What is a Force?

What is a Force?

 A Push or a Pull (in physics  no difference).

 Many different kinds of forces in nature (gravity, friction, air resistance +++).

 Weight is also a force  the force of gravity acting on an object (not be confused with mass which is the amount of matter an object has and is measured in kilograms)

 Forces can be measured with a spring scale .

 Units of force are called Newtons (N) (1 N = 1 kg·m/s

).

 Forces may cause masses to accelerate/decelerate.

 Newton’s 2

Law states “The amount of acceleration an

object experiences is directly related to the amount of force

acting on it” OR simply put: Force = mass x acceleration (F=ma)

F

= 30 N F

= 10 N

F

= 20 N

The object “feels” both forces, but feels an overall net force to the right.

 Forces are vector quantities (magnitude & direction).

 Therefore we need to figure out resultant or net force.

 Forces do not always cause an object to move.

 Equilibrium occurs when the net force equals zero.

What is a Force?

Draw the net force for each object.

5N

10 N

5N 10 N

5N 5 N

Monkey Joe pulled on a crate of bananas.

The crate feels the following forces: 25 N up, 50 N left, 35 N right, and 20 N down.

To the right  positive

F

= -50 N + 35 N F

= -15 N, to the left

a) What is the net-force in the x-direction (horizontal)?

b) What is the net-force in the y-direction (vertical)?

Up  positive

F

= 25 N - 20 N

F

= 5 N, up

c) What is the magnitude & direction of the overall net-force ?

-15 N

5 N F

F

= F

+ F

F

= (-15 N)

+ (5 N)

F

= 250 N

F

= 15.8 N; 18.4 ° above the –y axis

Magnitude Direction

 = ?

Tan θ = opposite adjacent

Tan θ = 5N/15N Tan θ = 0.33

F

= 15.8 N θ = 18.4°

Contact Forces vs. Field Forces

Contact Force: A force that must be touching an object to have an effect.

ex: you pushing on a chair

Field Force: A force that can have an effect even without

direct contact .

ex: force of gravity (i.e. weight)

An object in motion will stay in motion An object at rest will stay at rest AND

UNLESS

Acted on by a net external force

In other words, all things want to keep

doing what they’re already doing

Newton’s 1

Law is AKA “The Law of Inertia”

Inertia

o Not a measurable quantity, but a tendency

o Varies directly with the amount of mass an object has

o The more the mass, the more of a tendency of that object to want to keep on doing what it’s already doing

Demo with glass of water (add water) and piece of paper + coins on elbow

Examples:

1.Standing in the back of moving bus driver slams breaks 2.From a dead stop  driver hits the gas

3.Being struck by a car from behind  neck injuries 4.In a moving car  sharp right turn

5.Shoveling snow

A net force causes a mass to accelerate

F _net = ma

Can also be used for determining Weight

F _g = mg

F

= Net Force (N) m = mass (kg)

a = acceleration (m/s

)

Since Weight (F

) is the force of gravity acting on an object, Newton’s 2

Law may be re-written as

weight = mass x acceleration due to gravity

Where g = +10 m/s

+ for Force problems

F = ma

Mass = kg

Acceleration = m/s ² Force = kg·m/s ²

Newton (N) = kg·m/s ²

How much does a 1 kg object weigh on the earth’s surface?

F _g = mg

F _g = (1kg)(10 m/s ² )

F

= 10 N

weight is always a vector pointing

down… so if we call up positive,

then the weight would be negative.

What is the gravitational acceleration on a planet where a 2.0 kg mass has a weight of 16 N on the planet’s surface?

F _g = mg

16 N = (2 kg)g

g = 8 m/s ²

How much force must a 30,000 kg plane

develop to achieve an acceleration of 1.5 m/s

.

F = ma

F = (30,000 kg)(1.5 m/s

) F = 45,000 kg•m/s

F = 45,000 N

1a) What is the weight of a 17.3 kg TV set?

1b) What is the mass of an 87 N man?

Practice Questions

1c) What is the weight of a 2-gram metal washer?

1a) 173 N

1b) 8.7 kg

1c) 0.02 N

Only draw forces acting on the object

F

Note: Forces should be drawn from the center of the object.

F

(normal force)

Normal Force

• A normal force is the force of a surface on an object; it only exists when there is a surface in direct contact with the object.

Ex. table, floor, wall, ….etc

•The word “Normal” means perpendicular

•Don’t confuse Normal Force ( F

) ^with

Newtons (N), which is a unit of force.

Tension

w

• Tension is just the name of a

force in a rope/string/wire. T

• Tension is the same throughout any single piece of rope.

•Different ropes will have different tensions.

• Treat it exactly as you would

any force.

Draw a free-body diagram for the following objects.

The box sits motionless on the ground.

The box sits on the ground with Monkey Joe on it.

F

F

F

F

F

Net Force Equations

Use free-body diagrams to create (an)

equation(s) for the forces on an object in a specific direction (horizontal or vertical).

F

F

F

Vertical (y) direction:

up is positive F

= ma

F

– F

– F

= ma

Is it accelerating?

F

– F

– F

= 0

NO!

Draw a free-body diagram for the following objects.

A box is suspended

from the ceiling. A box is suspended from the ceiling.

F

F

T T T

T = F

2T = F

5 kg F

= 30 N F

= -10 N

A force of 30 N acts to the right on a 5-kg box atop a table, and a force of 10 N acts to the left on the same box. Find the acceleration of the box.

F

= ma F

+ (-F

) = ma 30 N + (-10 N) = (5 kg) a 20 N = (5 kg) a a = 4 m/s

to the right

F

F

VERTICAL Since F

= F

, there will not be any vertical movement or acceleration.

HORIZONTAL

A 5-kg box is lifted vertically by a pulley. The

rope has a tension of 80 N. Find the acceleration of the box. Draw a free-body diagram.

m = 5 kg T = 80 N

The Net force is only in the vertical direction

The 2 forces acting on the box are the Tension (T) up & the Weight (F_G) down

F

= T - F

Since the box is being lifted up, T > F_G

F

= T - mg

F_net= 80 N – (5 kg x 10 m/s²)

F

= 30 N

F

= ma

First, find F

Use F

to find “a”

a = F

/m

a = 30 N/5 kg

a = 6 m/s

1. Monkey Joe is pulling his 6 kg wagon full of bananas with a force of 30 N

horizontally to the right.

a) Draw a free-body diagram.

b) What is the acceleration of the wagon?

Practice Questions

F

F

F

= 30 N ^F

^{= F}

^{= 30 N}

No vertical movement means F

= F

F

= m

a a = F

/m

a = 5 m/s

to the right

2. Monkey Jane doesn’t want Monkey Joe to

take the wagon, so she applies a force of 42 N in the opposite direction as Monkey Joe.

a) Draw a free-body diagram.

b) Write the net-force equation in the y-direction.

Find the magnitude of the Normal Force?

c) Write the net-force equation in the x-direction.

Find the acceleration of the wagon?

F

F

F

= 30 N F

= 42 N

= mg = 6 kg x 10 m/s

= 60 N F

– F

= ma a = -2 m/s

, (to the left)

F

= F

3. Monkey Joe (m = 50 kg) stands on a bathroom scale in an elevator. What is the reading on the scale when:

Practice Questions

b) the elevator is moving down at a constant velocity a) the elevator is stopped between floors

c) the elevator is moving down at 2 m/s

d) the elevator is moving up at 2 m/s

Stopped means no acceleration  MJ’s actual weight = mg = 500 N

Constant velocity also means no acceleration  MJ’s actual weight = mg = 500 N

Acceleration down Scale Reading = Actual Weight – F_Net = 500 N - 100 N = 400 N

Acceleration up  Scale Reading = Actual Weight + F_Net = 500 N + 100 N = 600 N

A Martian lander is approaching the surface.

It is slowing its descent by firing its rocket motor. Which is the correct free-body

diagram for the lander?

Which will hit the ground first, a 1 kg mass, or a piece of chalk?

Make an argument for why the 1 kg mass would hit first.

Make an argument for why the piece of

chalk would hit the ground first.

For every

action, there is an equal

and opposite

reaction.

Forces always exist in pairs and occur simultaneously.

If every force has an equal an opposite reaction, why don’t those forces just cancel out?

(i.e. If I apply a force on a box and that box applies the same size force on me, then why can I move the box?) Because the action and reaction forces always act on different objects which have different masses.

A on B (action) B on A (reaction)

Although both forces have the same magnitude, the motion of the box is affected only by the force

acting on it  that is why I’ll be able to accelerate it.

Do the cannon and cannon ball have the same acceleration? Explain.

Do the cannon and cannonball experience

the same size force? Explain.

Car B is stopped for a red light. Car A, which has the same mass as car B, doesn’t see the red light and

runs into the back of B. Which of the following statements is true?

1. B exerts a force on A but A doesn’t exert a force on B.

2. B exerts a larger force on A than A exerts on B.

3. B exerts the same amount of force on A as A exerts on B.

4. A exerts a larger force on B than B exerts on A.

5. A exerts a force on B but B doesn’t exert a force

on A.

A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the

mass of the car. Which of the following statements is true?

1. The car exerts a force on the truck but the truck doesn’t exert a force on the car.

2. The car exerts a larger force on the truck than the truck exerts on the car.

3. The car exerts the same amount of force on the truck as the truck exerts on the car.

4. The truck exerts a larger force on the car than the car exerts on the truck.

5. The truck exerts a force on the car but the car

doesn’t exert a force on the truck.

When a gun is fired, how does the size of the force of the gun on the bullet compare to the force of the bullet on the gun?

How do the accelerations of the rifle and bullet compare? Defend your answers.

Same Force on each – Newton’s 3

Law Different Accelerations:

 gun has a small acceleration.

 bullet has a large acceleration.

A constant force of 6 Newtons is applied

horizontally to a 2 kg toy train on a horizontal surface. The toy train is initially at rest.

Practice Questions

a) Draw a free body diagram of the toy train with all the forces acting on it. Label each vector.

F

F

F = 6 N

b) Find the acceleration of the toy train.

F = ma; a = F/m = 6 N/2 kg = 3 m/s

c) Find the velocity of the toy train at 5 seconds.

d) Find the distance traveled by the toy train from 0 to 5 seconds.

Horizontal v

= 0 m/s v

= ? m/s a = 3 m/s

t = 5 s

v

= v

+ at

v

= 0 m/s + (3 m/s

x 5 s) v

= 0 m/s + 15 m/s

v

= 15 m/s

 _x

_{= ?}

 x

= v

t + ½at

 x

= 0 m + ½(3 m/s

)(5 s)(5 s)

 x

= 37.5 m

A 24-N force (F) is applied to the three blocks.

a) Find the acceleration of the system.

F = ma = m

x a; a = F/m

= 24 N/3.0 kg = 8 m/s

b) Find tension T

.

F = m

a = 0.5 kg x 8 m/s

= 4 N to the right c) Find tension T

.

F = m

a = (0.5 kg + 1 kg) x 8 m/s

= 12 N to the right

Practice Questions

A force that opposes the motion of an object.

Examples:

• Friction between the tires on your car and the road

•Air resistance (drag)

• leads to a terminal velocity

F

F

Static Friction : F

= μ

F

Friction is a force vector pointing in the

opposite direction as the motion of an object.

Static Friction  equal & opposite in direction to the applied force on an object (keeps it at rest)

F

= Force of static friction (Newtons) F

= Normal force (in Newtons)

μ

= coefficient of static friction

- represents how rough or smooth a surface is.

- has NO UNITS…it’s just a ratio.

Kinetic Friction (or Sliding Friction)

F

= μ

F

The force of static friction ^is

always larger than the force of

. You can have EITHER static friction (at rest) OR kinetic friction (moving) at one time, never both at the same instant .

F

= Force of kinetic friction F

= Normal force

μ

= coefficient of kinetic friction

(or the frictional force on a moving object)

Draw a free-body diagram for the following objects.

Monkey Joe pushes the box to the right, but it does not move.

Monkey Joe pushes the box to the right, and it moves at a

constant velocity.

F

F

F

F

F

F

F

F

1a) Monkey Joe pushes a 3 kg box of

bananas with a force of 25 N. If there is a frictional force of 10 N, what is the

acceleration of the box?

3 kg a = __ m/s 5

b)What would happen if Monkey Joe

reduced his force and only pushed with a force of 10 N?

3 kg a = __ m/s 0

MJ would not be able to

overcome static friction

Air Resistance (aka drag)

Air resistance is a type of friction that is due to air molecules colliding with a moving object.

• molecules apply a force on the object

The faster an object moves, the more air

resistance it feels.

Terminal Velocity

As the force of air resistance increases ( F

) , the acceleration of a falling object decreases.

When the force of air resistance pushing up is equal to the force of gravity pulling down, the object has reached terminal velocity.

- If the forces up equal the forces down, what

is the acceleration of the object? 0 m/s

Draw a free-body diagram for the following objects.

An object is falling and accelerating at

10 m/s

.

An object is falling and has reached terminal velocity.

F

F

F

F

Where F

= air

resistance force

Factors That Affect Friction

1. The type of surfaces in contact.

•Sandpaper causes more friction than ice.

• Rubber on concrete has more friction than rubber on steel.

2. How much the surfaces are pressed together.

• Heavier objects have more friction

than light objects.

1. A 2 kg object is on a flat horizontal surface. 

= 0.2 and 

= 0.1 ^F

F

a) Draw a free body

diagram for the block.

b) How much horizontal force will start the block moving?

F

= μ

F

c) How much horizontal force will keep it moving at a constant speed.

F

= F

= mg = 2 kg x 10 m/s

= 20 N

= 0.2 x 20 N = 4 N

F

= μ

F

= 0.1 x 20 N = 2 N Practice Questions

(no applied forces, so no friction yet)

2. A constant force of 16 Newtons (F

) is applied to the right on a block sitting on a horizontal

surface for 8 seconds. The block has a mass of 3 kg and is initially at rest.

Practice Questions

b) Find the acceleration of the block.

F

F

F

= 16 N



= 0.3 & 

= 0.2

F

a = F

/m F

= F

– F

= F

– (μ

F

) = F – (μ

mg)

F

= 16 N – (0.2 x 3 kg x 10 m/s

) = 10 N

a) Draw a free body diagram of the block.

a = 10 N/3 kg

a = 3.33 m/s

c) Find the velocity of the block at 5 sec.

d) Find the distance traveled by the block from the 5 to 8 second mark.

Horizontal v

= 0 m/s v

= ? m/s

a = 3.33 m/s

t = 5 s

v

= v

+ at

v

= 0 m/s + (3.33 m/s

x 5 s) v

= 0 m/s + 16.65 m/s

v

= 16.67 m/s

Horizontal

v

= 16.67 m/s a = 3.33 m/s

t = 8 – 5 = 3 s

 x

= ? m

 x

= v

t + ½at

 x

= (16.67)(3) + ½(3.33)(3)(3)

 x

= 50 m + 15 m

 x

= 65 m