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Momentum:- Momentum of a particle (p) is equal to the mass of the particle (m) times velocity of the particle (v). So, p = mv
First Law (Law of Inertia): Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled by some external force to change that state.
Second Law: The rate of change of momentum of a body is directly proportional to the impressed force and takes place in the direction of the force.
So, F= dp/dt
This results force acting a body F is equal to the mass of the body m times acceleration of the body a.
So, F = ma
Third Law:
To Every action there is an equal and opposite reaction.
F_{AB }= - F_{BA}
So, J = Ft
= mv – mu
(a) The lift possesses zero acceleration (fig-1): W = mg
(b) The lift moving upward with an acceleration a (fig-2):
W = mg + ma
= mg + mg
= 2 mg
(c) The lift moving downward with an acceleration a (fig-3):
W = mg – ma
= mg – mg
= 0
So, m_{i} = F/a
Note: When the velocity of a body is comparable to the velocity of light, inertial mass changes with velocity in accordance with following formula,
m_{i} = m_{i}_{0}/√(1-v^{2}/c^{2})
Here m_{i0} is the rest mass of body, v is the velocity of body and c is the velocity of light.
In an isolated system (no external force), the algebraic sum of the momentum of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other.
Equilibrium of concurrent forces:-
?Forces are said to be concurrent if they meet at a common point. Below figure shows three concurrent forces , and .
Here F is the force and d is the distance of line of action.
Work done by a couple: (W)-
W = × θ
Here is the angular displacement.
(a) Driven body moving vertically:-
Acceleration of the system, a = (M-m/M+m) g
Tension in the string, T = (2Mm/M+m) g
The force on the pulley, F = (4Mm/M+m) g
(b) Driven body moving horizontally:-
Acceleration of the system, a = (M /M+m) g
Tension in the string, T= (Mm/M+m) g
The force on the pulley, F= [√2 (Mm) /M+m] g
Here g is the free fall acceleration.
(a) Acceleration, a = F/(m + M)
(b)
(c) Contact force, F_{12} = (m/m+M) F = F_{21}
Friction:- Whenever a body tends to slide over another’s surface, an opposing force, called force of friction comes into play. This force acts tangentially to the interface of two bodies.
Static friction:- Static friction is the force of friction between two surfaces so long as there is no relative motion between them. It is always equal to the applied force. The static frictional forces are incorporated in the following inequality.
f_{s }≤ µ_{s}N
f_{s,max} = µ_{s}N
Here µ_{s }is the coefficient of kinetic friction and N is the normal force.
So, coefficient of static friction, µ_{s}= f_{s,max} /N
So, f_{k} = µ_{k }N
Here µ_{k} is the coefficient of kinetic friction.
Thus, coefficient of kinetic friction, µ_{k} = f_{k}/N
So, f_{r} = µ_{r }N
Here µ_{r} is the coefficient of rolling friction and N is the normal force.
The angle made by the resultant reaction force with the vertical (normal reaction) is known as the angle of the friction.
Now, in the triangle OAB
AB/OB = cotθ
So, OB = AB/ cotθ
= AB tanθ
Or, tanθ = OB/AB
= f / N
So, tanθ = f / N = µ_{s}
Consider a body of mass m resting on an inclined plane of inclinationq. The forces acting on the body are shown – F_{f} being the force of friction. If friction is large enough, the body will not slide down.
along x: mg sin θ – f = 0 …(1)
Along y: N –mg cosθ = 0 …(2)
i.e. N = mg cos θ and f = mg sin θ
Thus, gives,
mg sin θ ≤ mg cos θ
So, tan θ ≤ . This signifies, the coefficient of static friction between the two surfaces, in order that the body doesn’t slide down.
When q is increased, then tan θ > . Thus sliding begins, and the angle θ_{r} = tan^{-1} . This angle is known as the angle of repose.
(Example: centrifugal force = mv^{2}/r = mrω^{2})
m_{g} = F /g
= F_{G }r^{2}/GM
Here, g is the free fall acceleration, F_{G} is the gravitational force and G is the gravitational constant.
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