physics 211: lecture 5, pg 1 physics 211: lecture 5 today’s agenda l more discussion of dynamics ...

Physics 211: Lecture 5, Pg 1

Physics 211: Lecture 5Physics 211: Lecture 5

Today’s AgendaToday’s Agenda

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)

Original: Mats A.SelenModified: W.Geist


MaterialsMaterials

Bring strings and springs. Two spring powered carts on track:

Use equal and unequal masses.

Have students predict the outcome (i.e. accelerations and final speeds)


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


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


Lecture 5, Lecture 5, Act 1Act 1Mass 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


Lecture 5, Lecture 5, Act 1Act 1SolutionSolution

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.



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


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:


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



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



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

FFP,W

FFP,F FFP,E


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


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


Example...Example...

Draw a picture showing all of the forces

FFFFB,F

FFF,BFFB,E

FFE,B

y y

x x


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


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


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


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


Example RecapExample Recap

FX

N = mg

mg

aX = FX / m y y

x x


Lecture 5, Lecture 5, Act 2Act 2Normal Force Normal Force

(Check your answer: Similar to mass (Check your answer: Similar to mass attached to string) attached to string)

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



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


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.


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



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

2 skateboards



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


Lecture 5, Lecture 5, Act 3Act 3Force 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 kgFor what object do you draw the FBD?


Lecture 5, Lecture 5, Act 3Act 3Solution: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.

..


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 |


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


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


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.


relaxed position

FX = -kx > 0

xx 0


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 = - kx < 0

xx > 0

relaxed position

Horizontalsprings


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

Spring/string


m m m

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

(1) (2)

?

Lecture 5, Lecture 5, Act 4Act 4Force 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?

Scale on a skate



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



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

m m m

T T T T

TTT


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

More discussion of dynamics (Read Chapter 4 in text)

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

» Ropes & Pulleys (tension)

» Hooke’s Law (springs).

physics 211: lecture 5, pg 1 physics 211: lecture 5 today’s agenda l more discussion of dynamics ...

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