Exam (elaborations)
NEWTON’S LAWS OF MOTION Newton’ 1st law or Law of Inertia
NEWTON’S LAWS OF MOTION Newton’ 1st law or Law of Inertia
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K.V. SilcharNEWTON’S LAWS OF MOTION
Newton’ 1st law or Law of Inertia
Every body continues to be in its state of rest or of uniform motion
until and unless and until it is compelled by an external force to change its state
of rest or of uniform motion.
Inertia
The property by virtue of which a body opposes any change in its
state of rest or of uniform motion is known as inertia. Greater the mass of the
body greater is the inertia. That is mass is the measure of the inertia of the
body.
Numerical Application
If, F = 0 ; u = constant
Physical Application
1. When a moving bus suddenly stops, passenger’s head gets jerked in the
forward direction.
2. When a stationery bus suddenly starts moving passenger’s head gets jerked
in the backward direction.
3. On hitting used mattress by a stick, dust particles come out of it.
4. In order to catch a moving bus safely we must run forward in the direction of
motion of bus.
5. Whenever it is required to jump off a moving bus, we must always run for a
short distance after jumping on road to prevent us from falling in the forward
direction.
Key Concept
In the absence of external applied force velocity of body remains
unchanged.
Newton’ 2nd law
Rate of change of momentum is directly proportional to the applied
force and this change always takes place in the direction of the applied force.
dp F
dt
28
,K.V. Silchar
or, dp =F (here proportionality constant is 1)
dt
putting, p = mv
F = dp
dt
or, F = dmv
dt
or, F = mdv + vdm
dt dt
or, F = mdv (if m is constant dm/dt = 0)
dt
or, F = ma
Note :- Above result is not Newton’s second law rather it is the conditional result
obtained from it, under the condition when m = constant.
Numerical Application
a = FNet
m
Where FNet is the vector resultant of all the forces acting on the body.
F1
F2
F6 m F3 m FNet
F5 F4
Where, FNet = F1 + F2 + F3 + F4 + F5 + F6
29
,K.V. Silchar
Physical Application
Horizontal Plane
i) Case - 1 N
Body kept on horizontal plane is at rest.
For vertical direction
N = mg(since body is at rest)
mg
ii) Body kept on horizontal plane is accelerating horizontally under single horizontal
force.
N
For vertical direction a
N = mg (since body is at rest) F
For horizontal direction
F = ma mg
iii) Body kept on horizontal plane is accelerating horizontally towards right under two
horizontal forces. (F1 > F2) N
a
For vertical direction
N = mg (since body is at rest) F2 F1
For horizontal direction
F1 - F2 = ma mg
iv) Body kept on horizontal plane is accelerating horizontally under single inclined
force FSinθ F
N
For vertical direction a
N + FSinθ = mg (since body is at rest) θ FCosθ
For horizontal direction
FCosθ = ma
mg
v) Body kept on horizontal plane is accelerating horizontally towards right under an
inclined force and a horizontal force. F1Sinθ
a N F1
a
For vertical direction
N + F1Sinθ = mg (since body is at rest) F2 θ F1Cosθ
For horizontal direction
F1Cosθ – F2 = ma
mg
30
, K.V. Silchar
vi) Body kept on horizontal plane is accelerating horizontally towards right under two
inclined forces acting on opposite sides.
N F1Sinθ F1 a
For vertical direction
N + F1Sinθ = mg + F2 SinФ
(since body is at rest) F2CosФ
Ф θ
For horizontal direction F1Cosθ
F1Cosθ – F2CosФ = ma F2 F2SinФ
mg
Inclined Plane
i) Case - 1 N
Body sliding freely on inclined plane.
a
Perpendicular to the plane
N = mgCosθ (since body is at rest) mgSinθ θ
Parallel to the plane mgCos θ
mgSinθ = ma θ mg
ii) Case - 2
Body pulled parallel to the inclined plane.
N a F
Perpendicular to the plane
N = mgCosθ (since body is at rest)
mgSinθ
Parallel to the plane θ
F - mgSinθ = ma
mgCos θ
mg
θ
iii) Case - 3
Body pulled parallel to the inclined plane but accelerating downwards.
N
Perpendicular to the plane F
N = mgCosθ (since body is at rest)
a
Parallel to the plane mgSinθ θ
mgSinθ - F = ma
mgCos θ
θ mg
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