i d Newton’s First & Second LLaw

Physics 1Physics 1

• Unit is the NEWTON(N)• Is by definition a push or a pull• Is by definition a push or a pull• Can exist during physical

contact(Tension, Friction, Applied Force)Force)

• Can exist with NO physical contact, called FIELD FORCES ( gravitational, electric, etc)electric, etc)

INERTIA – a quantity of matter, also called MASS. Unit for MASS = kilogram.

Weight or Force due to Gravity is how your MASS is effected by gravity. mgWy g y mgW =

NOTE: MASS and WEIGHT are NOT the same thing. MASS never changesNOTE: MASS and WEIGHT are NOT the same thing. MASS never changesWhen an object moves to a different planet.

What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s/s)?

NWNWmgW

MARS 96.272)2.3)(3.85(94.835)8.9)(3.85(

====→=

MARS ))((

An object in motion remains in motion in a jstraight line and at a constant speed OR an object at rest remains at rest, UNLESS acted j ,upon by an EXTERNAL (unbalanced) Force.There are TWO conditions here and one constraintThere are TWO conditions here and one constraint.

Condition #1 – The object CAN move but must be at a CONSTANT SPEEDCondition #2 – The object is at RESTjConstraint – As long as the forces are BALANCED!!!!! And if all the forces are balanced the SUM of all the forces is ZERO.

The bottom line: There is NO ACCELERATION in this case AND the object The bottom line: There is NO ACCELERATION in this case AND the object must be at EQILIBRIUM ( All the forces cancel out).

∑ =→= 00 Facc

A i t i l t ti f f l tA pictorial representation of forces complete with labels.

•Weight(mg) – Always drawn from the center, straight down

TFN

straight down•Force Normal(FN) – A surface force always drawn perpendicular to a surface.

TFf

W1,Fg1or m1g

perpendicular to a surface.•Tension(T or FT) – force in ropes and always drawn AWAY from object.j•Friction(Ff)- Always drawn opposing the motion.

m2g

FNFf

mg

Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.

TIPS for solving problems• Draw a FBD• Resolve anything into COMPONENTSy g• Write equations of equilibrium

S l f k• Solve for unknowns

A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50Nright at a constant speed with a force of 50N.

a) Calculate the Force of Frictionb) Calculate the Force Normal NFF fa 50==

FNFa NFmg n 98)8.9)(10( ===

mg

Ff

g n ))((

g

Suppose the same box is now pulled at an angle of 30 degrees above the horizontaldegrees above the horizontal.

a) Calculate the Force of Friction

b) Calculate the Force Normal

NFFNFF

axf

aax

3.433.4330cos50cos

===== θ

FN Fa

Naxf 3.3

mgFN !≠

Ff30

Fax

Fay

FmgF

mgFFmgF

N

ayN

N

30sin50)8.9)(10(

!

−→−=

=+≠

mgNF

FmgF

N

ayN

73

30sin50)8.9)(10(

=

→

If an object is NOT at rest or moving at a constant speed, that means the FORCES are UNBALANCED. One force(s) in a certain direction over power the others.p

THE OBJECT WILL THEN ACCELERATETHE OBJECT WILL THEN ACCELERATE.

The acceleration of an object is directly proportional to the NET FORCE andinversely proportional to the mass.

1 ∑= FFNET

FFm

aFa

NET

NET1αα ∑FFNET

Tips:maFm

a NETNET =→= Tips:

•Draw an FBD•Resolve vectors into components•Write equations of motion by adding and •Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.•Solve for any unknownsSolve for any unknowns

A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.p

FNFa maFF

maFNet =In which direction, is this object accelerating?

mg

Ff

101250 a

maFF fa

=−

=−accelerating?

The X direction!g

2/8.3101250

smaa

=So N.S.L. is worked out using the forces in the “x” direction in the x direction only

A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable.

maFT

FN

amTgmmaFNet

22 =−=

TamT 1=amamgm =−

m1g

122

122

212

)( mmagmamamgmamamgm

+=+==

m2g 2

21

2 /7.714

)8.9)(11( smmmgma =→

+=

maF 123))(3(amTgm

maFNet

22 =−= NT 1.23)7.7)(3( ==

amT 1=

mFmaF NETN t =→=

RiseSlope

ma

maFNet

=

→

RunSlope =

Where does the calculus fit in?

dtxdm

dtdvmamF

2

===rr There could be situations where you are

given a displacement function or velocity function. The derivative will need to bedtdt function. The derivative will need to be taken once or twice in order to get the acceleration. Here is an example.

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72-kg. The elevator starts from rest and travels upward with a speed that varies with time

220.03)( tttv +=according to:

When t = 4.0s , what is the reading on the bathroom scaleWhen t 4.0s , what is the reading on the bathroom scale (a.k.a. Force Normal)?

+=+

== 4003)20.03( 2

tttddva +=→=−=net

mgmaFmamgFmaF

=+=

+===

)4(40.03)4(

40.03

a

tdtdt

a

4.6 m/s/s=+=+=→=

)6.4)(72()8.9)(72(N

NN

FmgmaFmamgF

1036.8 N