Transcript Chapter 6

CHAPTER-6
Force and Motion-II
Force and Motion-II
Motion under frictional and centripetal
forces:
Force of friction ( required for car motion
out of a pit and out of a curve)
Centripetal force (Radial force) (required to
turn car in a circle
Ch 6-2 Friction
 An external force F applied to a
block resting on a rough surface
 No motion till F is less than static
frictional force fs :

fs=sFN where FN =Fg
 Body slides if F fs
 When body start sliding static
frictional force fs reduces to kinetic
frictional force fk
 fk=kFN where FN =Fg
 If lFl=lfkl body moves at constant
speed
 If lFl  lfkl body accelerates
Checkpoint 6-1
 A block lies on floor.
(a) What is the magnitude of the
frictional force on it from the
floor?
(b) If a horizontal force of 5 N is
now applied to the block, but the
block does not move, what is the
magnitude of the frictional force
on it?
(c) If the maximum value fs,max of
the static frictional force on the
block is 10 N, will the block move
if the magnitude of the
horizontally applied force is 8 N?
(d) If it is 12 N?
(e) What is the magnitude of the
frictional force in part (c)
Ans:
(a) Zero
(b) 5 N
(c) No
(d) Yes
(e) 8 N
Checkpoint 6-2
 In the figure, horizontal force F1
of magnitude 10 N is applied to a
box on a floor, but the box does
not slide. Then, as the magnitude
of vertically applied force F2 is
increased from zero but before
the box begins to slide, do the
following quantities increase,
decrease, or stay the same.
(a) magnitude of the frictional force
on the box
(b) the magnitude of the normal
force on the box from the floor
(c) the maximum value of fs,max of
the static frictional force on the
box?
Answer
(a) magnitude of the frictional
force on the box remains
same
(b) the magnitude of the normal
force on the box from the
floor decreases
(c) the maximum value of fs,max
of the static frictional force
on the box decreases
Example-Problem-6-2
 Loaded sled being
pulled at constant
speed, find T








Since v is constant, then
l Tcos  l= lfkl = kFN but
FN+Tsin =Fg=mg
FN = mg-Tsin 
Then
Tcos = fk= k FN
= k (mg-Tsin )
Solve for T
Sample-Problem-6-3
 Coefficient of static s
friction for coin on verge of
sliding down
fs =s FN; s =fs / FN
 Fx= fs –Fgsin=0
fs = Fg sin =mg sin
  Fy= FN –Fg cos=0
FN = Fg cos=mg cos
s =fs / FN
= mg sin /mg cos
s = tan 
Ch 6-5: Uniform Circular Motion
(Radial force and friction force)
 Circular motion under the
effect of centripetal force
 Centripetal force FR accelerates a
body by changing the direction of
the body’s velocity v without
changing the body’s speed
 Constant centripetal acceleration
aR, directed toward the center of the
circle with radius R
aR=v2/R
 Centripetal force FR=maR
=m (v2/R)
 For puck tied to a string with
tension T and moving in a circle
FR=T=m (v2/R)
Checkpoint 6-3
Near the ground, is
the speed of large
raindrops greater
than, or the same as
the speed of small
raindrops, assuming
that all raindrops are
spherical and have
the same drag
coefficient?
Answer:
It takes longer time to
attain the terminal
velocity after falling for
the larger raindrops as
compared to the small
drops. Under a constant
acceleration, the large
raindrops achieve larger
value of terminal velocity
than small raindrops.
Sample-Problem-6-8
 Static frictional force fs
prevent her from sliding down
fs=Fg
 fs=sFN and normal force FN
is provided by centripetal
force FR
 Then FN = FR=m(v2/R)
 fs= s FN =sm(v2/R)=Fg=mg
v2 =Rg/ s
Example
(Motion of car on a un-banked curved road)
 Coefficient of static
friction s when car is
verge of sliding out of
track
 -FR=-mv2/R= - fS= -s N

-mv2/R= - smg

s = v2/Rg
Example-Problem-6-10
(Motion of car on a Banked Curved Road)
Calculate banking angle 
prevent sliding
 FX = -FR=-mv2/R= - Nr = -N sin 
-mv2/R= -N sin ……..(1)
 FY = NY -mg =0
NY = N cos  = mg ….(2)
Then
(mv2/R)/(mg)= (N sin )/(N cos )
v2/Rg= tan 
 = tan-1 (v2/Rg)