1. D YNAMICS T HE CAUSES OF MOTION Mr. Rockensies Regents Physics

2. I NTRODUCTION Motion is caused by Force. Force – any push or pull on an object Symbol - F F = ma, where F is a Force, m is a mass, and a is an acceleration

3. T YPES OF F ORCES Contact Forces – touch the object 1. Normal Force (F N ) – contact force which is perpendicular to the contact surface 2. Friction Force (F f ) – contact force which is parallel to the contact surface 3. Tension (F T ) – force through a rope, cable, chain, etc. can only pull Remote Forces – act from a distance 1. Gravity (F g ) – Weight 2. Static Electricity (F E ) 3. Magnetism (F B or F M ) 4. Nuclear Forces

4. U NITS Forces are measured in Newtons, N 1 kg weighs about 10 N, 100 g weighs 1 N Remember back to the lab we did

5. Forces are vectors – have magnitude and direction Just like with projectile motion, we will need to break down resultants into its components, and find the resultant when given components.

6. Mass – amount of matter Weight – force experienced by that matter Mass ≠ Weight Mass – measured on a balance Force – measured on a scale using a spring Equilibrium – “equal forces”, forces which are balanced or add to zero

7. A breeze blowing to the east pushes a sailboat along a calm ocean with a force of 400 N. The boat has a mass of 1000 kg. What is the acceleration felt by the boat? F = 400 N m = 1000 kg F = ma 400 N = (1000 kg)a a = 400 N/1000 kg a = 0.4 m/s 2 A man pushes a cart with a mass of 50 kg along an even, frictionless surface. The cart accelerates at a rate of 4 m/s 2. What is the force the man pushes the cart with? m = 50 kg a = 4 m/s 2 F = ma F = (50 kg)(4 m/s 2 ) F = 200 N

8. H OW DO WE DERIVE THE U NITS ?? If we start with the equation, F = ma, we can replace the variables with their appropriate units. F = (kg)(m/s 2 ) F = kgm = N s 2

9. F ORCE V ECTORS Resultant – the addition of two vectors For Force, we use the symbol F NET OR F R When there is a net force, F NET, that is not equal to zero, it is said to be unbalanced. This indicates that an object is accelerating because there is a force acting on it. When we look at the forces acting on an object, we draw them concurrently.

10. Resultant vectors are found using either: a)Pythagorean Theorem + SOHCAHTOA for right angles b)by drawing vectors to scale for any other angles (Like in the Combining Forces Lab (#9) F NET F1F1 F2F2 E E = Equilibrant – a single force which brings about equilibrium. It is equal to the resultant but opposite in direction

11. M AXIMUM R ESULTANT Forces in the same direction or zero degrees apart 7 N 5 N 12 N equals

12. M INIMUM R ESULTANT Opposite direction, 180° apart 5 N 7 N equals 2 N Every number between 2 and 12 is a possible resultant. The equilibrant then would also be between 2 and 12, but in the opposite direction.

13. O NLINE E XAMPLES Resultant versus Equilibrant Drawing Components Calculate the Magnitude of the Resultant

14. C OMPONENTS All forces can be resolved into horizontal and vertical components. Case 1: Inclined Force on an object on a level surface F x = Fcosθ F y = sin θ F F x moves the box F y lifts the box

15. Case 2: Block on an inclined plane F Normal F Parallel F Perpendicular F Gravity F parallel (F || ) causes sliding F || = F g sinθ F perpendicular (F | ) = F g cosθ F N = F |

16. P RACTICE P ROBLEM 1 ( PART 1) A 50 kg sled is pulled by a boy across a smooth, icy surface. If the boy is pulling the sled 500 N at 30° above the horizontal, what is the horizontal component of the force? sled 500 N 30°

17. P RACTICE P ROBLEM 1 ( PART 2) sled 500 N What is the acceleration experienced by the sled? Does the vertical component of the force affect the acceleration?

18. A NSWERS A x = Acosθ F x = (500 N)cos(30°) F x = 430 N F NET = ma 430 N = (50 kg)a a = 8.6 m/s 2 The vertical component doesn’t affect acceleration. It only causes the object to lift off the ground, rather than move it backward or forward.

19. P RACTICE P ROBLEM 1 ( PART 3) sled 500 N Using the same sled from the previous example, what is the weight of the sled? Find the vertical component of the force pulling the sled. Would this force cause the sled to lift of the ground? Why?

20. P RACTICE P ROBLEM 1 ( PART 4) sled 500 N What is the Normal Force felt by the sled?

21. A NSWERS Weight is equal to F g. F g = mg. Since the mass of the sled is 50 kg, we can find weight by plugging in the numbers and solving for F g. F g = (50 kg)(9.81 m/s 2 ) F g = 490 N down A y = Asinθ F y = 500 N sin(30°) F y = 250 N up This would not be enough to lift the sled because the force due to gravity is much greater.

22. A NSWERS The Normal Force felt by the sled would be equal in magnitude to the Gravitational Force (Weight), but in the opposite direction (perpendicular to the surface). Therefore, F N = 490 N up

23. P RACTICE P ROBLEM 2 ( PART 1) A block with a mass of 100 kg is at rest on an inclined plane with an angle of 30°. What is the weight of the block? What is the parallel force of the block? What is the perpendicular force of the block? 100 kg 30°

24. D RAWING A F ORCE D IAGRAM 100 kg 30° Green Vector represents the force due to Gravity (Weight = mg) Blue Vectors represent the components of the weight (Perpendicular and Parallel Forces) Red Vector represents the Normal Force of the incline pushing up on the box

25. A NSWERS The Weight of the block is equal to mass times acceleration due to gravity. W = mg W = (100 kg)(9.8 m/s 2 ) W = 980 N F || = F g sinθ F || = (980 N)sin(30°) F || = 490 N F | = F g cosθ F | = (980 N)cos(30°) F | = 850 N

26. P RACTICE P ROBLEM 2 ( PART 2) What is the acceleration of the block as it slides down the inclined plane? What is the Normal Force felt by the block? As the angle of the inclined plane increases, what happens to the parallel and perpendicular forces? what happens to the parallel and perpendicular forces 100 kg 30°

27. A NSWERS Normal Force is equal to the Perpendicular Force, but opposite in direction. F N = 850 N To find the acceleration of the block as it slides down the incline, we need to use the parallel force. F || = ma 490 N = (100 kg)a a = 4.9 m/s 2