Welcome back to Physics 211

1
Welcome back to Physics 211

• Todays agenda
• Friction block on plane
• Internal forces
• Tension
• Circular motion

2
Announcements

• MP Homework 3 due Friday 12 pm

3
Block on plane revisited
F
N
W
q
4
Initially at rest

• What is largest angle before slips ?
• Resolve perpendicular to plane ?
  NWcosq
• resolve parallel ? FWsinq
• since F lt mN we have
• Wsinq lt mWcosq ie
• tanq lt m

5
Angle gt tan-1m

• Resolve along plane
• Wsinq-mKWcosqa
• Or
• ag(sinq-mKcosq)

6
Friction demo

• Static friction depends on surface and normal
  force for pulled block
• Kinetic friction generally less than maximal
  static

7
Summary of friction

• 2 laws of friction static and kinetic
• Static friction tends to oppose motion and is
  governed by inequality
• Fs ltmsN
• Kinetic friction is given by equality FKmKN

8
Internal Forces

• So far replaced macroscopic bodies by points
  why is this ok ?
• Specifically, such body composed of (very many)
  parts neglected all internal forces of these
  parts on each other
• Also neglected rotational motion -- later

9
Simple example
A
B
Const v
B
NAG
A
NBG
NAB
PAH
NBA
FAG
FBG
WBE
WAE
10
Newtons 2nd Law

• Newtons 2nd law (NSL) for A (a0)
• NAGWAE0 PAHNABFAG0
• NSL for B
• NBGWBE0 NBAFBG0
• add ?
• NAGNBGWAEWBE0
• PAH(NABNBA)FAGFBG0

11
Newtons 3rd law

• NAGNBGWAEWBE0
• ?N (AB),GW(AB),G0
• and
• PAH(NABNBA)FAGFBG0 ?
• PAHF(AB),G0

12
Composite system

• NTL for A and B imply that can consider
• combined system CAB in which NAB etc
• do not appear internal forces

NCGNAGNBG
PCH
FCFAGFBG
WCEWAEWBE
13
Internal forces summary

• Can apply Ns laws to composite body
• Can ignore internal forces of one part of body on
  another since cancel (NTL)
• Justifies treating macroscopic bodies as
  point-like

14
Tension
15
Two blocks are connected by a heavy rope. A
hand pulls block A in such a way that the blocks
move upward at increasing speed. The (downward)
tension force on the upper block by the rope is
1. less than 2. equal to 3. greater than the
(upward) tension force on the lower block by the
rope. 4. Answer depends on which block is heavier.
16
Hand pulls block A so blocks move up at
increasing speed.
17
Notice for mR0, the tension forces exerted at
either end are the same.

• The term tension in the string is therefore
  often used as a short-hand for the tension forces
  exerted on or by the string at either end.

18
Blocks A and C are initially held in place as
shown. After the blocks are released, block A
will accelerate up and block C will accelerate
down. The magnitudes of their accelerations are
the same. Will the tension in the string be
1. equal to 1.0 N (i.e., the weight of
A), 2. between 1.0 N and 1.5 N, 3. equal to 1.5
N (i.e., the weight of C), or 4. equal to 2.5 N
(i.e., the sum of their weights)?
19
Free-body diagram for block A.
Free-body diagram for block C.
20
Pulleys etc
2 pulleys 2TW FTW/2
T
N pulleys FW/N!
F
W
21
Forces in circular motion

• .

22
Two identical balls are connected by a string and
whirled around in circles of radius r and 2r at
constant speed. The acceleration of ball B is
1. four times as great 2. twice as great 3. equal
to 4. one half as great as the acceleration of
ball A.
23
The two balls are whirled around in a circle as
before. Assume that the balls are moving very
fast and that the two strings are massless. The
tension in string P is
1. less than 2. equal to 3. greater than the
tension in string R.
24
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25
Motion of car on banked circular track
car
N
a
q
R
W
Speed v
Horizontal forces Vertical
26
Motion on loop-the-loop
what is normal force on car at top and bottom of
loop ? Neglect friction Assume moves with speed
vB at bottom and vT at top
car
27
Free body diagrams
At bottom
At top
Newton
28
Apparent weight ?
What is criteria to just make it over loop ?
29
Another example
m2 falls coefficient of kinetic friction on
plane is m
m1
m2
q
30
Free body diagrams
Block 2
Block 1
31
Newtons 2nd law
32
Criteria for Motion ?
What is largest value of m which supports motion
? (assume q300, m11.0 kg, m22.0 kg)