P.1 Book 2 Section 3.3 Addition and resolution of forces 3.3 Addition and resolution of forces Confrontation Adding forces Check-point 3 Resolving forces into components Check-point 4

P.2 Book 2 Section 3.3 Addition and resolution of forces Confrontation Why does the dog not move? How do forces acting on the dog combine together? There is no net force acting on the dog since all the forces cancel out each other.

P.3 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces Resultant force – a single force representing the sum of two or more forces acting on an object – providing the same total effect of the original forces – with stronger or weaker magnitude than the original forces

P.4 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces Forces can be added using the ‘tip-to-tail’ method, just like displacements. E.g. adding F 1 and F 2 F 1 + F 2 F2F2 F1F1

P.5 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces Parallelogram of forces method: Draw F 1 and F 2 as 2 as adjacent sides of a parallelogram. The diagonal of the parallelogram is the resultant force. F1F1 F2F2 F 1 + F 2

P.6 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces Expt 3b Addition of forces Video 3.5 Addition of forces Simulation 3.1 Addition of vectors

P.7 Book 2 Section 3.3 Addition and resolution of forces Experiment 3b Addition of forces 1 Stretch a rubber band with a spring balance. Note the magnitude and direction of F. 2 Stretch the rubber band by the same amount using 2 spring balances. Note the magnitude and direction of F 1 and F 2.

P.8 Book 2 Section 3.3 Addition and resolution of forces Experiment 3b Addition of forces 3 Check if the resultant of F 1 and F 2 is equal to F in magnitude and direction. Video 3.6 Expt 3b - Addition of forces

P.9 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces a Finding resultant forces graphically With correct scale and direction on a grid or graph paper, we can find resultant force graphically. E.g. 1 5 N2 N 7 N scale: 1 cm = 1 N

P.10 Book 2 Section 3.3 Addition and resolution of forces a Finding resultant forces graphically E.g. 2 Example 3 A stretched bow 2 N 5 N 53  scale: 1 cm = 1 N 6.4 N 39 

P.11 Book 2 Section 3.3 Addition and resolution of forces Example 3 A stretched bow Tension of bowstring = 34 N Total force on arrow = ? 1 cm = 10 N Length of vector of resultant force = 4.8 cm Resultant force = 4.8  10 = 48 N (inclined upwards by 45  to the horizontal)

P.12 Book 2 Section 3.3 Addition and resolution of forces 1 Adding forces Resultant of 2 parallel forces 2 forces in the same direction Magnitude: algebraic sum Case 1: Direction: same as the 2 forces b Finding resultant forces algebraically

P.13 Book 2 Section 3.3 Addition and resolution of forces Resultant of 2 parallel forces Case 2: 2 forces in the opposite direction Magnitude: algebraic difference Direction: same as the larger force b Finding resultant forces algebraically

P.14 Book 2 Section 3.3 Addition and resolution of forces b Finding resultant forces algebraically Resultant of 2 perpendicular forces 2 perpendicular forces can be added algebraically using Pythagoras’ theorem. F1F1 F2F2 F 1 + F 2 E.g. adding F 1 and F 2 Example 4 Resultant of two perpendicular forces Simulation 3.2 Addition of forces

P.15 Book 2 Section 3.3 Addition and resolution of forces Example 4 Resultant of two perpendicular forces Magnitude and direction of resultant force on the small fish = ? Draw a rectangle as shown. Magnitude = 53.9 N The resultant force is 53.9 N acting at 68.2  to the 20-N force from the big fish.

P.16 Book 2 Section 3.3 Addition and resolution of forces Magnitude = ________ Check-point 3 – Q1 Draw and find the resultant force. 4.1 N Direction = ____________________ 76.0  to the horizontal 76 

P.17 Book 2 Section 3.3 Addition and resolution of forces Check-point 3 – Q2 Tension in each rope = 120 N Magnitude of resultant force = ? By Pythagoras’ theorem, magnitude of resultant force = ____________ = ________  170 N

P.18 Book 2 Section 3.3 Addition and resolution of forces 2 Resolving forces into components Resolving forces into components means: Usually, we take components that are  to each other. splitting one force into two whose total effect is the same as the original one. Simulation 3.3 Resolving forces

P.19 Book 2 Section 3.3 Addition and resolution of forces 2 Resolving forces into components a Resolving a force graphically E.g. Resolving along x-axis and y-axis F is the diagonal of the rectangle. F x and F y can be measured directly.

P.20 Book 2 Section 3.3 Addition and resolution of forces 2 Resolving forces into components b Resolving forces algebraically Example 5 Resolving a force algebraically  F x = F cos   F y = F sin 

P.21 Book 2 Section 3.3 Addition and resolution of forces Example 5 Resolving a force algebraically A force of 20 N pulls a suitcase. Horizontal component = ? F x = F cos 30  = 20  cos 30  = 17.3 N

P.22 Book 2 Section 3.3 Addition and resolution of forces b Resolving forces algebraically Finding the resultant force of two inclined forces algebraically Example 6 In adding 2 inclined forces that are not perpendicular to each other: -first resolve the 2 forces into their components, -then find the resultant.

P.23 Book 2 Section 3.3 Addition and resolution of forces Example 6 Finding the resultant force of two inclined forces algebraically Resultant force acting on the small fish = ? Resolve the 50-N force into 2 components: x-axis: F x = 50 cos 53  – 20 = 10.1 N y-axis: F y = 50 sin 53  = 39.9 N

P.24 Book 2 Section 3.3 Addition and resolution of forces Example 6 Finding the resultant force of two inclined forces algebraically = 41.2 N The resultant force is 41.2 N at 75.8  to the horizontal on the right of the small fish.

P.25 Book 2 Section 3.3 Addition and resolution of forces b Resolving forces algebraically To air clothing Example 7

P.26 Book 2 Section 3.3 Addition and resolution of forces Example 7 To air clothing A and B are fixed ends of the same level. C is the mid-point. Total weight of T-shirt and hanger = 5 N Tension in the rope = ?

P.27 Book 2 Section 3.3 Addition and resolution of forces Example 7 To air clothing In the vertical direction, net force = 0 T cos 80  + T cos 80  – 5 = 0 2T cos 80  = 5 T = 14.4 N The tension in the rope is 14.4 N.

P.28 Book 2 Section 3.3 Addition and resolution of forces b Resolving forces algebraically A bicycle on a slope Example 8

P.29 Book 2 Section 3.3 Addition and resolution of forces Example 8 A bicycle on a slope A 150-N bicycle is at rest on a slope at 30  to the horizontal. (a) Find the components of weight along the road surface and normal to it.

P.30 Book 2 Section 3.3 Addition and resolution of forces Example 8 A bicycle on a slope Component of W = W cosθ = 150  cos 30  = N Component of W = W sinθ = 150  sin 30  = 75 N Along the road surface: Normal to the road surface:

P.31 Book 2 Section 3.3 Addition and resolution of forces Example 8 A bicycle on a slope (b) Find the friction f and normal force R. The bicycle is at rest Along the surface: f = W sinθ  f = 75 N Normal to the surface: R = W cosθ  R = N  net force = 0

P.32 Book 2 Section 3.3 Addition and resolution of forces Check-point 4 – Q1 A dog tied by a string to a railing is trying to get away. Which of the following correctly shows the free-body diagram of the dog? B C A

P.33 Book 2 Section 3.3 Addition and resolution of forces Check-point 4 – Q2 What would happen if the mass is released? The tension in the string is _______. The component force F parallel to the table = _______  _______ W W cosθ

P.34 Book 2 Section 3.3 Addition and resolution of forces Check-point 4 – Q2 If F is larger than the ________ acting on the toy, the toy slides towards the edge of the table. As the toy slides closer to the edge of the table, angle  (increases/decreases), while the component F (increases/decreases). friction

P.35 Book 2 Section 3.3 Addition and resolution of forces Check-point 4 – Q3 A man just able to push a load of weight 500 N up a slope of 20  to the horizontal. Friction = 20 N Assume: pushing force parallel to the ramp Magnitude of the force = ? Pushing force F = W sin  + f = 500 sin 20  + 20 = 191 N

P.36 Book 2 Section 3.3 Addition and resolution of forces The End