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Dynamics
Dynamics is the study of the forces acting upon moving particles.
NEWTON’S FIRST LAW
EVERY BODY WILL REMAIN AT REST OR CONTINUE TO MOVE IN A STRAIGHT LINE AT
CONSTANT SPEED UNLESS AN EXTERNAL FORCE ACTS UPON IT.
So, for any system where there is a zero resultant force then there is no change in velocity
of the system (and vice versa).
We say that the system is in equilibrium.
EXAMPLE
A train of total mass 300 tonnes is travelling along a straight horizontal track at a constant
speed of 54 km h-1. The resistances to motion are 50 Newtons per tonne.
Calculate the tractive force of the engine.
EXAMPLE
A particle of weight 52 N is in equilibrium on a rough slope inclined at an angle of 30º to
the horizontal.
Find the magnitude of both the friction force and the normal reaction.
NEWTON’S SECOND LAW
THE ACCELERATION OF A BODY IS PROPORTIONAL TO THE RESULTANT FORCE AND ACTS IN
THE DIRECTION OF THAT FORCE.
The constant of proportionality is found to be the mass of the object.
F  ma
DEFINITION
1 Newton is the force required to accelerate a mass of 1 kg at 1 ms-2.
EXAMPLE
Find the acceleration and the normal reaction force R in the system below.
R
10 N
50 N
60º
10g N
a
EXAMPLE
Find the acceleration and the angle θ in the following situation.
54 N
49 N
a
10g
θ
EXAMPLE
A particle of mass 20 kg is given an initial velocity of (3i + 4j) ms-1. It travels for 12
seconds until it reaches a velocity of (39i - 8j) ms-1.
Find the acceleration and the magnitude of the resultant force on the particle.
NEWTON’S THIRD LAW
WHEN TWO BODIES INTERACT THEY EXERT FORCES ON ONE ANOTHER WHICH ARE EQUAL IN
MAGNITUDE AND OPPOSITE IN DIRECTION.
So, for every action there is an equal and opposite reaction.
EXAMPLE
A woman of mass 60 kg is standing in a lift.
a) Draw a diagram to show the forces acting on the woman.
Find the normal reaction when the lift is moving :b) upwards with constant speed 3 ms-1.
c) upwards with constant acceleration of 2 ms-2.
d) downwards with constant acceleration of 2 ms-2.
e) downwards with constant retardation of 2 ms-2.
In order to calculate the maximum number of occupants that can be safely carried, the
following assumptions are made: the lift has mass 300 kg, all resistances to motion can be
neglected, the mass of each occupant is 75 kg, and the tension in the supporting cable
should not exceed 12 000 N.
f) What is the greatest number of occupants that may be safely carried if the magnitude of
the acceleration does not exceed 3 ms-2?
CONNECTED PARTICLES
When two or more particles are connected together we usually model each particle
separately and use simultaneous equation to solve.
EXAMPLE
A car of mass 1 tonne tows a trailer of mass 400 kg along a road with a constant
acceleration of 2 ms-2. If the tractive force due to the car’s engine is 4000 N and the car
experiences a resistance of 1000 N, find the tension in the tow-bar and the resistive force
exerted on the trailer.
EXAMPLE
A light, inextensible string passes over a smooth, weightless pulley. A 10 kg mass is
attached to one end of the string and a 6 kg mass to the other. The system is released.
a) Calculate the acceleration of the two masses and the tension in the string.
b) Find the force exerted on the pulley.
EXAMPLE
Light, inextensible strings AC and DF are attached to each side of a block of mass 11 kg
which is on a rough horizontal table. The string sections BC and DE are parallel to the
table and the strings pass over smooth pulleys at B and E. Objects of mass 5 kg and 12 kg
are attached to the free ends A and F, respectively, of the strings and hang freely. The
coefficient of friction between the block and the ground is 25 .
77
The system is released from rest.
a) Show that the acceleration of the system is 1.2 ms-2, and find the tensions in the string.
When the system reaches a speed of 4 ms-1 the string DF breaks.
b) Calculate the total distance travelled by the block before it comes to rest.
IMPULSE AND MOMENTUM
In general when two bodies collide, the mutual action between the bodies has a very large
value for a very short time.
Forces of this nature are called impulsive forces.
If a constant force F acts on a body for a time t seconds, then the impulse I is given by :-
I  Ft
Impulse is a vector quantity with units of Newton seconds (Ns)
If a body of mass m kg is travelling at a velocity of v ms-1, then its momentum is given by
Momentum = mv
Momentum is a vector quantity with units kg m s-1 or Ns.
EXAMPLE
A particle of mass 2 kg is given an acceleration of 4i + 3j by an impulsive force over a
period of 3 seconds.
Find the magnitude of the impulse of this force.
IMPULSE-MOMENTUM EQUATION
Consider a particle of mass m, acted upon by a constant force F for a short time t.
Suppose the particle has initial velocity u and final velocity v.
As
Then
So
Or
v  u  at
v u
a
t
 v u 
F  ma  m 

 t 
Ft  mv  mu
Impulse = Change in Momentum
This rule also works for variable acceleration.
EXAMPLE
An artillery shell of mass 50 kg is fired with a speed of 560 ms-1.
Given that the shell is in the barrel of the gun for 0.05s, calculate the average force, in
kN, exerted on the shell by the explosive charge.
EXAMPLE
A body of mass 4 kg, initially at rest on a smooth horizontal surface, experiences a force
of 5i -12j N for 2 seconds. Find the final velocity of the body and hence its speed.
EXAMPLE
A smooth peg of negligible diameter is fixed at a height of 3l above a horizontal table and
a light inextensible string of length 4l hangs over the peg.
Particles of mass m and 2m are attached to either end of the string. The system is held at
rest with the particles hanging at the same level, distance l from the table.
When the system is released, find the speed u with which the 2m mass hits the table.
Given that this particle rebounds from the table with speed ½u, find in terms of m, g and l
the magnitude of the impulse on the table.
CONSERVATION OF MOMENTUM
When two particles collide they exert an equal and opposite impulse upon each other.
Hence the changes in the momentum are also equal and opposite.
m1v1  m1u1  (m2v2  m2u2 )
m1v1  m2v2  m1u1  m2u2
Final Momentum = Initial Momentum
This is the principle of conservation of momentum. Formally written :If no external force acts upon a system in a particular direction then there can be no
change in the total momentum of the system in that direction.
EXAMPLE
A pile driver of mass 2 tonnes, moving with velocity 7 ms-1 hits a stationary pile of mass
0.5 tonnes. Find their common velocity after the impact.
EXAMPLE
A shell of mass 20 kg is travelling horizontally at 100 ms-1 when it explodes into three
pieces A, B and C of masses 12, 6 and 2 kg respectively. A travels in the original
direction of motion at v ms-1, B at 30º above the horizontal at 60 ms-1 and C backwards at
an angle θ below the horizontal at a speed of 120 ms-1. Find θ and v.