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Physics 111: Lecture 5 Today’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). Review: Newton's Laws. - PowerPoint PPT Presentation
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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
The Free Body Diagram
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
The Free Body Diagram...
Pisahkan palang dari Sistem lingkunagnnya.
FW,P
FP,W
FP,F FP,E
FF,P FE,P
9Hukum-Hukum Newton
The Free Body Diagram...
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
11Hukum-Hukum Newton
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
12Hukum-Hukum Newton
Example... Gambarkan semua gaya
yang bekerja pada balok
FFB,F
FF,BFB,E
FE,B
y
x
13Hukum-Hukum Newton
Example... Uraikan gaya-gaya yang bekerja
FFB,F
FF,BFB,E =mg
FE,B
y
x
14Hukum-Hukum Newton
Example... Gambarkan diagram
FFB,F
mg
y
x
15Hukum-Hukum Newton
Example... Selesaikan persamaan (Hk) Newton pada
masing-masing komponen. FX = maX
FB,F - mg = maY
FFB,F
mg
y
x
16Hukum-Hukum Newton
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
17Hukum-Hukum Newton
Example Recap
FX
N = mg
mg
aX = FX / m y
x
18Hukum-Hukum Newton
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
19Hukum-Hukum Newton
Lecture 5, Act 2Solution
Semua gaya bekerja dalam arah y, so use:
Ftotal = ma
N - mg = maN = ma + mg
therefore N > mg
m
N
mg
a
20Hukum-Hukum Newton
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
21Hukum-Hukum Newton
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
22Hukum-Hukum Newton
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
23Hukum-Hukum Newton
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
24Hukum-Hukum Newton
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 !
25Hukum-Hukum Newton
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
...
26Hukum-Hukum Newton
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
27Hukum-Hukum Newton
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
28Hukum-Hukum Newton
Springs... relaxed position
FX = -kx > 0
x
x 0
FX = - kx < 0
xx > 0
relaxed position
29Hukum-Hukum Newton
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?
30Hukum-Hukum Newton
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
31Hukum-Hukum Newton
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
32Hukum-Hukum Newton
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
33Hukum-Hukum Newton
Review
Discussion of dynamics.
Review Newton’s 3 Laws
The Free Body Diagram
The tools we have for making & solving problems:» Ropes & Pulleys (tension)» Hooke’s Law (springs)
34Hukum-Hukum Newton
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
35Hukum-Hukum Newton
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
36Hukum-Hukum Newton
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
37Hukum-Hukum Newton
Lecture 6, Act 1Solution
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
38Hukum-Hukum Newton
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
39Hukum-Hukum Newton
Accelerometer...
Draw a free body diagram for the mass: What are all of the forces acting?
m
T (string tension)
mg (gravitational force)
i
40Hukum-Hukum Newton
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
41Hukum-Hukum Newton
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
42Hukum-Hukum Newton
Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:T
mg
FTOT
m
T (string tension)
mg (gravitational force)
43Hukum-Hukum Newton
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)
mg (gravitational force)
44Hukum-Hukum Newton
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
45Hukum-Hukum Newton
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
46Hukum-Hukum Newton
Inclined plane...
Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.
m
a
i
j
47Hukum-Hukum Newton
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
48Hukum-Hukum Newton
Inclined plane...
Alternative solution using vectors:
m
mgN
a = g sin iN = mg cos
j
i
j
49Hukum-Hukum Newton
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
51Hukum-Hukum Newton
Lecture 6, Act 2Solution
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
52Hukum-Hukum Newton
Lecture 6, Act 2Solution
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 ....
53Hukum-Hukum Newton
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
54Hukum-Hukum Newton
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
55Hukum-Hukum Newton
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
56Hukum-Hukum Newton
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
57Hukum-Hukum Newton
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
58Hukum-Hukum Newton
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
59Hukum-Hukum Newton
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
60Hukum-Hukum Newton
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
61Hukum-Hukum Newton
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
63Hukum-Hukum Newton
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
64Hukum-Hukum Newton
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