the total moment = Fd Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment in single support situations (C) Most: calculate the moment in twin support situations (B) Some: determine the moment of a couple (torque) (A) x d F F

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Couples Show that the total moment = Fd Keywords: Force Moment Couple Torque COM Weight Turning All: calculate the moment in single support situations (C) Most: calculate the moment in twin support situations (B) Some: determine the moment of a couple (torque) (A) x d F F Clockwise moment = Fx Clockwise moment = F(d-x) Total = Fx + F(d-x) Total = Fx + Fd – Fx Total = Fd

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Keywords: Velocity Acceleration Displacement Initial Final Time Constant All: make calculations using the equations of motion (C) Most: link algebraic and graphical methodologies (B) Some: derive the equations of motion from 1st principles (A) The equations of motion Only work when the acceleration is constant!

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(Take acceleration due to gravity as = 9.8 ms -2 unless otherwise stated) 1.A car accelerates from rest at 3 ms -2 for 4s. How far has it travelled? 2.A stone is dropped from a cliff: (a)How far will it have fallen in 4s? (b)What will its velocity be at that point? (c)What is the average velocity during the 4s? 3.Starting from rest a car travels for 2 minutes with a uniform acceleration of 0.3 ms -2 after which its speed is kept constant until the car is brought to rest with a uniform decceleration of 0.6ms -2 if the total distance travelled is 4500m how long did the journey take? 4.A stone is thrown upward with a velocity of 12 ms -1 (a)How far will it have risen in 1s? (b)What will its velocity be at that paint? (c)What is the maximum height that it will reach before coming down again? Lesson 225

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Lesson A car accelerates from rest at 3 ms -2 for 4s. How far has it travelled? Distance travelled = ½ at 2 = ½x3x4 2 = 24 m 2.A stone is dropped from a cliff: (a)How far will it have fallen in 4s? (a) s = ½ gt 2 = ½ x 9.8 x 4 2 = 78.4 m (b)What will its velocity be at that point? (b) v = at = 9.8x4 = 39.2 ms -1 (c)What is the average velocity during the 4s? (c) average velocity = 39.2/2 = 19.6 ms -1

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Lesson Starting from rest a car travels for 2 minutes with a uniform acceleration of 0.3 ms -2 after which its speed is kept constant until the car is brought to rest with a uniform retardation of 0.6ms -2 if the total distance travelled is 4500m how long did the journey take? Initial acceleration time = 2 minutes = 120s (note conversion to seconds) Distance travelled in that time = ½ 0.3x120 2 = 2160 m Speed after this initial acceleration = 0.3x120 = 36 ms -1 Distance travelled during final deceleration = 36 2 /2x0.6 = 1080 m Time of final deceleration = 1080/18 = 60 s Distance of constant velocity section = 4500 – 2160 – 1080 = 1260 Time of constant velocity section = 1260/36 = 35 s Therefore total time for the journey = = 215 s = 3 min 35 s 4.A stone is thrown upward with a velocity of 12 ms -1 (a)How far will it have risen in 1 s? (b)What will its velocity be at that paint? (c)What is the maximum height that it will reach before coming down again? (a) s = ut – ½ gt 2 = 12 – ½ x9.8x1 = 7.1 m (b) v = u –gt = 12 – 9.8x1 = 2.2 ms -1 (c) maximum height (h) when vertical velocity = 0 v 2 = 0 = 12 2 – 2x9.8xhh = 7.5m

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Projectile : an object acted upon only by the force of gravity Putting our SUVAT equations to good use

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Starting points We need to consider the vertical and horizontal components of motion separately. 2.The acceleration involved is always g (downwards) & only affects the vertical component 3.Any horizontal velocity is constant and unaffected by g 4.We’ll consider up as positive and down as negative

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Note : Horizontally there is no change in velocity, (1 square per photo all the way across) Note : Vertically, the distance travelled between each photo increase (acceleration) Note : Vertically, the motion of the dropped ball and the thrown ball is identical

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A worked example.... An object is projected at a horizontal velocity of 15 ms -1 from the top of a 35m tower. Calculate the following a)How long it takes to reach the ground b)How far it travels horizontally c)It’s speed at impact

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A worked example.... An object is projected at a horizontal velocity of 15 ms -1 from the top of a 35m tower. Calculate the following a)How long it takes to reach the ground (Using S=ut + ½ at 2 = 2.67s) a)How far it travels horizontally (Using S = ½ (u + v) t or S = velocity x time = 40m) a)It’s vertical speed at impact (Using V 2 = U 2 + 2aS = 26.2 ms -1 )

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An example.... A ball is projected horizontally at 0.52 ms -1 across the top of an inclined board which is 60cm wide (horizontally) and 120cm high (vertically in the direction of the incline). The ball reaches the bottom of the board in 0.9s. Calculate the following: a)The distance travelled across the board [0.468m] b)Its acceleration on the board [2.96ms -2 ] c)Its speed at the bottom of the board [2.8ms -1 ]

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