dynamic forces equations of motion. 1, acceleration (rate of change of velocity) velocity is the...
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DYNAMIC FORCESEquations 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 u = initial velocity (in m/s)v= final velocity (in m/s)t = time (in seconds (s))a = m/s2
Can be rearranged
tov = u + at
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
20 – 10 5
= 2m/s2
v = 20m/s u =10m/s t = 5 sec
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)
v = u + at
v = 10 + 2 x 4 = 18m/s
Can be re-arranged to
u = 10m/s a = 2m/s2 t = 4 sec
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)
v – u = a v = a 14 = 1.4m/s2
t t 10
acceleration
Force due to acceleration
F = ma
Tension
Weight mg
U = 0 from standing start
First find the acceleration
Sample question (answer)
F = ma F = 800 x 1.4 = 1120N Wt of box mg = 800 x 9.81 = 7848N (7.8kN)
Tension in cable = 1120 + 7848 = 8938N (8.9kN)
acceleration
Force due to acceleration
F = ma
Tension
Weight mg
Sample question
• If the stone is lowered at the same rate of acceleration the tension is less than the weight
of the box
Tension in cable = 7848 - 1120 = 6728N (6.7kN)
Force due to acceleration
F = ma
Tension
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
Force due to acceleration
F = ma
Weight mg
F = ma F = 60 x 2 = 120NWt 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
v =20m/s u = 10m/s t = 4 sec
s = (20 + 10) x 4 2
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
u =0 so s = v x t 2
t = 2s v
t = 16 = 4 seconds 4
3, Displacement related to acceleration
s = ut + at2
2
With zero acceleration (constant velocity) a = 0 s = utFrom a standing start
ut = 0s = 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
2s = a t2
80 = a122
80 = a144
a = 0.56m/s2
Finding v
a = v – u t
From a standing start u = 0
a = v t
v = a x t
v = a x t v = 0.56 x 12 = 6.7 m/s
Or alternatively•
• U = 0 from standing start
s = (v + u) x t 2
s = v x t 2
v = 2s t
v = 80 = 6.7m/s 12
4, final velocity, initial velocity acceleration and displacement
•v2 = u2 + 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)
v2 = u2 + 2 as
V2 = 102 + (2 x 2 x 10)
= 100 + 40 = 140m/s2
D’Alembert’s Principle
Acceleration (a)
Applied Force (F)
Inertia Force (Fi)
F + Fi = 0 so F = -Fi
F = ma
so ma = - Fi
or Fi = - ma
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
mass
D’Alembert’s Principle
Acceleration
Friction Force
20N
Applied Force
200N
Spring Force
150N
Mass
20Kg
Fr (Friction)
20N
W
(weight)
Mg =
20 x 9.81
=196.2N
F (applied)
200N
Fs (spring)
150N
Free body diagram
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)
10m
5Kg
GPE = mass x gravity x height
(mgh)=
5 x 9.81x10
490.5joules
Work done (in lifting the mass) = force
x distance49.05 x 10
= 490.5joules
(F = mg)
Work done = energy gained
Kinetic energy
10m
5Kg
GPE = mgh = 490.5 joules
KE (at the bottom) =490.5J =
mv2 2
(v = velocity)
Neglecting friction, all the GPE at the top of the slope converts to kinetic energy
at the bottom
Kinetic energy
10m
5Kg
GPE = mgh = 490.5 joules
The velocity at the bottom v =√(2Ke÷m)
= √ (2 x 490.5 ÷ 5)√196.2
= 14m/s
Neglecting friction, all the GPE at the top of the slope converts to kinetic energy
at the bottom
Work and energy
Force
distance
Work done (in joules) = Force x distance moved (in direction of force)
Work done = energy used
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
1
10
Tan of the angle = opposite/adjacent =
1/10= 0.1Tan-1 0.1 = 5.7o
Example
• Then find the height of the slope • sin 5.7o = opposite/hypotenuse
• Opposite = hypotenuse x sine5.7o
• 30 x sine 5.7o
• =2.98mht
30m
5.7o
Example 1 (conservation of energy method)
• Find the gravitational potential energy of the car at the top of the slope
• 800 x 9.81 x 2.98• = 23.4kN
ht
30m
5.7o
Example 1 (conservation of energy method)
• Work done against friction = force x distance = 50N x 30m = 1.5kj
ht
30m
5.7o
Example 1 (conservation of energy method)
• Kinetic energy of the car at the bottom of the slope
mv2/2 = 23.4 -1.5 = 21.9kjv= √(2 x Ke/m)
v= √(2 x 21900/800) v= √54.75
v = 7.4 m/s •
ht
30m
5.7o
Example 2 (Resolving forces method)
• Work out the forces involvedWeight of car (force
acting vertically downward) = mass x
gravity= 800x 9.81
7848N
ht
30m
5.7o
Example 2 (Resolving forces method)
• Work out the forces involved
5.7o
5.7o
Fn
Fs
wt
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
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 = 779.5N (Force down the slope)
5.7o
5.7o
Fn
Fs
wt
Example 2 (Resolving forces method)
• Work out the forces involved
Resultant force down the slope = Fs – friction force
779.5 – 50 =749.5N
5.7o
5.7o
Fn
Fs
wt
Example 2 (Resolving forces method)
Acceleration down the slope a= F/m
749.5/800= 0.94m/s2
5.7o
5.7o
Fn
Fs
wt
Example 2 (Resolving forces method)
Velocity at the bottom of the slope
v2 = u2 + 2as (u2 = 0) so
v2 = 2asv2 = 2 x 0.94 x 30
v =√56.4v = 7.5m/s
5.7o
5.7o
Fn
Fs
wt
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)
• Velocity of the hammer just before impact v2 = (9810)/0.5 x m• (9810)/0.5 X 200 =98.1• v = √98.1 = 9.9m/s
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
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