physics 211: lecture 5, pg 1 physics 101 l more discussion of dynamics recap newton's laws free...

Physics 211: Lecture 5, Pg 1

Physics Physics 101101

More discussion of dynamics

Recap Newton's Laws

The Free Body DiagramFree Body Diagram

The tools we have for making & solving problems:Ropes & Pulleys (tension)Hooke’s Law (springs)

Physics 211: Lecture 5, Pg 2

Review: Newton's LawsReview: Newton's Laws

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FFNET = maa

Where FFNET = FF

Law 3: Forces occur in action-reactionaction-reaction pairs, FFA ,B = - FFB ,A.

Where FFA ,B is the force acting on object A due to its interaction with object B and vice-versa.

m is “mass” of object

Physics 211: Lecture 5, Pg 3

Gravity:Gravity:Mass vs WeightMass vs Weight

What is the force of gravity exerted by the earth on a typical physics student?

Typical student mass m = 55kg g = 9.81 m/s2. Fg = mg = (55 kg)x(9.81 m/s2 )

Fg = 540 N = WEIGHT

FFE,S = -= -mg g

FFS,E = F = Fg = = mg g

Physics 211: Lecture 5, Pg 4

Mass vs. WeightMass vs. Weight An astronaut on Earth kicks a bowling ball straight ahead

and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon in the same manner with the same force.

His foot hurts...

(a) more

(b) less (c) the same

Physics 211: Lecture 5, Pg 5

SolutionSolution The masses of both the bowling ball and the astronaut remain

the same, so his foot will feel the same resistance and hurt the same as before.

Ouch.

Physics 211: Lecture 5, Pg 6

SolutionSolution However the weights of

the bowling ball and the astronaut are less:

Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth.

W = mgMoon gMoon < gEarth

Physics 211: Lecture 5, Pg 7

The Free Body DiagramThe Free Body Diagram

Newton’s 2nd Law says that for an object FF = maa.

Key phrase here is for an objectfor an object.. Object has mass and experiences forcesObject has mass and experiences forces

So before we can apply FF = maa to any given object we isolate the forces acting on this object:

Physics 211: Lecture 5, Pg 8

The Free Body Diagram...The Free Body Diagram...

Consider the following case as an example of this…. What are the forces acting on the plank ? Other forces act on F, W and E. focus on plank

P = plank

F = floor

W = wall

E = earthFFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

Physics 211: Lecture 5, Pg 9

The Free Body Diagram...The Free Body Diagram...

Consider the following case What are the forces acting on the plank ?

Isolate the plank from

the rest of the world.

FFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

Physics 211: Lecture 5, Pg 10

The Free Body Diagram...The Free Body Diagram...

The forces acting on the plank should reveal themselves...

FFP,W

FFP,F FFP,E

Physics 211: Lecture 5, Pg 11

Aside...Aside...

In this example the plank is not moving... It is certainly not accelerating! So FFNET = maa becomes FFNET = 0

This is the basic idea behind statics, which we will discuss in a few weeks.

FFP,W + FFP,F + FFP,E = 0

FFP,W

FFP,F FFP,E

Physics 211: Lecture 5, Pg 12

ExampleExample

Example dynamics problem:

A box of mass m = 2 kg slides on a horizontal frictionless floor. A force Fx = 10 N pushes on it in the xx direction. What is the acceleration of the box?

FF = Fx ii aa = ?

m

y y

x x

Physics 211: Lecture 5, Pg 13

Example...Example...

Draw a picture showing all of the forces

FFFFB,F

FFF,BFFB,E

FFE,B

y y

x x

Physics 211: Lecture 5, Pg 14

Example...Example...

Draw a picture showing all of the forces. Isolate the forces acting on the block.

FFFFB,F

FFF,BFFB,E = mgg

FFE,B

y y

x x

Physics 211: Lecture 5, Pg 15

Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram.

FFFFB,F

mgg

y y

x x

Physics 211: Lecture 5, Pg 16

Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newton’s equations for each component.

FX = maX

FB,F - mg = maY

FFFFB,F

mgg

y y

x x

Physics 211: Lecture 5, Pg 17

Example...Example... FX = maX

So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.

FB,F - mg = maY

But aY = 0 So FB,F = mg.

The vertical component of the forceof the floor on the object (FB,F ) isoften called the Normal Force Normal Force (N).

Since aY = 0 , N = mg in this case.

FX

N

mg

y y

x x

Physics 211: Lecture 5, Pg 18

Example RecapExample Recap

FX

N = mg

mg

aX = FX / m y y

x x

Physics 211: Lecture 5, Pg 19

Normal ForceNormal Force A block of mass m rests on the floor of an elevator that is

accelerating upward. What is the relationship between the force due to gravity and the normal force on the block?

m

(a)(a) N > mgN > mg

(b)(b) N = mgN = mg

(c)(c) N < mgN < mg

a

Physics 211: Lecture 5, Pg 20

SolutionSolution

m

N

mg

All forces are acting in the y direction, so use:

Ftotal = ma

N - mg = ma

N = ma + mg

therefore N > mg

a

Physics 211: Lecture 5, Pg 21

Tools: Ropes & StringsTools: Ropes & Strings

Can be used to pull from a distance. TensionTension (T) at a certain position in a rope is the magnitude of the

force acting across a cross-section of the rope at that position. The force you would feel if you cut the rope and grabbed the

ends. An action-reaction pair.

cut

TT

T

Tension doesn’t have a direction. When you hook up a wire to an object the direction is determined by geometry of the hook up.

Physics 211: Lecture 5, Pg 22

Tools: Ropes & Strings...Tools: Ropes & Strings...

Consider a horizontal segment of rope having mass m: Draw a free-body diagram (ignore gravity).

Using Newton’s 2nd law (in xx direction): FNET = T2 - T1 = ma

So if m = 0 (i.e. the rope is light) then T1 =T2 T is constant anywhere in a rope or string

as long as its massless

T1 T2

m

a x x

Physics 211: Lecture 5, Pg 23

Tools: Ropes & Strings...Tools: Ropes & Strings...

An ideal (massless) rope has constant tension along the rope.

If a rope has mass, the tension can vary along the rope For example, a heavy rope

hanging from the ceiling...

We will deal mostly with ideal massless ropes.

T = Tg

T = 0

T T

Physics 211: Lecture 5, Pg 24

Tools: Ropes & Strings...Tools: Ropes & Strings...

What is force acting on box by the rope in the picture below? (always assume rope is massless unless told different)

mg

T

m

Since ay = 0 (box not moving),

T = mg

Physics 211: Lecture 5, Pg 25

Force and accelerationForce and acceleration A fish is being yanked upward out of the water using a fishing line that breaks when the tension

reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s2. What is the mass of the fish?

m = ?a = 12.2 m/s2

snap ! (a) 14.8 kg

(b) 18.4 kg

(c) 8.2 kg

Physics 211: Lecture 5, Pg 26

Solution:Solution: Draw a Free Body Diagram!!

T

mg

m = ?a = 12.2 m/s2

Use Newton’s 2nd lawin the upward direction:

FTOT = ma

T - mg = ma

T = ma + mg = m(g+a)

mT

g a

kg28

sm21289

N180m

2.

..

Physics 211: Lecture 5, Pg 27

Tools: Pegs & PulleysTools: Pegs & Pulleys

Used to change the direction of forces

An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:

FF1 ideal peg

or pulley

FF2

| FF1 | = | FF2 |

Physics 211: Lecture 5, Pg 28

Tools: Pegs & PulleysTools: Pegs & Pulleys

Used to change the direction of forces

An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:

mg

T

m T = mg

FW,S = mg

Physics 211: Lecture 5, Pg 29

SpringsSprings

Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.

FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.

relaxed position

FX = 0

x

Physics 211: Lecture 5, Pg 30

Springs...Springs...

Hooke’s Law:Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.

FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.

relaxed position

FX = -kx > 0

xx 0

Physics 211: Lecture 5, Pg 31

Springs...Springs...

Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.

FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.

FX = - kx < 0

xx > 0

relaxed position

Physics 211: Lecture 5, Pg 32

Scales:Scales:

Springs can be calibrated to tell us the applied force. We can calibrate scales to read Newtons, or... Fishing scales usually read

weight in kg or lbs.

02468

1 lb = 4.45 N

Physics 211: Lecture 5, Pg 33

m m m

(a)(a) 0 lbs. (b)(b) 4 lbs. (c)(c) 8 lbs.

(1) (2)

?

Force and accelerationForce and acceleration

A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs?

Physics 211: Lecture 5, Pg 34

Solution:Solution: Draw a Free Body Diagram of one

of the blocks!!

Use Newton’s 2nd Lawin the y direction:

FTOT = 0

T - mg = 0

T = mg = 4 lbs.

mg

T

m T = mg

a = 0 since the blocks are stationary

Physics 211: Lecture 5, Pg 35

Solution:Solution:

The scale reads the tension in the rope, which is T = 4 lbs in both cases!

m m m

T T T T

TTT

Physics 211: Lecture 5, Pg 36

Recap of today’s lecture..Recap of today’s lecture..

More discussion of dynamics

Recap The Free Body DiagramFree Body DiagramThe tools we have for making & solving problems:

» Ropes & Pulleys (tension)

» Hooke’s Law (springs).