Newton’s 1 st Law: The Law of Inertia “Every object maintains a state of rest or uniform motion...

Newton’s 1st Law:The Law of Inertia

“Every object maintains a state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force.”

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Newton’s 1st Law: The Law of Inertia

• Inertia– Depends on the mass of an object– Inertia is a resistance to a change in

motion– Large mass = Large inertia– Small mass = Small inertia

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• Simply put: – Objects like to stay where they are or

to keep moving as they have been– The heavier an object is, the harder it

is to change its state of motion.– Think of inertia as

• Lazy• “I’m a keep on doing what I’m doing”

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Newton’s 1st LawThe Law of Inertia

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If the net force acting on an object is zero, the object will stay at rest or continue to move at constant velocity

Example:– A computer sitting on a desk (at rest)

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“At rest” means Velocity = 0


– A hockey puck moving across smooth ice at a constant velocity

– A body in a vehicle keeps moving forward when the car stops suddenly• The seatbelt applies a force in the opposite

direction to stop the body from going forward

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“At rest” means Velocity = 0

Note: Be safe and always wear your seatbelt when driving or as a passenger!

Newton’s 2nd Law

“An unbalanced force causes an object to accelerate”

“If the net external force on an object is not zero, the object will accelerate in the direction of this net force”

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Newton’s 2nd Law• An object’s acceleration (a) is:

– Directly proportional to the applied unbalanced force (F)

– inversely proportional to the object’s mass (m)

F = ma (N) (kg) (m/s2)

F - NET FORCE in Newtons (N = kgm/s2 )m - mass is in kga - net acceleration is in m/s2

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Newton’s 2nd Law

• Newton’s 2nd Law tells us– The heavier an object is (i.e. more massive),

the greater the force is required to accelerate it• Example: a pickup truck requires a much larger

engine (larger force) compared to a small compact car.

– To increase the rate of acceleration requires a greater force as well• Example: consider the acceleration of a Smart

Car compared to a Lamborghini.– The Lamborghini’s bigger engine (bigger force) means it

has bigger acceleration

F = ma

What do you think?

Newton’s 2nd Law Example 1

Find the net force acting on a 10 kg object that is accelerating at 5.0m/s2 south.

G: = 5.0 m/s [S] m = 10 kg

R: Net force

A/S:

FNET = maFNET = (10)(-5)FNET = -50 N

= 50 N [S] P: The net force is 50 N South.

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Draw a FBD

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+N+E

- S

-W

Label Directions


Find the acceleration of a 15 kg object that is experiencing a net force of 30 N to the right.

G: = 30 N [R] m = 15 kg

R: net acceleration

A/S:

FNET = ma a = FNET /m a = +30 /15 a = +2.0 m/s2

∴ = 2.0 m/s2 [R] P: The acceleration is 2 m/s to the right.

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Draw a FBD

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+up+right

- down

-left

Label Directions


Determine the net force on the following object.

There is only movement in the x direction.

Perform the vector sum of the forces in the x direction

= (+7.0) + (-5.0) = 7.0 – 5.0 = 2.0 N

= 2.0 N [R] R means “right”

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+y+x

- y

-x

Label Directions7.0 N5.0 N



There is only movement in the y direction.

Perform the vector sum of the forces in the y direction

= (+5.0) + (-1.0) = 5.0 – 1.0 = 4.0 N

= 4.0 N [U] U means “up”

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+y+x

- y

-x

Label Directions

1.0 N

5.0 N

If the mass is 0.50 kg, use Newton’s 2nd Law to find the acceleration.

FNET = ma a = FNET /m a = 4.0/0.50 a = 8.0 m/s2 ∴ = 8.0 m/s2 [U]



There is movement in both x and y!

x direction = 7.0 – 5.0 = 2.0 N

y direction = 4.0 – 1.0 = 3.0 N

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+y+x

- y

-x

Label Directions

1.0 N

4.0 N

5.0 N 7.0 N

and are the components of the net force . Draw a right angle triangle:Fy points up (+), Fx points right (+).

NET is assumed

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Ѳ

F2 = Fx2 + Fy

2

F = F = F = 3.6 N

tan Ѳ = Fy / Fx

Ѳ = tan -1 (Fy / Fx)

Ѳ = tan -1 (3 /2) Ѳ = 56 ̊

Use Pythagoras & SOH CAH TOA

= 3.6 N [R56̊U]

Important Announcements

• Tuesday April 5th

– Automotive Safety Report (due Monday April 11th)

• Wednesday April 6th

– Review• Thursday April 7th

– Forces Unit Test• Friday April 8th

– PA Day!

Coefficient of Friction Tables


“Every object maintains a state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force.”

From Yesterday

Newton’s 2nd Law• An object’s acceleration (a) is:

– Directly proportional to the applied unbalanced force (F)

– inversely proportional to the object’s mass (m)

F = ma (N) (kg) (m/s2)

F - NET FORCE in Newtons (N = kgm/s2 )m - mass is in kga - net acceleration is in m/s2

From Yesterday

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x direction = 7.0 – 7.0 – 3.0 = -3.0 N

y direction = 3.0 + 4.0 - 4.0 = 3.0 N positive y direction

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+y+x

- y

-x

Label Directions

Draw a right angle triangle:Fy points +y. Fx points -x.

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�⃗� 𝑋

�⃗� F2 = Fx2 + Fy

2

F = F = F = 4.2 N

tan Ѳ = Fy / Fx

Ѳ = tan -1 (Fy / Fx)

Ѳ = tan -1 (3.0

/3.0) Ѳ = 45 ̊

= 4.2 N [L45 ̊U]

4.0 N

4.0 N

7.0 N 7.0 N

3.0 N

3.0 N

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For the tanѲ calculation, use positive FX, FY

numbers and then work out [L45 ̊U]

Newton’s 3rd Law:Action-Reaction

“Every action has an equal and opposite reaction”

– If object A exerts a force on object B, then object B exerts the exact same force back on object A but in the opposite direction.

A = B

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Newton’s 3rd Law:Action-Reaction

If you exert 10N of force on an object, the object will exert 10N of force back on you! • You punch a wall, it punches back with an

equal but opposite force

• A shotgun recoils. The gun exerts a force on the bullet to move it forward. The bullet exerts and equal but opposite force against the gun, causing it to recoil madly!

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Newton’s 3rd LawAction-Reaction

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Problem SolvingWith Newton’s Laws

A box of mass 40 kg is pushed horizontally to the right across a floor with a force of 185N. μK = 0.30.

a) Determine the weight of the box?Weight = G Given Newton’s 2nd Law, force is F = ma

We have:G = ma “a” is acceleration

On Earth for “a”, physics uses g = 9.81 m/s2

G = m = (40)(-9.81) = - 392N = 392N [Down]

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+U+R

- D

-LN

G

A = 185NF

FBD!


b) Determine the size of the normal force.

Normal force is in the y direction.

= mY from Newton’s 2nd Law

As there is no movement in the y direction: Y = 0

= mY = N + G

0 = N + (-392) N = + 392 N N = 392 N [Up]

Notice that FN = FG as we have previously discussed. As shown, Newton’s Law supports this.

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+U+R

- D

-L

N

G

A = 185NF

FBD!


c) Determine the size of the frictional force.

FF = µKFN FF = (0.3) (392)FF = 118N

∴F = 118N [LEFT]

+U+R

- D

-LN

G

A = 185NF

FBD!

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d) Determine the resulting acceleration of the box.

From the problem’s description, acceleration is along the x axis

= mX from Newton’s 2nd Law

= mX = F + A

(40)X = (-118) + (185) X = X = + 1.675 m/s2 X = 1.675 m/s2 [right]

+U+R

- D

-LN

G

A = 185NF

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Newton’s 3rd Law Example

Standing on perfectly smooth ice, Eric pushes Andrew with a force of 20N to the right. Eric’s mass is 75kg. Andrew’s mass is 65kg. Find:

a) The reaction force that Andrew applies against Ericb) Andrew’s accelerationc) Eric’s Acceleration

G: EonA = 20 N [R] mE = 75 kg mA = 65 kg

R: a) AonE b) A c) E

A/S: a) Since Eric pushes Andrew with a force of 20 N to the right, then Andrew produces a reaction force of 20N to the left (Newton’s 3rd Law)

AonE = - EonA

= - 20 N [R] = 20 N [L]

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Newton’s 3rd Law Example

b)

aA = F/mA

aA = (+20)/(65)aA= + 0.31 m/s2

= 0.31 m/s2 [R]

c)

aE = F/mA

aE = (-20)/(75)aA = - 0.267 m/s2

= 0.267 m/s2 [L]

Andrew and Eric both experienced the same force; however, their different masses mean they will have different accelerations.

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EonA = 20 N [R]

AonE = 20 N [L]

Two people are pushing on a boat of mass 200 kg. They both push with a force of 300 N at an angle of 25̊ to each side of the boat. Sketch a FBD and calculate the net force and net acceleration of the boat.

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++

300 N

300 N

25̊

25̊

We will use the Trigonometric Method to solve this

= Net Force = the vector sum of all forces acting on the objectWe add the given force vectors head to tail

Redraw the Force vectors so they make a triangle

300 N 300 N

25̊

25̊

300 N 300 N25̊

25̊

25̊ 25̊Ѳ Ѳ = 180 ̊ - 25 ̊-

25̊Ѳ = 130 ̊

Forces at Different AnglesNewton’s 2nd Law

25̊ uses the Z Rule

Forces at Different AnglesNewton’s 2nd Law (cont’d)

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To Find the “Net Force” we will use the cosine lawc2 = a2 + b2 – 2 abcos(C) or c =

300 N 300 N25̊

25̊ 25̊130 ̊

FNET

a = 300 N b = 300 N

25̊C =130 ̊

c = FNET

c = FNET = FNET = 543 N

FNET = m aNET

aNET = FNET/m aNET = 543 N/200 kg aNET = 2.72 m/s2

A person of mass 70 kg is sitting on a 20 kg toboggan. If two people are pulling her with two different ropes, find the person`s net force and net acceleration. The force of person 1 is 50N [E40̊N] and the force of person 2 is 60N[E25̊S] as seen from the top.

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N+E+

50 N

60 N

40̊

25̊

Forces at Different Angles & Newton’s 2nd Law

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N+E+

50 N

60 N

40̊

25̊

Use the Trigonometric Method to solve this

= Net Force

Redraw the Force vectors so they make a triangleVector addition is head to tail

40 ̊ is due to z rule

C=180̊-̊40̊-̊25̊C =115̊

Forces at Different Angles Newton’s 2nd Law

C

25̊

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c = FNET = FNET = 92.9 N

F = m a

a = F/m a = 92.9 N/(70 kg + 20 kg) a = 1.0̊3 m/s2

C = 115 ̊c =

a= = b

2nd Law WorksheetGarfield #3

G: Weight = G = 5106 N [DOWN] R: Y =? ENGINE = 2.5107 N [UP] Just lifting off means no Normal Force

A/S: Using Newton’s 2nd Law, consider X and Y directions

XX = m X =0 no x direction movement.

YY = m Y = +G vector sum of all Y forcesShow the known vectors using the correct signs given the directions m Y = 2.5107 - 5106

m Y = 2107

Y = (EQU I)

m is unknown! This is where we must look at the given data and think about all the force relationships we have discussed in this unit.

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ENGINE

G

2nd Law WorksheetGarfield #3 (cont`d)

We were given G = 5106 NWe know G = mg

Thus:5106 = mg5106 = m (9.81)m = 5.1105 kg

Substitute m = 5.1105 kg into Y = (EQU I)

Y = Y =39 m/s2

or Y =39 m/s2 [up]

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ENGINE

G

2nd Law WorksheetGarfield #4

G: Weight = G = 19600 N [DOWN] R: AIR = ? Y = 8 m/s2

MOTOR = 37600 N [UP]

A/S: Using Newton’s 2nd Law, consider X and Y directions

XX = m X =0 no x direction movement

YY = m Y = MOTOR +G + AIR vector sum of all Y forcesShow the known vectors using the correct signs given the directions m(8)= + AIR

8m= + AIR

m is unknown! Once again, this is where we must look at the given data and think about all the force relationships we have discussed in this unit…

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MOTOR

G

AIR

2nd Law WorksheetGarfield #4 (cont`d)

We were given G = 19600 NWe know G = mg

Thus:19600 = mg19600 = m (9.81)m = 1997.96 kg

Substitute m = 1997.96 kg into 8m= 18000 + AIR

(8)(1997.96) = 18000 + AIR

= 18000 + AIR AIR = -20̊16 N

or AIR = 20̊16 [DOWN]

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MOTOR

G

AIR

Solving Newton Problems• Read the problem and identify what’s given: FN, FG, FF, FA,

m, aX, aY, µ etc. Then identify what’s required and missing.

• Draw a FBD of all Forces. Label conventional directions.• Apply Newton’s 2nd Law

– Consider X direction• = mX = vector sum of all the x forces

– Solve for all possible values: FX or aX or m, Ff, Fa etc.

– Consider Y direction• = mY = vector sum of all the y forces

– Solve for all possible values: FY or aY ,m, or Fn, Fa etc.

– Note! If there is no x or y movement in the problem, then aX or aY

are 0!• This method usually yields 2 equations and 2 unknowns which we can

solve (ie. Find Ff, m or u etc.). • If still values are missing, try relating them to FF=uFN, FG=mg etc.

Look for clues in the problem (ie. “starting to move horizontally” then we know aX=0)

• IF! there is both non zero net Fx and Fy forces– We find the net force with:

• F2 = Fx2 + Fy

2 tan Ѳ = Fy / Fx (use absolute values)

F=ma

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ѲOften one of Fy or Fx is 0 so we don’t’ need to do this.

Newton’s 1 st Law: The Law of Inertia “Every object maintains a state of rest or uniform motion...

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