Download - Physics 111: Lecture 8 Today’s Agenda
Physics 111: Lecture 8, Pg 1
Physics 111: Lecture 8
Today’s Agenda Friction Recap Drag Forces
Terminal speed Dynamics of many-body systems
Atwood’s machineGeneral case of two attached blocks on inclined planesSome interesting problems
Physics 111: Lecture 8, Pg 2
Friction Review:
Friction is caused by the “microscopic” interactions between the two surfaces:See discussion in text
Physics 111: Lecture 8, Pg 3
Model for Friction
Arah vektor gaya gesek fF adalah tegak lurus dengan vektor gaya normal N, dan berlawanan arah dengan gaya yang bekerja pada sistem
Gesekan Kinetic (sliding): Besarnya vektor gaya gesek berbanding lurus dengan besarnya gaya normal N.
fF = KN
Gesekan Static: Gaya gesekan menyeimbangkan total yang yang bekerja pada sistem sehingga sistem tidak bergerak. Besarnya gaya gesek statis maksimum berbanding lurus dengan gaya normal N.
fF SN
Physics 111: Lecture 8, Pg 4
Kinetic Friction:
K koefisien gesek kinetik.i : F KN = maj : N = mg
so F Kmg = ma
maF
mg
N
i
j
KN
Physics 111: Lecture 8, Pg 5
Static Friction:
Koefisien gesek statis S, menentukan gaya gesek statis maksimum, SN, yang timbul antara persentuhan dua buah benda.
S dapat ditentukan dengan cara memperbesar F sampai benda mulai bergeser:
FMAX - SN = 0 N = mg FMAX = S mg S FMAX / mg
FMAX
mg
N
i
j
S N
Physics 111: Lecture 8, Pg 6
Lecture 8, Act 1Two-body dynamics
Sebuah balok bermassa m, jika diletakkan pada permukaan bidang miring yang kasar ( > 0) dan diberikan gaya sesaat, akan bergerak turun pada bidang miring dengan kecepatan konstan. Jika balok yang sejenis(same ) denganmassa 2m diletakkan pada bidang miring yang sama kemudian diberikan gaya sesaat, maka balok tersebut akan:
(a) berhenti (b) dipercepat (c) Bergerak dengan kecepatan tetap
m
Physics 111: Lecture 8, Pg 7
Lecture 8, Act 1Solution
Gambarkan diagram benda bebas dan tentukan gaya total dalam arah X
q
i
j
mg
N
q
KN
FNET,X = mg sin q Kmg cos q
= ma = 0 (kasus balok pertama)
Jika massa digandakanmaka semua komponen menjadi dua kaliLebih besar. Tetapi percepatan tetap nol
Speed will still be constant!
Physics 111: Lecture 8, Pg 8
Drag Forces:
When an object moves through a viscous medium, like air or water, the medium exerts a “drag” or “retarding” force that opposes the motion of the object.
Fg = mg
FDRAG
j v
Physics 111: Lecture 8, Pg 9
Drag Forces:
This drag force is typically proportional to the speed v of the object raised to some power. This will result in a maximum (terminal) speed.
Fg = mg
FD = bvn
j
v feels like n=2
Parachute
Physics 111: Lecture 8, Pg 10
Terminal Speed:
Suppose FD = bv2. Sally jumps out of a plane and after falling for a while her downward speed is a constant v.What is FD after she reaches this terminal speed?What is the terminal speed v?
FTOT = FD - mg = ma = 0.
FD = mg
Since FD = bv2
bv2 = mg Fg = mg
FD = bv2
j
v
bmg
v =
Physics 111: Lecture 8, Pg 11
Many-body Dynamics
Systems made up of more than one object
Objects are typically connected:
By ropes & pulleys today
By rods, springs, etc. later on
Physics 111: Lecture 8, Pg 12
Atwood’s Machine:
Find the accelerations, a1 and a2, of the masses.
What is the tension in the string T ?
Masses m1 and m2 are attached to an ideal massless string and hung as shown around an ideal massless pulley.
Fixed Pulley
m1
m2
j
a1
a2
T1 T2
Physics 111: Lecture 8, Pg 13
Atwood’s Machine... Draw free body diagrams for each object Applying Newton’s Second Law: ( j -components)
T1 - m1g = m1a1
T2 - m2g = m2a2
But T1 = T2 = T since pulley is ideal
and a1 = -a2 = -a.since the masses are connected by the string
m2gm1g
Free Body Diagrams
T1 T2
ja1 a2
Physics 111: Lecture 8, Pg 14
T - m1g = -m1 a (a) T - m2g = m2 a (b)
Two equations & two unknownswe can solve for both unknowns (T and a).
subtract (b) - (a): g(m1 - m2 ) = a(m1+ m2 )
a =
add (b) + (a): 2T - g(m1 + m2 ) = -a(m1 - m2 ) =T = 2gm1m2 / (m1 + m2 )
Atwood’s Machine...
21
221
mm)mm(
g+-
-
g)mm()mm(
21
21
+-
Physics 111: Lecture 8, Pg 15
Atwood’s Machine...
m1
m2
j
a
a
TT
So we find:
a m mm m
g
( )( )
1 2
1 2
g)mm(
mm2T
21
21
+=
Atwood’s Machine
Physics 111: Lecture 8, Pg 16
Is the result reasonable? Check limiting cases!
Special cases:i.) m1 = m2 = m a = 0 and T = mg. OK!ii.) m2 or m1 = 0 |a| = g and T= 0. OK!
Atwood’s machine can be used to determine g (by measuring the acceleration a for given masses).
a)mm()mm(
g12
12 +=
-
a m mm m
g
( )( )
1 2
1 2
gmmmm2
T21
21
)(
Physics 111: Lecture 8, Pg 17
Attached bodies on two inclined planes
All surfaces frictionless
m1m2
smooth peg
q1 q2
Physics 111: Lecture 8, Pg 18
How will the bodies move?
From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law:
Taking “x” components:
1) T1 - m1g sin q1 = m1 a1X
2) T2 - m2g sin q2 = m2 a2X
But T1 = T2 = Tand a1X = -a2X = a(constraints)
m1
y x
T1
N
m1g
m2
m2g
T2
Nx y
q1q2
Physics 111: Lecture 8, Pg 19
Solving the equationsUsing the constraints, solve the equations.
T - m1gsin q1 = -m1 a (a)T - m2gsin q2 = m2 a (b)
Subtracting (a) from (b) gives: m1gsin q1 - m2gsin q2 = (m1+m2 )a
So: a m m
m mg
1 1 2 2
1 2
sin sinq q
Physics 111: Lecture 8, Pg 20
Special Case 1:
m1m2
q1 q2
m1 m2
If q1 = 0 and q2 = 0, a = 0.
Boring
a m m
m mg
1 1 2 2
1 2
sin sinq q
Physics 111: Lecture 8, Pg 21
Special Case 2:
If q1 = 90 and q2 = 90, a m mm m
g
( )( )
1 2
1 2
m2
TT
m1
Atwood’s Machine
m1m2
q1 q2
a m m
m mg
1 1 2 2
1 2
sin sinq q
Physics 111: Lecture 8, Pg 22
Special Case 3:
If q1 = 0 and q2 = 90, a mm m
g
2
1 2( )
m1
m2
Lab configuration
m1m2
q1 q2
a m m
m mg
1 1 2 2
1 2
sin sinq q
Air-track
Physics 111: Lecture 8, Pg 23
Lecture 8, Act 2Two-body dynamics
In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless.
(a) Case (1) (b) Case (2) (c) same
m
10kga
m
a
F = 98.1 N
Case (1) Case (2)
Physics 111: Lecture 8, Pg 24
Lecture 8, Act 2 Solution
m
10kga
Add (a) and (b):
98.1 N = (m + 10kg)a
kg10mN198a
.
Note:kg10m
mN198T
.
(a)
(b)
T = ma (a)(10kg)g -T = (10kg)a (b)
For case (1) draw FBD and write FNET = ma for each block:
Physics 111: Lecture 8, Pg 25
Lecture 8, Act 2 Solution
The answer is (b) Case (2)
T = 98.1 N = mam
N198a . For case (2)
m
10kga
m
a
F = 98.1 N
Case (1) Case (2)
kg10mN198a
.m
N198a .
Physics 111: Lecture 8, Pg 26
Problem: Two strings & Two Masses onhorizontal frictionless floor:
m2 m1T2 T1
Given T1, m1 and m2, what are a and T2?T1 - T2 = m1a (a) T2 = m2a (b)
Add (a) + (b):
T1 = (m1 + m2)a a
a
i
21
1
mmT+
=
Plugging solution into (b):
21
212 mm
mTT
+=
Physics 111: Lecture 8, Pg 27
Lecture 8, Act 3Two-body dynamics
Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings?
(a) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3
T3 T2 T13m 2m m
a
Physics 111: Lecture 8, Pg 28
Lecture 8, Act 3Solution
Draw free body diagrams!!T33mT3 = 3ma
T3 T22mT2 - T3 = 2ma
T2 = 2ma +T3 > T3
T2 T1mT1 - T2 = ma
T1 = ma + T2 > T2
T1 > T2 > T3
Physics 111: Lecture 8, Pg 29
Lecture 8, Act 3 Solution
Alternative solution:
T3 T2 T13m 2m m
a
Consider T1 to be pulling all the boxes
T3 T2 T13m 2m m
a T2 is pulling only the
boxes of mass 3m and 2m
T3 T2 T13m 2m m
a
T3 is pulling only the box of mass 3m
T1 > T2 > T3
Physics 111: Lecture 8, Pg 30
Problem: Rotating puck & weight.
A mass m1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m2.What is the tension (T) in the string?What is the speed (v) of the sliding mass?
m1
m2
v
R
Physics 111: Lecture 8, Pg 31
Problem: Rotating puck & weight...
Draw FBD of hanging mass:Since R is constant, a = 0.
so T = m2g
m2
m2g
T
m1
m2
v
RT
Physics 111: Lecture 8, Pg 32
Problem: Rotating puck & weight...
Draw FBD of sliding mass:
m1
T = m2g
v gR mm
2
1
m1g
N
m1
m2
v
RT
Use F = T = m1a
where a = v2 / R
m2g = m1v2 / R
T = m2g
Puck
Physics 111: Lecture 8, Pg 33
Recap of today’s lecture
Friction Recap. (Text: 5-1) Drag Forces. (Text: 5-3)
Terminal speed.
Dynamics of many-body systems. (Text: 4-7)Atwood’s machine.General case of two attached blocks on inclined planes.Some interesting special cases.
Look at Textbook problems Chapter 6: # 3, 7, 21, 75