7 problems of newton law

Physics 111: Lecture 5, Pg 1

Physics 111: Lecture 5Physics 111: Lecture 5

Today’s AgendaToday’s Agenda

More discussion of dynamics

Recap

The Free Body DiagramFree Body Diagram

The tools we have for making & solving problems:

» Ropes & Pulleys (tension)

» Hooke’s Law (springs)


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.


Gravity:Gravity:

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

Typical student mass m = 55kgg = 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 and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force.

His foot hurts...

(a) more

(b) less (c) the same

Ouch!


Lecture 5, Lecture 5, Act 1Act 1SolutionSolution

Ouch!

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.



Wow!

That’s light. 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..

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 caseWhat are the forces acting on the plank ?

P = plank

F = floor

W = wall

E = earthFFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P



Consider the following caseWhat 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/s 2.

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 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



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


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

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



The direction of the force provided by a rope is along the direction of the rope:

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 kg


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...



relaxed position

FX = -kx > 0

xx 0


Springs...Springs...



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


Problem: AccelerometerProblem: Accelerometer

A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g.

a

i i


Accelerometer...Accelerometer...

Draw a free body diagram for the mass: What are all of the forces acting?

m

TT (string tension)

mgg (gravitational force)

i i



Using components Using components (recommended):

ii: FX = TX = T sin = ma

jj: FY = TY mg

= T cos mg = 0 TT

mgg

mmaa

jj

ii

TX

TY



Using components Using components :

ii: T sin = ma

jj: T cos - mg = 0

Eliminate T :

mgg

m

maaT sin = ma

T cos = mgtan

a

g

TX

TY

jj

ii

TT



Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):

Find the total vector force FFNET:

TT

mgg

FFTOT

m

TT (string tension)




Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):

Find the total vector force FFNET: Recall that FFNET = ma:

So

maa

tan ma

mg

a

g

TT

mgg

tan a

g

m

TT (string tension)




Let’s put in some numbers:

Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s.Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 .

= arctan (a/g) = 15.6 deg

tan a

g

a

Cart w/ accelerometer


Problem: Inclined planeProblem: Inclined plane

A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ?

ma


Inclined plane...Inclined plane...

Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.

ma

ii

jj



Consider x and y components separately: ii: mg sin =ma. a = g sin

jj: N - mg cos = 0. N = mg cos

mgg

NN

mg sin

mg cos

maa

ii

jj

Incline



Alternative solution using vectors:

m

mgN

a = g sin ii

N = mg cos jj

ii

jj


Angles of an Inclined planeAngles of an Inclined plane

ma = mg sin

mgN

The triangles are similar, so the angles are the same!


Lecture 6, Lecture 6, Act 2Act 2Forces and MotionForces and Motion

A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1?

(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm

x = 0

M

k

x 1 = ?

Mk

= 30o



x 1

Mk

x

y

Choose the x-axis to be along downward direction of ramp.

Mg

FBD: The total force on the block is zero since it’s at rest.

N

Fx,g = Mg sin

Force of gravity on block is Fx,g = Mg sin Consider x-direction:

Force of spring on block is Fx,s = -kx1

F x,s = -kx 1



x 1

Mk

x

y

Since the total force in the x-direction must be 0:

F x,g = Mg sin

F x,s = -kx 1

Mg sinkx1κ

sin Μgx

1

θ

m20mN125

50sm189kg15x

2

1.

...


Problem: Two Blocks Problem: Two Blocks

Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2?

mm11 mm22

F


Problem: Two BlocksProblem: Two Blocks

Realize that F = (mm11+ mm22) a : Draw FBD of block mm22 and apply FNET = ma:

F2,1F2,1 = mm22 a

F / (mm11+ mm22) = a

m22,1

m2m1

FF

Substitute for a :

mm22

(m1 + m2)m2F2,1 F


Problem: Tension and AnglesProblem: Tension and Angles

A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope?

mm


Problem: Tension and AnglesProblem: Tension and Angles

Draw a FBD:

T1 T2

mg

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0

T1sin T2sin

T2cos T1cos

jj

ii

Fx,NET = T1cos - T2cos = 0 T1 = T2

2 sin mg

T1 = T2 =Fy,NET = T1sin + T2sin - mg = 0


Problem: Motion in a CircleProblem: Motion in a Circle

A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v.What is the tension T in the string at the top of the rock’s

trajectory?

R

v

TT

Tetherball


Motion in a Circle...Motion in a Circle...

Draw a Free Body Diagram (pick y-direction to be down): We will use FFNET = maa (surprise) First find FFNET in y direction:

FFNET = mg +T

TTmgg

y y


Motion in a Circle...Motion in a Circle...

FFNET = mg +T

Acceleration in y direction:

maa = mv2 / R

mg + T = mv2 / R

T = mv2 / R - mg

R

TT

v

mgg

y y

F = ma


Motion in a Circle...Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?

i.e. find v such that T = 0.

mv2 / R = mg + T

v2 / R = g

Notice that this doesnot depend on m.

R

mgg

v

T= 0

v Rg

Bucket


Lecture 6Lecture 6, , Act 3Act 3Motion in a CircleMotion in a Circle

A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground?

R

mgg NN

vv

(a) (b) (c) Rgv =mRgv =mRg

v =

Trackw/ bump


Lecture 6Lecture 6, , Act 3Act 3SolutionSolution

mv2 / R = mg - N

For N = 0:

R

v

mgg NN

v Rg


Recap of Today’s lecture:Recap of Today’s lecture:

The Free Body DiagramFree Body Diagram

The tools we have for making & solving problems:

» Ropes & Pulleys (tension)

» Hooke’s Law (springs)

Accelerometer Inclined plane Motion in a circle

7 problems of newton law

Education

plank f

forces ff ffb

force fx

typical physics student

f mgg yy

object b

object ff

external forces