DYNAMICSTHE CAUSES OF MOTIONMr. 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 FORCESContact Forces – touch the object1. Normal Force (FN) –

contact force which is perpendicular to the contact surface

2. Friction Force (Ff) – contact force which is parallel to the contact surface

3. Tension (FT) – force through a rope, cable, chain, etc. can only pull

Remote Forces – act from a distance1. Gravity (Fg) – Weight

2. Static Electricity (FE)

3. Magnetism (FB or FM)

4. Nuclear Forces

UNITSForces are measured in Newtons,

N1 kg weighs about 10 N, 100 g

weighs 1 NRemember 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 matterWeight – force experienced by that matterMass ≠ Weight

Mass – measured on a balanceForce – 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 Nm = 1000 kgF = ma400 N = (1000 kg)aa = 400 N/1000 kga = 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 kga = 4 m/s2

F = maF = (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 vectorsFor 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:a) Pythagorean Theorem + SOHCAHTOA for right

anglesb) by drawing vectors to scale for any other angles

(Like in the Combining Forces Lab (#9)

FNET

F1

F2

E

E = Equilibrant – a single force which brings about equilibrium. It is equal to the resultant but opposite in direction

MAXIMUM RESULTANT

Forces in the same direction or zero degrees apart

7 N

5 N

12 N

equals

MINIMUM RESULTANT

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.

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

Fx = Fcosθ

Fy = sin θF

Fx moves the boxFy lifts the box

Case 2: Block on an inclined plane

FNormal

FParallel

FPerpendicular

FGravity Fparallel (F||) causes sliding

F|| = Fgsinθ

Fperpendicular(F|) = Fgcosθ

FN = F|

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?

sled

500 N

30°

PRACTICE PROBLEM 1 (PART 2)

sled

500 N

What is the acceleration experienced by the sled?

Does the vertical component of the force affect the acceleration?

ANSWERSAx = AcosθFx = (500 N)cos(30°)Fx = 430 N

FNET = ma430 N = (50 kg)aa = 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)

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?

PRACTICE PROBLEM 1 (PART 4)

sled

500 N

What is the Normal Force felt by the 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

Ay = AsinθFy = 500 N sin(30°)Fy = 250 N up

This would not be enough to lift the sled because the force due to gravity is much greater.

ANSWERS

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, FN = 490 N up

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

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

ANSWERSThe Weight of the block is equal to mass times acceleration due to gravity.W = mgW = (100 kg)(9.8 m/s2)W = 980 N

F|| = Fg sinθF|| = (980 N)sin(30°)F|| = 490 N

F| = Fg cosθF| = (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

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

To find the acceleration of the block as it slides down the incline, we need to use the parallel force.F|| = ma490 N = (100 kg)aa = 4.9 m/s2