Physics 111: Lecture 5 Today’s Agenda

Physics 111: Lecture 5Today’s Agenda

More discussion of dynamics

The Free Body Diagram

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

1Hukum-Hukum Newton

Review: Newton's Laws

Law 1: Sebuah benda yang tidak sedang mengalami gaya luar dikatakan berada pada keadaan bergerak atau bergerak dengan kecepatan tetap jika ditinjau dari suatu kerangka acuan inersial (diam)

Law 2: Untuk sembarang benda, FNET = ma

Dimana FNET = F

Law 3: Pasangan gaya aksi-reaksi adalah, FA ,B = - FB ,A.

FA ,B adalah gaya yang bekerja pada benda A sebagai hasil interaksinya dengan benda B

2Hukum-Hukum Newton

Berapakah besarnya gaya gravitasi yang ditimbulkan oleh bumi terhadap seorang mahasiswa?

Massa mahasiswa m = 55kg g = 9.81 m/s2. Fg = mg = (55 kg)x(9.81 m/s2 ) Fg = 540 N = BERAT

Gravitasi:

FE,S = -mg

FS,E = Fg = mg

3Hukum-Hukum Newton

Lecture 5, Act 1Massa dan Berat

Seorang astronot menendang bola di bumi sehingga kakinya terluka. Setahun kemudian, astronot tersebut menendang bola yang sama di permukaan bulan dengan gaya yang sama besarnya. Bagaimana kondisi luka yang akan dialaminya…

(a) Lebih parah

(b) Kurang (c) Sama

Ouch!

4Hukum-Hukum Newton

Lecture 5, Act 1Solution

Massa bola dan astronot tetap sama (di bulan maupun di bumi), sehingga kakinya akan mengalami luka dengan kondisi yang sama.

5Hukum-Hukum Newton

Lecture 5, Act 1Solution Berat bola dam

astronot menjadi berkurang di bulan

Jadi astronot tersebut akan lebih mudah mengangkat bola di bulan dari pada di bumi.

W = mgBulan gBulan < gBumi

Wow!That’s light.

6Hukum-Hukum Newton


Hukum II Newton menyatakan bahwa gaya yang dilakukan oleh sebuah benda F = ma.

Kata kunci: sebuah benda.

Sebelum menerapkan persamaan F = ma terhadap sebuah benda, maka gaya-gaya yang bekerja pada benda tersebut harus diuraikan.

7Hukum-Hukum Newton

The Free Body Diagram...

Perhatikan contoh di bawah ini:Gaya-gaya apa saja yang bekerja pada

palang?

P = plankF = floorW = wallE = earth

FW,P

FP,W

FP,F FP,E

FF,P FE,P

8Hukum-Hukum Newton


Pisahkan palang dari Sistem lingkunagnnya.

FW,P

FP,W

FP,F FP,E

FF,P FE,P

9Hukum-Hukum Newton


Gaya yang bekerja pada palang seimbang satu sama lain...

FP,W

FP,F FP,E

10Hukum-Hukum Newton

Pada contoh ini, palang tersebut tidak bergerak Tidak mengalami percepatan! FNET = ma menjadi FNET = 0

Ini merupakan konsep mendasar statika, yang akan dibahas lebih lanjut dalam beberapa minggu berikut.

FP,W + FP,F + FP,E = 0

FP,W

FP,F FP,E


Contoh

Contoh masalah dinamika:

Massa sebuah balok m = 2 kg tergelincir pada permukaan tanpa gesekan. Gaya Fx = 10 N mendorongnya dalam arah x. Berapakah percepatan yang dialami balok?

F = Fx i a = ?m

y

x


Example... Gambarkan semua gaya

yang bekerja pada balok

FFB,F

FF,BFB,E

FE,B

y

x


Example... Uraikan gaya-gaya yang bekerja

FFB,F

FF,BFB,E =mg

FE,B

y

x


Example... Gambarkan diagram

FFB,F

mg

y

x


Example... Selesaikan persamaan (Hk) Newton pada

masing-masing komponen. FX = maX

FB,F - mg = maY

FFB,F

mg

y

x


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.

Komponen vertikal gaya dari lantai terhadap benda (FB,F ) sering disebut gaya normal (N).

Karena aY = 0 , berarti N = mg in this case.

FX

N

mg

y

x


Example Recap

FX

N = mg

mg

aX = FX / m y

x


Lecture 5, Act 2Normal Force

Sebuah balok bermassa m diletakkan pada lantai lift yang sedang bergerak dipercepat ke atas. Bagaimana hubungan antara gaya gravitasi dengan gaya normal yang bekerja pada balok? (a) N > mg

(b) N = mg(c) N < mg

m

a



Semua gaya bekerja dalam arah y, so use:

Ftotal = ma

N - mg = maN = ma + mg

therefore N > mg

m

N

mg

a


Tools: Ropes & Strings Dapat digunakan untuk menarik benda dari

jarak jauh. Tegangan (T) pada posisi tertentu pada tali

adalah besarnya gaya yang bekerja pada seluruh bagian penampang lintang tali pada posisi tersebut. Gaya tersebut dapat dirasakan jika tali

dipotong dan kita memegang kedua ujungnya

An action-reaction pair.cut

TT

T


Tools: Ropes & Strings... Segmen horisontal dari sebuah tali dengan

massa m: Gambarkan diagram gaya (ignore gravity).

Gunakan Hk. II Newton (in x direction): FNET = T2 - T1 = ma

Jika m = 0 (i.e. the rope is light) maka T1 =T2

T1 T2

m

a x


Tools: Ropes & Strings... Seutas tali ideal (massa diabaikan) mempunyai

tegangan yang konstan pada setiap titik sepanjang tali

Jika tali memiliki massa, tegangan dapat berubah pada posisi tertentu Misalnya tali yang tergantung

dari langit-langit

We will deal mostly with ideal massless ropes.

T = Tg

T = 0

T T


Tools: Ropes & Strings... Gaya yang bekerja pada tali searah

dengan gerak tali itu sendiri

mg

T

m

Since ay = 0 (box not moving),

T = mg


Lecture 5, Act 3Force and acceleration Seekor ikan ditarik dari air dengan

menggunakan pancing yang akan patah jika tegangannya mencapai 180 N. Tali pancing akan putus jika percepatan gerak ikan melebihi 12.2 m/s2. Berapakah massa ikan tersebut (a) 14.8 kg

(b) 18.4 kg(c) 8.2 kg

m = ?a = 12.2 m/s2

snap !


Lecture 5, Act 3Solution:

Gambarkan diagram gayaT

mg

m = ?a = 12.2 m/s2 Gunakan Hk. II Newtondalam arah y

FTOT = ma

T - mg = ma

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

m Tg a

kg28sm21289

N180m2

...


Tools: Pegs & Pulleys Digunakan untuk mengubah arah

gaya Katrol ideal dapat mengubah arah gaya

tanpa mengubah besarnya gaya tersebut.

F1 ideal peg or pulley

F2

| F1 | = | F2 |mg

T

m T = mg

FW,S = mg


Springs Hukum Hooke: Gaya yang dilakukan

pegas setara dengan jarak regangan atau kompressi pegas dari posisi normalnya. FX = -k x Dimana x adalah jarak

pergeseran dari posisi normal dan k adalah konstanta keseimbangan.relaxed position

FX = 0

x


Springs... relaxed position

FX = -kx > 0

x

x 0

FX = - kx < 0

xx > 0

relaxed position


m m m

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

(1) (2)

?

Lecture 5, Act 4Force and acceleration

Balok beban yang bermassa 4 lbs digantungkan dengan tali yang terpasang pada penunjuk skala. Penunjuk skala menunjukkan gaya sebesar 4 lbs jika dikaitkan pada suatu dinding beton. Berapakah penunjukan skala jika dua buah beban masing-masing 4 lbs digantung bersebelahan pada penunjuk skala tersebut?


Lecture 5, Act 4Solution:Gambarkan giagram gaya pada salah satu

beban!! Gunakan Hk. II Newton dalam arah y:

FTOT = 0

T - mg = 0

T = mg = 4 lbs.

mg

T

m T = mg

a = 0 karena beban tidak bergerak


Lecture 5, Act 4Solution: Penunjuk skala akan membaca

besarnya tegangan pada tali, jadi skala akan tetap terbaca 4 lbs!

m m m

T T T T

TTT


Recap of today’s lecture.. More discussion of dynamics.

Recap (Text: 4-1 to 4-5)The Free Body Diagram (Text: 4-6)The tools we have for making & solving problems:

» Ropes & Pulleys (tension) (Text: 4-6)» Hooke’s Law (springs). (Text: 4-5, ex. 4-6)

Look at Textbook problems Chapter 4: # 45, 49, 63, 73


Review

Discussion of dynamics.

Review Newton’s 3 Laws


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


Review: 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: The tension is the same on both sides!F1 = -T i

ideal peg or pulley

F2 = T j

| F1 | = | F2 |

massless rope


Review: Springs 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 Where x is the displacement from the equilibrium and k is the constant of proportionality.

relaxed position

FX = 0x


Lecture 6, Act 1Springs A spring with spring constant 40 N/m has a

relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring? x = 0

Mk k

Mx = 0 x = 1.5

(a) -20 N (b) 60 N (c) -60 N

x = 1



Recall Hooke’s law:

FX = -kx Where x is the displacement from equilibrium.FX = - (40) ( .5)

FX = - 20 N

(a) -20 N (b) 60 N (c) -60 N


Problem: 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


Accelerometer...

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

m

T (string tension)

mg (gravitational force)

i


Accelerometer...

Using components (recommended):

i: FX = TX = T sin = ma

j: FY = TY - mg = T cos - mg = 0

T

mg

m

ma

j

i

TX

TY


Accelerometer...

Using components :

i: T sin = ma

j: T cos - mg = 0

Eliminate T :

T sin = maT cos = mg mg

m

ma

ga

tan

TX

TY j

i

T


Accelerometer...

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

Find the total vector force FNET:T

mg

FTOT

m

T (string tension)



Accelerometer...

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

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

So

ma

ga

mgma

tan

Tmg

ga

tan

m

T (string tension)



Accelerometer... 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

ga

tan


Problem: 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 ?

m

a


Inclined plane...

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

m

a

i

j


Inclined plane... Consider x and y components

separately: i: mg sin =ma. a = g sin

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

mg

Nmg sin

mg cos

ma

i

j


Inclined plane...

Alternative solution using vectors:

m

mgN

a = g sin iN = mg cos

j

i

j


Angles of an Inclined plane

ma = mg sin

mgN

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

Lecture 6, Act 2Forces 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

Mk

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 sinConsider x-direction:

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

F x,s = -kx 1



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

Mg sin-kx10 κsin Μgx1

θ

x 1

Mk

x

y

F x,g = Mg sin

F x,s = -kx 1

m20mN125

50sm189kg15x2

1 ....


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?m1 m2

F


Problem: Two Blocks Realize that F = (m1+ m2) a :

Draw FBD of block m2 and apply FNET = ma:

F2,1F2,1 = m2 a

F / (m1+ m2) = a

m22,1 ÷÷øö

ççèæ

m2m1

FF

Substitute for a :

m2

(m1 + m2)m2F2,1 F


Problem: 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?

m


Problem: 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

j

i

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

2 sin mg

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


Problem: 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

vT


Motion in a Circle...

Draw a Free Body Diagram (pick y-direction to be down):

We will use FNET = ma (surprise) First find FNET in y direction:

FNET = mg +T

Tmg

y


Motion in a Circle...

FNET = mg +T Acceleration in y direction:

ma = mv2 / R

mg + T = mv2 / R

T = mv2 / R - mg

R

T

v

mg

y

F = ma


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

mg

v

T= 0

Rgv


Lecture 6, Act 3Motion 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

mg N

v

(a) (b) (c)

Rgv =mRgv =mRg

v =62Hukum-Hukum Newton

Lecture 6, Act 3Solution mv2 / R = mg – N For N = 0:

R

v

mg N

Rgv


Recap of Today’s lecture:

Example Problems

Accelerometer Inclined plane (Text: example 6-

1) Motion in a circle (Text: 5-2, 9-

1)

Look at textbook problems Chapter 4: # 47 Chapter 5: # 51, 95


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Physics 111: Lecture 5 Today’s Agenda

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