Forces and the Laws of

Motion

Newton’s Laws

Newton’s First Law(Law of inertia)

• An object at rest remains at rest, and an object in motion continues in motion with constant velocity unless the object experiences a net external force.

Inertia• Inertia is the tendency of an object not to

accelerate.• In other words, Newton’s First Law says that

when the net external force on an object is zero, the object’s acceleration (change in velocity) is zero.

Net force• The sum of forces action on an object is called

the net force.

A car moving at a constant velocity has a vector sum of zero.

F ground

F forward

F gravity

F resistance

Force Pairs

• Forces always come in pairs1. Applied Force…force ON an object2. Frictional Force…opposes motion3. Normal Force…upward forces,

opposes weight4. Gravity/Weight…acts downward

• Fa Fn Ff Fg

Inertia• Inertia is proportional to an objects mass. • The greater the mass, the less the body

accelerates under an applied force.• Ex. Basketball and bowling ball – Which has more

inertia?• Bowling Ball! Why? Because it has more mass!• Therefore, mass can be defined as the amount of

matter in an object and also as a measure of the inertia of an object.

Equilibrium

• Definition: objects that are either at rest or moving with constant velocity are said to be in equilibrium. Equilibrium means the net force acting on a body in equilibrium must be equal to zero.

• It the net force is zero, the body is in equilibrium. IF there is a net force, a second force equal and opposite to this net force will put the body in equilibrium.

Law of Acceleration – Newton’s Second Law

o Relates acceleration of an object to the force applied on it and the mass of the object

o A is directly proportional to Fo A is indirectly proportional to Mo Newton’s 2nd law of motion – which states:

The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass. Force = mass x acceleration

Formula

•F = ma

• Units: m….kg a….m/s2

F….kg-m/s2 or N (Newton)

Mass vs. Weight

• Mass…the amount of matter• Weight…the gravitational attraction for

matter• 1 N is the “weight” of one medium apple!• So, W = mg where g= 9.8 m/s2 • Is a vector … + is up - is down

VI. Net Force = Fnet = MAnet

A. Fnet is the sum of all of the forces acting on an object

VERTICAL: Fnet

= Fup

+ Fdown

= Fa

+ W = Fa

+ mg

HORIZONTAL:

Fnet

= Fforward

+ Ffriction

= Fa

+ μW = Fa

+ μmg

Ex. 13. A 1090 N person, standing on a scale, is moving upward in an elevator at a constant velocity. Find the reading on the scale.

+ 1090 N

Fnet = Fup + Fdown MAnet = Fscale + W Fscale = MAnet – W Fscale = 0 – ( - 1090) = + 1090 N

Ex. 14. A 88.2 kg person standing on a scale in an elevator is moving upward with an acceleration of 1.55 m/s2. Find the reading on the scale.

 

+ 1.00 x 103 N

Fnet = Fup + Fdown MAnet = Fscale + mg Fscale = MAnet – MG = M (a-g) = 88.2 [1.55 – (-9.80)] = 88.2 (11.35) = 1001.07 = + 1.00 x 103 N

15. A 36 kg object is accelerated upward at 2.0 m/s2 . Calculate the force applied on the object.

420 N

Fnet = MAnet = Fup + W Fup = MAnet – W = MAnet – mg = m (a-g) = 36 [2.0 – (-9.80)] = 36 (11.8) = 424.8 = 420 N

16.. A rocketship ( m = 1.75 x106 kg) has a lift-off force of 3.50 x107 N. Calculate the net acceleration of the spacecraft.

+ 10.2 m/s2

Fnet = MAnet = Fup + W Anet = (Fup + W) / M = (Fup + mg) / m = [3.50 x107 N + (1.75 x106 kg)(-9.80)] / 1.75 x106 kg

19.. A 4.55 kg cat is shot upwards with an overall acceleration of 15.5 m/s2. Calculate the thrust needed for the launch.

115 N

Fnet = MAnet = Fup + W Fup = Fnet – W = ma – mg = m (a-g) = 4.55 [15.5 –(-9.80)] = 4.55(25.3) = 115 N

Fnet = MAnet = Fup + W Fup = MAnet – W = MAnet – mg = m (a-g) = 36 [2.0 – (-9.80)] = 36 (11.8) = 424.8 = 420 N

16.. A rocketship ( m = 1.75 x106 kg) has a lift-off force of 3.50 x107 N. Calculate the net acceleration of the spacecraft.

+ 10.2 m/s2

Fnet = MAnet = Fup + W Anet = (Fup + W) / M = (Fup + mg) / m = [3.50 x107 N + (1.75 x106 kg)(-9.80)] / 1.75 x106 kg

17.. A 36 kg rock is pushed against a surface where μ = 0.800. What force must be applied to give an net acceleration of 0.25 m/s2?

290 N

Fnet = MAnet = Fa + Ff MAnet = Fa + μmg Fa = MAnet – μmg = m (a – μg) = 36 [0.25 – (0.800)(-9.80)] = 36 [0.25 + 7.84] = 36(8.09) = 291.24 = 290 N

18.. A 788 kg rock is pushed along a road, using 6850 N of force to achieve 0.250 m/s2 of acceleration. Calculate the COF of the road and the rock.

0.862

Fnet = MAnet = Fa + μmg μmg = MAnet – Fa μ = (ma – Fa) / mg = [(788)(0.250) – 6850] / (788)(-9.80) = -6653/-7722.4 μ = 0.862

Friction

Friction…any force that opposes the motion of an object

• Due to bonding forces between the 2 surfaces…”microwelds”

Types of friction:

1. Static Friction…force necessary to start the motion of an object at rest

…break the microwelds2. Sliding Friction…F necessary to keep an

object in motion at a constant velocity …less than the static friction

3. Rolling Friction…friction between a rolling object and the surface it is rolling on

4. Fluid Friction…friction between a moving fluid and the surface of the object it is flowing within

…”fluids” include liquids and gases

Coefficient of Friction = μ

• The ratio of the applied force to the weight• High μ…high resistance to motion• Low μ…low resistance to motion

• μ = -Fa(Ff) / Fg or -Fa(Ff)= μFg

COF problems25.0 N are needed to maintain a constant velocity for a 45.0 N block of wood moving on carpet. Calculate the COF of the surfaces.

0.556

125 N are needed to move a 31.2 kg box on a concrete surface. Calculate the COF.

0.409

A box requires 1175 N to move it on surface where the COF is 0.222. Calculate the mass of the box.

540. kg

The COF of two surfaces is 0.70. Calculate the force needed to maintain a constant velocity for a 750.0 kg metal sled.

+5100 N

Frictional factors:

• Affected by the types of surfaces• Not affected by mass or surface area of the

object

A 1220 kg go-cart is pushed with a force of 13500 N on a road with a COF of 0.755. Calculate the net acceleration rate.

3.67 m/s2

Fnet = MAnet = Fa + μmg Anet = (Fa + μmg)/ m = [13500 + (0.755)(1220)(-9.80)] / 1220 = (13500 – 9026.78)/ 1220 = 4473.33/1220 = 3.66657 = 3.67 m/s2

A car is pushed by a force of 14750 N on a road with a COF of 0.712, reaching an acceleration of 2.88 m/s2. Calculate the mass of the car.

1.50 x 103 kg

Fnet = MAnet = Fa + μmg Ma – μmg = Fa M (a- μg) = Fa M = Fa / (a- μg) = 14750 / [2.88 – (0.712)(-9.80)] = 14750 / (2.88 + 6.9776) = 14750 / 9.8576 = 1496.3 = 1.50 x 103 kg

VII. Free-falling bodies

A. in a vacuum… a = g

B. in the atmosphere, air molecules resist the falling motion of an object

• called Air Resistance (or Drag)• a type of frictional force

Newton’s 3rd Law states that for every action, there is an equal and opposite reaction.

When two objects interact with one another, the forces that the objects exert on each other are called an action-reaction pair.

The force that object 1 exerts on object 2 is called the action force, while the force that object 2 exerts on object 1 is called the reaction force.

The action force is equal in magnitude and opposite in direction to the reaction force.

Force Pairs

• Forces always come in pairs1. Applied Force…force ON an object2. Frictional Force…opposes motion3. Normal Force…upward forces,

opposes weight4. Weight…acts downward

• Fa Fn Ff Fg

Identify the action-reaction pairs!

• A person takes a step• A snowball hits someone in the back• A baseball player catches a ball• A gust of wind strikes a window

Free-falling bodies

A…in a vacuum… a = g

B…in the atmosphere, air molecules resist the falling motion of an object

called Air Resistance (or Drag) …a type of frictional force

C. Air Resistance factors:

1. Surface area of the object….shape …more SA, more drag

2. Density of the air …dense air, more drag

3. Speed of the motion …faster moving object, more contact, more drag

D. Terminal Velocity…maximum velocity reached by a falling object

when the drag force equal the pull of the gravity

Fnet = Fg + Fdrag

When Fnet = 0, Fg = Fdrag (constant velocity)