Transcript Dynamic forces - Physics Champion

DYNAMIC FORCES
Equations of motion
1, Acceleration (rate of change of Velocity)
velocity is the vector version of speed, (speed
in a straight line)
v–u
t
=a
Can be
rearranged to
v = u + at
u = initial velocity (in m/s)
v= final velocity (in m/s)
t = time (in seconds (s))
a = m/s2
Sample question
• A car increases it’s velocity from 10 m/s to
20m/s in 5 seconds. What is it’s
acceleration?
Sample question (answer)
v–u
t
=a
v = 20m/s u =10m/s t = 5 sec
20 – 10
5
= 2m/s2
Sample question
A car moving at 10 m/s accelerates at
2m/s2 for 4 seconds. What is it’s final
velocity?
Sample question (answer)
Can be re-arranged to
v = u + at
u = 10m/s a = 2m/s2 t = 4 sec
v = 10 + 2 x 4 = 18m/s
Sample question
•A stone block with a mass of
800Kg is lifted from rest with a
uniform acceleration by a crane
such that it reached a velocity of
14m/s. after 10 seconds.
Calculate the tension in the
lifting cable
Sample question
• To do this calculation we need to use another equation
• Force = mass x acceleration
• F = ma
• Weight (force acting downwards) = mass x acceleration due to
gravity
• Wt = mg
Wt = m x 9.81 (acceleration due to gravity on earth)
Sample question (answer)
First find the
acceleration
Tension
acceleration
U = 0 from standing start
v–u =a
t
v
t
=a
14 = 1.4m/s2
10
Weight
mg
Force due to
acceleration
F = ma
Sample question (answer)
F = ma
F = 800 x 1.4 = 1120N
Tension
Wt of box mg = 800 x 9.81 = 7848N (7.8kN)
acceleration
Tension in cable = 1120 + 7848
= 8938N (8.9kN)
Weight
mg
Force due to
acceleration
F = ma
Sample question
• If the stone is lowered at the same rate of
acceleration the tension is less than the weight
of the box
Force due to
Tension
acceleration
F = ma
Tension in cable = 7848 - 1120
= 6728N (6.7kN)
acceleration
Weight
mg
Lift
You get the same effect when
you travel in a lift. You feel
heavier when the lift begins and
accelerates upwards and lighter
when the lift accelerates
downwards
A man standing on weighing scales in a lift has a
mass of 60 Kg. The lift accelerates uniformly at
a rate of 2m/s2 Calculate the reading on the
scales during the period of acceleration .
acceleration
Weight
mg
Force due to
acceleration
F = ma
F = ma F = 60 x 2 = 120N
Wt of man mg = 60 x 9.81 = 588.6N
Force acting on scale = 120 + 588.6
= 708.6N
The reading on the scale is 708.6 ÷ 9.81
= 72.23Kg
2, Displacement (s) is the vector version of distance
(distance in a straight line)
Displacement = average
velocity x time
s = (v + u) x t
2
Sample question
•A car moving at 10 m/s increases
it’s velocity to 20m/s in 4 seconds.
How far will it have travelled during
this time?
Sample question (answer)
s = (v + u) x t
2
s = (20 + 10) x 4
2
v =20m/s u = 10m/s t = 4 sec
S = 60m
Sample question
• How long will it take for an athlete to
accelerate from rest to 4 m/s over 8m ?
Sample question (answer)
s = (v + u) x t
2
t =
2s
v
u =0 so
t =
s= vxt
2
16
4
= 4 seconds
3, Displacement related to acceleration
s = ut + at2
2
With zero acceleration
(constant velocity)
a=0
s = ut
From a standing start ut = 0
s = at2
2
Example
•A car accelerates uniformly from
rest and after 12 seconds has
covered 40m. What are its
acceleration and its final velocity ?
Example
2s = at2
80 = a
122
80 = a
144
2s = a
t2
a = 0.56m/s2
Finding v
From a standing start u = 0
a=v–u
t
v= axt
a=v
t
v= axt
v = 0.56 x 12 = 6.7 m/s
Or alternatively
•
•
U = 0 from standing start
s = (v + u) x t
2
s=
v = 2s
t
vxt
2
v = 80 = 6.7m/s
12
4, final velocity, initial velocity acceleration and
displacement
2
•v
=
2
u
+ 2 as
Example
If a car travelling at 10m/s accelerates
at a constant rate of 2m/s2, what is it’s
final velocity after it has travelled 10m?
Sample Question (answer)
2
v
=
2
u
+ 2 as
V2 = 102 + (2 x 2 x 10)
= 100 + 40 =
2
140m/s
D’Alembert’s Principle
Acceleration
(a)
Applied
Force (F)
mass
F + Fi = 0 so F = -Fi
F = ma
so ma = - Fi
or Fi = - ma
Inertia
Force (Fi)
When an object is
accelerating the
applied force making it
accelerate has to
overcome the inertia.
This is the force which
resists the acceleration
(or deceleration) and is
equal and opposite to
the applied force. This
means that the total
force acting on the
body is zero
D’Alembert’s Principle
Applied
Force
200N
Mass
20Kg
Spring Force
150N
Acceleration
Friction Force
20N
Free body
diagram
F (applied)
200N
W
(weight)
Fs (spring)
150N
Mg =
20 x 9.81
=196.2N
Fr (Friction)
20N
D’Alembert’s Principle
• Downward forces are minus and upward forces are
positive.
• From the diagram
• -200N – 196.2N +20N +150N
• = -226.2N (which is downward as to be expected)
Conservation of energy (Gravitational potential energy)
GPE = mass x
gravity x height
(mgh)
=
5 x 9.81x10
10m
Work done
(in lifting the
mass) = force
x distance
49.05 x 10
= 490.5joules
490.5joules
(F = mg)
5Kg
Work done = energy gained
Kinetic energy
5Kg
10m
GPE = mgh = 490.5
joules
Neglecting friction, all the
GPE at the top of the slope
converts to kinetic energy
at the bottom
KE (at the bottom) =
490.5J =
mv2
2
(v = velocity)
Kinetic energy
5Kg
10m
GPE = mgh = 490.5
joules
Neglecting friction, all the
GPE at the top of the slope
converts to kinetic energy
at the bottom
The velocity at the bottom
v =√(2Ke÷m)
= √ (2 x 490.5 ÷ 5)
√196.2
= 14m/s
Work and energy
Work done (in joules) = Force x distance moved (in direction of force)
Work done = energy used
Force
distance
Example
• A car of mass 800Kg is stood on a uniform
1 in 10 slope when it’s handbrake is
suddenly released and it runs 30 metres to
the bottom of the slope against a uniform
frictional force of 50N. What is the car’s
velocity at the bottom of the slope?
Example
• First find the angle of the slope
Tan of the angle =
opposite/adjacent =
1/10= 0.1
Tan-1 0.1 = 5.7o
1
10
Example
• Then find the height of the slope
• sin 5.7o = opposite/hypotenuse
• Opposite = hypotenuse x sine5.7o
30m
ht
• 30 x sine 5.7o
5.7o
• =2.98m
Example 1 (conservation of energy method)
• Find the gravitational potential energy of the car at the top
of the slope
30m
ht
5.7o
• 800 x 9.81 x 2.98
• = 23.4kN
Example 1 (conservation of energy method)
• Work done against friction = force x distance = 50N x
30m = 1.5kj
30m
ht
5.7o
Example 1 (conservation of energy method)
• Kinetic energy of the car at the bottom of the slope
30m
ht
5.7o
mv2/2 = 23.4 -1.5 = 21.9kj
v= √(2 x Ke/m)
v= √(2 x 21900/800)
v= √54.75
v = 7.4 m/s
•
Example 2 (Resolving forces method)
• Work out the forces involved
30m
ht
5.7o
Weight of car (force
acting vertically
downward) = mass x
gravity
= 800x 9.81
7848N
Example 2 (Resolving forces method)
Wt = weight of car (7848N)
Fn is the normal reaction force of the
slope on the car
Fs is the force on the car down the
slope
Fs
wt
Fn
5.7o
5.7o
Example 2 (Resolving forces method)
• Work out the forces involved
Sine 5.7o = Fs ÷ wt
Fs = Wt x sin 5.7o
Fs = 7848 sin 5.7o
Fs
Fs = 779.5N (Force down the slope
wt
Fn
5.7o
5.7o
Example 2 (Resolving forces method)
• Work out the forces involved
Resultant force down the slope =
Fs – friction force
779.5 – 50
=749.5N
Fs
wt
Fn
5.7o
5.7o
Example 2 (Resolving forces method)
Acceleration down the slope
a= F/m
749.5/800
= 0.94m/s2
Fs
wt
Fn
5.7o
5.7o
Example 2 (Resolving forces method)
Velocity at the bottom of the
slope
v2 = u2 + 2as (u2 = 0) so
v2 = 2as
v2 = 2 x 0.94 x 30
v =√56.4
v = 7.5m/s
Fs
wt
Fn
5.7o
5.7o
Linear Momentum
and Collisions
• Conservation
of Energy
•
• Momentum = mass x velocity
• The total momentum remains the same
before and after a collision
• Momentum is a vector quantity
Linear Momentum
and Collisions
• A railway coach of mass 25t is moving along a
level track 36km/hr when it collides with and
couples up to another coach of mass 20t
moving in the same direction at 6km/hr. Both of
the coaches continue in the same direction
after coupling. What is the combined velocity
of the two coaches?
Linear Momentum
and Collisions
•Let the mass of the first coach be M1
and the mass of the second coach be
M2 and the velocity of the first coach
be V1 and the velocity of the second
be V2
Linear Momentum
and Collisions
• Before coupling the momentum of the first
coach is 25 x 36 = 900tkm/hr and the
momentum of the second is 20 x 6 =120
tkm/hr
• Which is a total of 1020tkm/hr
Linear Momentum
and Collisions
• After the coupling the momentum of both is the
same as before the coupling which is 1020
tkm/hr
• And the combined mass is 45t
Linear Momentum
and Collisions
• Velocity after coupling is momentum divided by
mass
• 1020÷45
• 22.6 km/hr
Example
A hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and
drives it 100mm into the ground
a) Calculate the loss of energy on the impact.
b) Calculate the work done by the resistance of the ground.
c) calculate the average resistance to penetration.
Example
• Before falling the GPE of the hammer is 200kg x 9.81 x 5m = 9810j
• Kinetic energy of the hammer just before impact = 9810j (0.5 x m x v2)
v2 = (9810)/0.5 x m
(9810)/0.5 X 200 =98.1
v = √98.1 = 9.9m/s
• Velocity of the hammer just before impact
•
•
Example
• Momentum of hammer just before impact = mass x velocity
•
= 200kg x 9.9m/s = 1980 kgm/s
•
Momentum before collision = momentum after collision
• mass of hammer plus pile after collision = 200kg + 300kg = 500kg
• thus 500kg x velocity after collision = 1980 kgm/s
• Velocity after collision = 1980 ÷ 500 = 3.96m/s
Example
• Kinetic energy after collision (0.5 x mass x v2) = 0.5
x 500 x 3.962
• = 3920j
• Loss of energy = 9810 – 3920 = 5890j
• To find deceleration of pile hammer
• v2 = u2 +2as (v2 =0)
• u2 = - 2as
•
Example
• a = - u2/2s
• 3.962/0.2
• - 78.4m/s2
• The minus sign means deceleration, Resistive force
of the ground = mass x deceleration
• 500 x78.4
• 39.2kN
Example
• Work done by ground = Force x distance
• 39.2kN x 0.1 m
• 3.92kj
(this agrees with the fact that fact that the kinetic energy of the
hammer and pile after impact was 3.92kj and zero when the pile
stopped moving in the ground