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Atwood’s machineGeneral case of two attached blocks on inclined planesSome interesting problems
Physics 111: Lecture 8, Pg 2
Friction Review: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 FrictionModel for Friction
The direction of the frictional force vector ffF is perpendicular to the normal force vector NN, in the direction opposing the net applied force.
Kinetic (sliding): Kinetic (sliding): The magnitude of the frictional force vector is proportional to the magnitude of the normal force N.
fF = KN
Static: Static: The frictional force balances the net applied forces such that the object doesn’t move. The maximum possible static frictional force is proportional to N.
fF SN
Physics 111: Lecture 8, Pg 4
Kinetic Friction:Kinetic Friction:
K is the coefficient of kinetic friction.
i : F KN = ma
j : N = mg
so F Kmg = ma
maaFF
mgg
NN
ii
j j
KN
Physics 111: Lecture 8, Pg 5
Static Friction:Static Friction:
The “coefficient of static friction,” S, determines maximum static frictional force, SN, that the contact between the objects can provide.
S is discovered by increasing F until the object starts to slide: FMAX - SN = 0 N = mg
A block of mass m, when placed on a rough inclined plane ( > 0) and given a brief push, keeps moving down the plane with constant speed.If a similar block (same ) of mass 2m were placed on the same incline and given a brief push, it would:
(a)(a) stop
(b) (b) accelerate
(c) (c) move with constant speed m
Physics 111: Lecture 8, Pg 7
Lecture 8, Lecture 8, Act 1Act 1SolutionSolution
Draw FBD and find the total force in the x-direction
ii
jj
mg
N
KN
FNET,X = mg sin Kmg cos
= ma = 0 (first case)
Doubling the mass will simplydouble both terms…net forcewill still be zero!
Speed will still be constant!Speed will still be constant!
Physics 111: Lecture 8, Pg 8
Drag Forces: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 j
v
Physics 111: Lecture 8, Pg 9
Drag Forces: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 j
v feels like n=2
Parachute
Physics 111: Lecture 8, Pg 10
Terminal Speed: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 = mgFg = mg
FFD = bv2
j j
v
bmg
v =
Physics 111: Lecture 8, Pg 11
Many-body DynamicsMany-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: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
T1T2
Physics 111: Lecture 8, Pg 13
Atwood’s Machine...Atwood’s Machine... Draw free body diagrams for each object Applying Newton’s Second Law: ( jj -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 unknowns
we 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...Atwood’s Machine...
21
221
mm
)mm(g
+
--
g)mm(
)mm(
21
21
+-
Physics 111: Lecture 8, Pg 15
Atwood’s Machine...Atwood’s Machine...
m1
m2
j
a
a
TT
So we find:
am m
m mg
( )
( )1 2
1 2
g)mm(
mm2T
21
21
+=
Atwood’s Machine
Physics 111: Lecture 8, Pg 16
Is the result reasonable?Is the result reasonable? Check limiting cases!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(g
12
12 +=
-
am m
m mg
( )
( )1 2
1 2
gmm
mm2T
21
21
)(
Physics 111: Lecture 8, Pg 17
Attached bodies on two inclined planesAttached bodies on two inclined planes
All surfaces frictionless
m1
m2
smooth peg
1 2
Physics 111: Lecture 8, Pg 18
How will the bodies move?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:
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)(a) Case (1) (b)(b) Case (2) (c)(c) same
m
10kga
m
a
F = 98.1 N
Case (1) Case (2)
Physics 111: Lecture 8, Pg 24
Lecture 8, Lecture 8, Act 2Act 2 SolutionSolution
m
10kga
Add (a) and (b):
98.1 N = (m + 10kg)a
kg10m
N198a
.
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, Lecture 8, Act 2Act 2 SolutionSolution
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)
kg10m
N198a
.
m
N198a
.
Physics 111: Lecture 8, Pg 26
Problem: Two strings & Two Masses onProblem: Two strings & Two Masses onhorizontal frictionless floor:horizontal frictionless floor:
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
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?