Physics. Session Opener A recent space shuttle accident occurred because of failure of heat...
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Transcript of Physics. Session Opener A recent space shuttle accident occurred because of failure of heat...
Physics
Session Opener
A recent space shuttle accident occurred because of failure of heat protecting devices.
How was this heat generated ?
Session Objectives
1. Friction & frictional force
2. Bodies connected in frictional surface
3. Kinetic friction
4. Static friction
5. Angle of friction
6. Coefficient of friction
7. Laws of friction
Session Objective
Friction and Frictional Force
Whenever an object moves or tends to move, contact surfaces of object resist the force tending to generate motion.
This property of the surface is friction
The resistive force is the friction force
An equal and opposite force f acts
Consider the experiment :
1. F applied.
‘A’ does not move.
2. ‘A’ still at rest on increased F.
f adjusts to match F
MB
F MA
A
Bodies Connected in Frictional Surface
f
3. increased further F
A has acceleration. f has a maximum limiting value
f during the motion is lower than at rest.
F
4. is reducedA moves uniformly
Bodies Connected in Frictional Surface
MB
F fF fF fF fF fF fF fF fF fF f
Friction retards motion
f adjusts to stay equal and opposite to F
f has a limiting value
Bodies Connected in Frictional Surface
moving friction force is less than friction force at rest.
Friction force (f) is zero if no external force ( ) existsF
Friction force is a contact force, independent of area of contact.
Bodies Connected in Frictional Surface
Friction force is a non-conservative force as mechanical energy changes to heat during friction.
That is why the space shuttle overheated and exploded
Ff
Kinetic Friction
appears when two objects in contact have relative motion.
Directed opposite the direction of motion
fk is a constant.
fk on B may start the motion of B.
F fk
A
B
F fk
fk
Static Friction
F has component along contact surface, but there is no motion.
f = Fcos till limiting friction.
At limiting friction f(=fs) is constant
fs > fk
f acts along the surface and opposite the force component
F N
mg
fsFcos
Class Exercise
Class Exercise - 1
Blocks A and B are pressed against a smooth wall and are in equilibrium by a horizontal force F as shown. Then friction force due to A on B
(a) upward
(d) system cannot be in equilibrium
(c) dependent on relative mass of A and B
(b) downward
Solution
A B
Wall
F
As wall is smooth, A and B will keep slipping down with acceleration g andcannot be in equilibrium.
Hence answer is (d).
Coefficients of Friction
Limiting static friction
ss s
f: cons tant f N
N
Kinetic friction
kk k
f: cons tant f N
N
are coefficient of friction
s k
0
f
F
Static region
Ffs
Graphically
Kinetic region
Nf kk
Nf smax,s Limiting friction
Coefficients of Friction
Laws of Friction
1. Bodies slip over each other
k kf N
2. Direction of kinetic friction is opposite to the velocity
3. Bodies do not slip over each other
s sf NN is called the limiting friction. s
do not depend upon the area of contact.
k sf or f4.
Class Exercise
Class Exercise - 4
m
Wall
F
An object of mass m is pressed against a wall with a force F and is in equilibrium. The coefficient of friction between the wall and the object is . Then F must be equal to
(c) mg (d) cannot be found
mg(a) mg
(b)
Solution
As equilibrium exists, wall presses against the object by a force of F (equal and opposite to force exerted by object against the wall)
FF
f
mg
f F
mg f F
mgF
At equilibrium
Class Exercise - 5
Fm a
mb m c
Smooth surface
Three blocks of masses ma,mb and mc are arranged on a smooth horizontal surface as shown. Surface between ma and mb is smooth and the coefficient of friction between mb and mc is .Then the minimum force F required to keep mb from sliding down is:
a b
c
m m gm
a a b c b
c
m m m m g
m
b
a b cm m m g
c
a b c b
c
m m m m g
m
d
Solution
F m a
m b m c
Smooth surface
a
Let the force F give an acceleration a to the system
To give an acceleration a to mc, mb presses against mc with force mca, so mc gives a normal reaction mc a (to left) on mb.
a b cF m m m a
b cm g m a
b
c
m ga
m
a b c b
c
m m m m gF
m
for mb not slipping down
Class Exercise - 6
A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is . The force required to initiate horizontal motion of the block is maximum when
(a) Lift is moving up with constant acceleration
(b) Lift is moving down with constant acceleration
(c) Lift is stationary
(d) Lift is in free fall
Solution
F is largest when N is maximum
N is maximum when lift moves up with constant acceleration a [N = m(g + a)].
So F is maximum at that condition.
Condition of motion of M: F N
f
N
W
F
Class Exercise - 8
A 70 Kg box is pulled along a horizontal surface by a 400 N force at an angle of 30° above horizontal. If the coefficient of friction is 0.50, what is the acceleration of the box? (g = 10 m/s2)
Solution
f
N
mg
F cos 30°
F sin F
Vertical: N = mg – F sin30
Horizontal: F cos 30 – f = ma
F cos30 N ma
F cos30 sin30 mg ma
3 0.5400 0.5 70 10 70 a
2 2
Solving for a: a = 1.37 m/s2
Angle of Friction
R
max,sf
f along surface (surface property)
N
N normal to surface
Angle of friction
ftan
N
Static > kinetic
Class Exercise
Class Exercise - 2
The angle between the resultant cont-act force and the normal reaction force exerted by a body on the other when they are one on top of the other is . Then, what is the coefficient of friction,
(a) = tan (b) > tan
(c) < tan (d) and are not related
Solution :
Hence answer is (a)
ftan
N
Class Exercise - 3
F
w
A pushing force F is applied to a body of weight W, placed on a horizontal table, at an angle . The angle of friction is . The magnitude of F to initiate motion in the body is
cos
sinWa.
cos
cosWb.
sin
sinWd.
cos
tanWc.
Solution
F cos
f
F sin N
W
Vertical: sinFWNfcosF Horizontal:
(a = 0 at initiation of motion)
f N tan W F sin tan
F cos W F sin tan
sinF cos sin . W tan
cos
cos cos sin sin WsinF
cos cos
F cos W sin Wsin
Fcos
Angle of Repose
Object at rest
sf
tanN
Object slips
increased .
is angle of repose
Nfs
mg
Angle of Repose
reduced
At , object moves uniformly.k
: Angle of repose (kinetic friction)k
kk
ftan
N
k s
Nfk
mg
Class Exercise
Class Exercise - 9
A block ‘A’ slides from rest down an incline with a 30° inclination with horizontal. It covers 3 m in 5 seconds. What is the value of coefficient of kinetic friction? (g = 10 m/s2)
A
30°
Solution
mg sin
mg cos
x
mg
fN y
Along y: N = mg cos
Along x : mg sin – f = ma
g sin cos a
2 21 1x ut at at
2 2
22x 1
gsin .cos .gt
10 2 3 2.
2 25 3 10
0.53
Class Exercise - 10
M
30°
A block of mass 2kg lies on a rough inclined plane of inclination 30° to the horizontal. Coefficient of friction between the block and the plane is 0.75. What minimum force will make the block move up the incline? (g = 10 m/s2)
Solution
mg cos
mg
mg sin
y N xF
Along y: N = mg cos
F is up the incline (along x)
F – (mg sin + f) = ma
To start the motion a is put equal to zero
mg sin mg cos
mg sin cos
1 3 32 10 . 22.98 N
2 4 2
minF mg sin f
Thank you