Transcript Chapter 5

CHAPTER-5
Force and Motion-1
Ch 5-1 Physics
Physics: Study of cause of acceleration
of an object
Cause: Force
Force is said to act on the object to
change its velocity
Ch 5-3 Newton’s First Law
 If no force acts on a body,
then the body’s velocity cannot change
i.e. the body cannot accelerate.
 If body is at rest, it stays at rest, if it
is in motion, it continues to move with
the same velocity
Ch 5-4 Force
 A force F is measured by
the acceleration a it
produces.
 Principal of superposition of
forces:
When two or more forces
are acting on a body
resultant force or net force
Fnet can be obtained by
adding the forces individually
Fnet = F
 If no net force acts on a
body (Fnet =0), the body
velocity cannot change.
Checkpoint 5-1
 Which of the figure’s six
arrangements correctly show the
vector addition of forces F1 and
F2 to yield the third vector,
which is meant to represent net
force Fnet?
 Ans: c, d and e
Ch 5-5 Mass
 Mass is an intrinsic characteristic of a body
 Mass of a body relates a force acting on a
body to the resulting acceleration
 The ratio of the masses of two bodies m0 and
mx is equal to inverse of their acceleration a0
and ax when same force is applied to both
bodies

mx/m0 = a0/ax
Ch 5-6 Newton’s Second Law
 The net force Fnet acting on a body is equal to
the product of the body mass m and its
acceleration a
Fnet = ma; a= Fnet / m
 Acceleration component along a given axis is
caused only by sum of forces component along
that axis
ax = Fnet,x /m ; ay = Fnet,y /m ; az = Fnet,z /m
 SI unit of force Newton (N)
1N= 1kg.m/s2
Checkpoint 5-2
The figure here shows two
horizontal forces acting on
a block on a frictionless
floor. If a third horizontal
force F3 also acts on the
block, what are the
magnitude and direction of
F3 when the block is:
(a) stationary
(b) moving to the left with a
constant speed 5 m/s?
F= F3
+5-3=0
Then F3 =-5+3= -2 N
in both cases (a) and (b)
Checkpoint 5-3
The figure shows overhead views of
four situations in which two forces
accelerate the same block across a
frictionless floor. Rank the situation
according to the magnitudes of
(a) the net force on the block and
(b) The acceleration of the block,
greatest first
Calculate Fnet= F1 + F2
(a) (1) Fnet=5+3=8
(2) Fnet=5-3=2
(3) Fnet= (52+32)=34~6
(4) Fnet=[(5+3cos)2+(3sin)2]
> 6
Ans: (1),(4), (3), (2)
(b) Acceleration of block a
a = Fnet/ constant block mass
The order of acceleration is
same as that of Fnet
Ans: (1),(4), (3), (2)
Ch 5-7: Some Particular Forces
Gravitational Force Fg:
The force with which a body is
pulled towards earth. Downward
force –Fg= m(-|g|)
Weight:
Equals the magnitude of a net
force required to prevent the body
from falling freely, as measured on
the ground.
weight |W| = |Fg|
Ch 5-7: Some Particular Forces (FN and T)
 Normal Force FN :
When a body presses against a
surface, the surface pushes
the body with a normal force
FN, that is  to the surface
For a block resting on a table
FN = mg
 Tension Force T:
It is the force applied to a
body through a cord, pulled
taut opposite to object. For a
mass less cord , the pulls at
both ends of the cord has
same magnitude T
Checkpoint 5-3
 In figure shown below, is the
magnitude of the normal force FN
greater than, less than, or equal to
mg if the block and table are in an
elevator moving upward
(a) at constant speed
(b) at increasing speed
(a) For constant speed
Fnet=0= FN – mg=0
FN = mg
(b) For increasing speed
Fnet=ma= FN – mg
FN = ma + mg
FN > mg
Checkpoint 5-4
 The suspended body in the figure (c) weighs
75 N. Is T equal to greater than, less than
75 N when body is moving upward at
(a) constant speed
(b) at increasing speed
(c) at decreasing speed
(a) For constant speed
Fnet=0= T – mg=0
T = mg = 75 N
(b) For increasing speed
Fnet=ma = T – mg
T = ma + mg
T > mg
(c) For decreasing speed
Fnet=-ma = T – mg
T = mg - ma
T < mg
Ch 5-8 Newton’s Third Law
 Newton’s Third Law:
When two bodies interact, the
forces on the bodies from each
other are always equal in
magnitude and opposite in
direction.
 For the book and crate
FBC=-FCB
 FBC and FCB are called third-law
force pair
FBC and FCB are also called
action and reaction forces pair
Ch 5-9 Applying Newton’s Laws
Problem-solving strategy:
 Draw a simple and neat diagram
of the system
 Isolate the object whose motion
is being analyzed. Draw freebody diagram for this object for
the external forces acting on
the object
 Establish convenient coordinate
axes for each object. Find
component of forces along these
axes. Apply Newton’s Second
Law F=ma in component form.
 Solve the component equations
for the unknowns
Ch 5-9 Applying Newton’s Laws
 For (a)
Fy=FN-Fgs=May=0; FN= Fgs
Fx=Max=T=Ma; T=Ma
 For (b)
Fx=0
Fy=T-FgH=T-mg=-ma
T=m(g-a) and T=Ma
Solve for a
a =mg/(m+M)
Ch 5-9 Applying Newton’s Laws
Sample Problem 5-6
 Find T1, T2 and T3 ?; For (a)
Fy=T3-Fg=0; T3= Fg=mg and
Fx=0
 For (b)
Fx=T2cos 47-T1cos 28=0
T2cos 47=T1cos 28 ……(1)
Fy= T2 sin 47+T1 sin 28-T3=0
T2 sin 47+T1 sin 28=T3
….(2)
Solve eq. (1) and (2) to get T1
and T2
Sample Problem 5-7
 Find T and FN if  = 27º
 Resolving force Fg=mg along and
the plane and perpendicular to
the plane
 T-mg sin = 0; T= mg sin 
 Also FN-mg cos  =0
Then FN= mg cos 
Checkpoint 5-7
 In the figure , horizontal force F is
appled to a block on a ramp.
(a) Is the component of F that is
perpendicular to the ramp Fcos or
Fsin
(b) Does the presence of F increases or
decreases the magnitude of the normal
force on the block from the ramp?
 Ans:
(a) Fsin  to ramp
(b) F  to ramp=
N-mgcos- Fsin =0
Then N= mgcos+ Fsin
Fsin increases the
magnitude of the
normal force on the
block from the ramp
Checkpoint 5-8
In the figure , what does the scale read
if the elevator cable breaks sp that
the cab falls freely ; that is what is
the apparent weight of the passenger
in free fall?
Answer
In free fall
F = N-Fg=-Fg
Apparent weight N
N= -Fg+Fg = 0
Sample Problem 5-9
Find a?
 Introduce action and
reaction par forces FAB
and FBA
 Fapp=(mA+mB)a
a= Fapp/(mA+mB)
 Then FBA =mBa