Dynamics The causes of motion Mr. Rockensies Regents Physics

Introduction 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

Types of Forces Contact Forces – touch the object Normal Force (FN) – contact force which is perpendicular to the contact surface Friction Force (Ff) – contact force which is parallel to the contact surface Tension (FT) – force through a rope, cable, chain, etc. can only pull Remote Forces – act from a distance Gravity (Fg) – Weight Static Electricity (FE) Magnetism (FB or FM) Nuclear Forces

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

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.

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

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/s2 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/s2. What is the force the man pushes the cart with? m = 50 kg a = 4 m/s2 F = ma F = (50 kg)(4 m/s2) F = 200 N

How do we derive the Units?? If we start with the equation, F = ma, we can replace the variables with their appropriate units. F = (kg)(m/s2) F = kg•m = N s2

Force Vectors Resultant – the addition of two vectors For Force, we use the symbol FNET OR FR When there is a net force, FNET, 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.

Resultant vectors are found using either: Pythagorean Theorem + SOHCAHTOA for right angles by drawing vectors to scale for any other angles (Like in the Combining Forces Lab (#9) F2 FNET F1 E = Equilibrant – a single force which brings about equilibrium. It is equal to the resultant but opposite in direction E

Maximum Resultant Forces in the same direction or zero degrees apart equals 12 N

• Minimum Resultant Opposite direction, 180° apart 7 N 5 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.

Online Examples Resultant versus Equilibrant Drawing Components Calculate the Magnitude of the Resultant

Components All forces can be resolved into horizontal and vertical components. Case 1: Inclined Force on an object on a level surface Fy = sin θ F Fx = Fcosθ Fx moves the box Fy lifts the box

Fparallel (F||) causes sliding F|| = Fgsinθ Case 2: Block on an inclined plane FNormal Fperpendicular(F|) = Fgcosθ FN = F| FParallel FPerpendicular FGravity Fparallel (F||) causes sliding F|| = Fgsinθ

Practice Problem 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? 500 N 30° sled

Practice Problem 1 (part 2) What is the acceleration experienced by the sled? Does the vertical component of the force affect the acceleration? 500 N sled

Answers Ax = Acosθ Fx = (500 N)cos(30°) Fx = 430 N FNET = ma 430 N = (50 kg)a a = 8.6 m/s2 The vertical component doesn’t affect acceleration. It only causes the object to lift off the ground, rather than move it backward or forward.

Practice Problem 1 (part 3) 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? 500 N sled

Practice Problem 1 (part 4) What is the Normal Force felt by the sled? 500 N sled

Answers Weight is equal to Fg. Fg = mg. Since the mass of the sled is 50 kg, we can find weight by plugging in the numbers and solving for Fg. Fg = (50 kg)(9.81 m/s2) Fg = 490 N down This would not be enough to lift the sled because the force due to gravity is much greater. Ay = Asinθ Fy = 500 N sin(30°) Fy = 250 N up

Answers The Normal Force PLUS the vertical force the boy applies balance out the Gravitational Force. So… Gravitational Force is 490 N down. The boy’s force of pull is 250 N up. FN + Fboy = Fg FN + 250 N = 490 N FN = 240 N

Practice Problem 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°

Drawing a Force Diagram Green Vector represents the force due to Gravity (Weight = mg) 100 kg Red Vector represents the Normal Force of the incline pushing up on the box 30° Blue Vectors represent the components of the weight (Perpendicular and Parallel Forces)

Answers The Weight of the block is equal to mass times acceleration due to gravity. W = mg W = (100 kg)(9.8 m/s2) W = 980 N F|| = Fg sinθ F|| = (980 N)sin(30°) F|| = 490 N F| = Fg cosθ F| = (980 N)cos(30°) F| = 850 N

Practice Problem 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? 100 kg 30°

Answers 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/s2 Normal Force is equal to the Perpendicular Force, but opposite in direction. FN = 850 N