GR11 Support pack for Mechanics

Study & Master Support Pack | Grade 11 CAPS Physical Sciences Mechanics This support pack for the Mechanics strand of the Physical Sciences Grade 11 CAPS curriculum provides revision summaries on the topic to help prepare for the examinations. Learners can work through these individually at home or these could form the basis of a catch-up class or online lesson. You have permission to print or photocopy this document or distribute it electronically via email or WhatsApp. Cambridge University Press Africa is a proudly South African publisher. For more information on our Study & Master CAPS-approved textbooks and valuable resource materials, visit www.cambridge.org Brighter thinking I Better learning www.cambridge.org

Unit 1: Vectors in two dimensions Summary • A vector is a quantity that has magnitude and a direction. • A scalar is a quantity that has magnitude but no direction. • Vectors are represented by a line with an arrowhead on a cartesian plane. • A resultant is a single vector that has the same effect as the original vectors combined. You can determine the resultant graphically with the tail-to-head or tail-to-tail method, or using calculations (Pythagoras’ Theorem). • Three forces are in equilibrium when the resultant is zero. • A vector can be resolved into two components that are perpendicular to each other. 2 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Revision exercises: Unit 1 – Vectors in two dimensions 1. Find the resultant (both magnitude and 3. Three construction workers wish to remove direction) of the forces shown in the a pole from the ground. They tie ropes to following diagrams: the pole and pull horizontally on the ropes. The diagram represents the forces exerted on a) N the pole. 110 N 76 N 50 N y W E 35˚ 80 N 40˚ x 55 N Pole P S b) N a) Determine the direction (as measured with the x-axis) in which the pole will move. b) What must be the magnitude and direction of the equilibrant force needed to keep the 67 N pole in equilibrium? W E 45 N S 2. A boat is heading straight across a river, which is 600 m wide. At the same time, the river current carries the boat 150 m downstream. Use a vector diagram to find the resultant displacement of the boat when it reaches the opposite bank of the river. MECHANICS 3

Unit 1: Answers 1. a) R 5 80 N − 50 N, south 5 30 N, south The angle made with the starting bank is b) The resultant vector can be obtained by 90° − 14° 5 76° using the head-to-tail method as shown 5 618,5 m, 76° with the starting bank in the following diagram, and then using the Theorem of Pythagoras to find its 3. The 76 N force makes an angle of magnitude. 90° − 35° 5 55° with the x-axis. N y 110 N FR 76 N 67 N W θ E 45 N 55˚ 40˚ x 55 N S FR 5 √(67 N)2 + (45 N)2 5 80,7 N 67 N tan θ 5 45 N 5 1,49 The components of the resultant force are: ⇒ θ 5 56° FRx 5 (110 N) cos 40° − (76 N) cos 55° − R 5 80,7 N, 56° north of west (55 N) cos 0° 5 −14,32 N FRy 5 (110 N) sin 40° 1 (76 N) sin 55° 1 2. The vector diagram for the motion of the boat (55 N) sin 0° 5 132,96 N is as follows: R 5 √ (−14,32 N)2 + (132,96 N)2 5 133,73 N 150 m 14,32 N opposite bank tan θ 5 132,96 N 5 0,1077 90˚ ⇒ θ 5 6,15° a) The pole will move in a direction of 6,15° 600 m above the negative x-axis. R b) The equilibrant force will have a θ magnitude of 133,73 N and its direction will be 6,15° below the positive x-axis. starting bank The magnitude of the resultant displacement is: R 5 √ (600 m)2 + (150 m)2 5 618,5 m 150 m tan θ 5 600 m 5 0,25 ⇒ θ 5 14° 4 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Unit 2: Newton’s Laws and application of Newton’s Laws Summary • Applied force (F) is the push or pull on an object. • Tension (T) is the force in a rope or string when it is stretched. • Normal force (N) is the force exerted by a surface on an object in contact with it. It is perpendicular to the surface. • Friction (f) is the force that opposes the motion of an object and makes it slow down: Static friction is the force that must be overcome to move a stationary object and kinetic friction is the force that makes an object slow down. • A force diagram is a drawing of the object with all the forces acting on it drawn as arrows. • A free-body diagram represents the object as a box or dot with forces acting on it as arrows pointing away from the box. • To calculate the forces on an incline, choose the x-direction as the direction of the incline and the y-direction as perpendicular to the incline. The weight components are: wx = mg sin θ and wy = mg cos θ • Inertia is the tendency of a body to maintain its state of rest or constant motion in a straight line. • Newton’s First Law: An object will remain at rest or continue moving at a constant velocity in a straight line, unless an external (unbalanced) net force acts on it. • Newton’s Second Law: When a net force is exerted on an object, it causes the object to accelerate in the direction of the force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. (F = ma) • Newton’s Third Law: When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. • One newton is the force that will give a mass of one kilogram acceleration of one metre per second squared. • Newton’s Law of Universal Gravitation: Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between Gm1m2 them. (F = ) r2 • Earth has a gravitational field around it; the gravitational field strength F = acceleration due to gravity g = m = 9,8 m·s−2 • The weight of an object (in N) is an indication of the strength of Earth’s attraction for that object (W = mg). Mass (in kg) is the quantity of matter of an object. MECHANICS 5

Revision exercises: Unit 2 – Newton’s Laws and application of Newton’s Laws 01 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 N 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1. Give one term for the following descriptions. 2. Samantha performs an experiment to verify Newton’s Second Law. A trolley is a) The agency that changes or tends to accelerated across a surface, as shown below. change the state of motion of an object To compensate for friction the runway is b) A single force whose effect is equivalent adjusted so that the trolley runs slightly to the combined effects of two or more downhill. But the adjustment is of such separate forces magnitude that the trolley moves uniformly c) The two perpendicular parts a vector can with zero acceleration. The trolley is be split into accelerated using small weights, as shown. d) The condition of a system in which all string pulley competing forces are balanced trolley e) This force is exerted by the surface on an object and always acts perpendicular to the surface f) A measure of the extent to which a body resists changes to its state of motion g) The net force that produces an acceleration of 1 m∙s−2 when acting on a 1 kg object h) The mathematical relationship between slotted weights mass and acceleration if the net force stays the same Samantha records the acceleration for i) The ratio of the net force to the mass of different weights and plots a graph of the a body results as shown. j) The rate of change in velocity k) The ratio of the frictional force to the (4,5;9,0) normal force acting on a body l) Frictional force between two surfaces that 8 are not moving relative to each other m) The angle between friction and motion (3,4;7,0) n) The gravitational force experienced by 6 Net force (N) objects near the surface of Earth o) The direction of the gravitational force on 4 an object (2,0;4,0) p) An electronic device that makes small dots on paper ribbon, which is processed to 2 determine time and distance (1,1;2,0) q) The law that determines the attractive forces that two masses exert on each other 0 0 1 2 3 4 when they are at rest Acceleration (m·s–2) r) The amount of matter in an object s) The ‘feeling’ experienced when a person is a) Describe the most important fixed variable in free fall for this investigation. t) The sensation of weight that is felt due to b) Formulate a suitable investigative question. contact forces pressing against the body c) Use the graph to describe the relationship between the two variables on the axes. 6 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

d) Determine the gradient of the graph line. 6. (Assume the weight is in kilograms.) e) Use your answer in d) to determine the 15 kg 5 kg mass of the trolley. f) If the trend shown by the graph continues, what should the acceleration be if the net Two wooden blocks of masses 15 kg and force is 20 N? 5 kg lie on a horizontal surface. The blocks g) On the same grid as the graph above, draw are pushed across the surface by a person a graph of force versus acceleration if applying a horizontal force F on the 15 kg the mass of the trolley is doubled and the block. The frictional forces on the 15 kg and trolley is subjected to the same net forces. 5 kg blocks are 9 N and 3 N respectively. 3. A 10 kg block is pulled horizontally to the right a) State Newton’s Second Law in words. by an 8 N force, as shown in the diagram below. b) Sketch a force diagram for the: i) 15 kg block 8N ii) 5 kg block. 10 kg c) If, during the motion, the 5 kg block exerts a force of 20 N on the 15 kg block, If a constant frictional force of 3 N acts on the determine the magnitude of the force F block, calculate the acceleration of the block. exerted by the person on the 15 kg block. 4. A 40 N force, acting at 20° to the horizontal 7. Two objects, each with mass 5 3 102 kg, exert pulls a 16 kg crate to the right, as shown. a gravitational force of 5,4 3 10−6 N on each A constant frictional force of 8 N acts on other. What is the distance between them? the crate. 8. Two objects, A and B, of equal mass exert a gravitational force of 1,47 N on each other. 40 N If they are 1 m apart, what is the mass of 20˚ object A? 16 kg 9. Two objects with masses 2m and 3m exert a force F on each other when they are a certain a) Draw a force diagram, showing all the distance apart. What will the force between forces acting on the crate. them be if the distance between them remains b) Calculate the net horizontal force acting on the same but their masses change to 5m and the crate. 6m respectively? c) Calculate the acceleration of the crate. 5. The system shown below is released from rest. 3 kg 1 kg If the pulley is frictionless, calculate the: a) acceleration of the system b) magnitude of the tension in the cable. MECHANICS 7

Unit 2: Answers 1. a) Force g) b) Resultant c) Components (4,5;9,0) d) Equilibrium e) Normal force 8 f) Inertia (3,4;7,0) g) 1 Newton 6 h) Inversely proportional i) Acceleration Net force (N) j) Acceleration 4 (2,0;4,0) k) Coefficient of friction l) Static friction m) 180° 2 (1,1;2,0) n) Weight o) Downward, towards Earth 0 p) Ticker-timer 0 1 2 3 4 q) Newton’s Law of Universal Gravitation Acceleration (m·s–2) r) Mass 3. Take the direction to the right as positive. s) Weightlessness t) Apparent weight Net force in the horizontal direction is given by: net 5 app 1 f 2. a) Mass of the trolley 5 +8 N 1 (−3 N) b) What is the relationship between the net 5 +5 N force and acceleration of a trolley with net 5m constant mass? +5 N 5 10 kg 3 c) Acceleration is directly proportional to the 5 +0,5 m∙s−2 net force. 5 0,5 m∙s−2 to the right d) Do not use coordinates that are not on the line graph. 4. a) 40 N (F – F ) N Gradient 5 (a 2 − a1) 2 1 20˚ (8N − 6N) 5 (4m·s−2 − 3m·s−2) 5 2 kg f e) Mass 5 gradient 5 2 kg W f) 10 m·s−2 b) Take the direction to the right as positive. Fx 5 F cos θ 5 40 N cos 20° 5 37,59 N Fnet 5 Fapp 1 Ff 5 +37,59 N 1 (−8 N) 5 +29,59 N c) Fnet 5 ma +29,59 N 5 16 kg 3 a 5 +1,85 m∙s−2 5 1,85 m∙s−2 to the right 8 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

5. Take the clockwise direction as positive. c) Take the direction to the right as positive. For B: Fnet 5 ma 5 FAB 1 f a) Fnet on system: 5 kg 3 a 5 +20 N 1 (−3 N) Fnet 5 W(3 kg) 1 W(1 kg) a 5 +3,4 m∙s−2 5 3 kg(9,8 m∙s−2) 1 (−1 kg)(9,8 m∙s−2) For A: Fnet 5 ma 5 FBA 1 Ff 1 F 5 +19,6 N F +19,6 N 15 kg 3 (+3,4 m∙s−2) 5 −20 N 1 (−9 N) 1 F a 5 mnet 5 4 kg 5 +4,9 m∙s−2 F 5 +80 N T 5 4,9 m∙s−2 clockwise m1m2 7. r2 5 G F b) Consider the 3 kg mass. (5 3 102 kg)2 5 6,67 3 10−11 N·m2·kg−2 (5,4 3 10−6 N) 5 3,09 m2 r 5 1,76 m 3 kg Fr2 1,47 N 3 (1 m)2 8. mAmB 5 G 5 6,67 3 10−11 N·m2·kg−2 W m2 5 2,2 3 1010 kg2 m 5 1,48 3 105 kg m1m2 2m 3 3m 6m2 Fnet 5 ma 5 T 1 W 9. F 1 5 G r2 5G r 2 5 G 3 r2 5m 3 6m 30m 2 3 kg(+4,9 m∙s−2) 5 T 1 3 kg(1 9,8 m∙s−2) F2 5 G 5 G 3 r2 r2 T 5 −14,7 N Therefore, the magnitude of the tension in Therefore F2 is 5 times stronger than F1 5 5F1. the cable is 14,7 N. 6. a) When a net force acts on an object, it produces an acceleration in the direction of the net force. This acceleration is directly proportional to the net force and inversely proportional to the mass of the object. b) Let the 15 kg block be block A and the 5 kg be block B. i) N FBA F f W ii) N FAB f W MECHANICS 9