dynamic forces equations of motion. 1, acceleration (rate of change of velocity) velocity is the...

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?


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)


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


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


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)


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


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?


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 ?


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


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


•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


• Work done against friction = force x distance = 50N x 30m = 1.5kj

ht

30m

5.7o


• 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


• 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


• 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


• 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


Acceleration down the slope a= F/m

749.5/800= 0.94m/s2

5.7o

5.7o

Fn

Fs

wt


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

dynamic forces equations of motion. 1, acceleration (rate of change of velocity) velocity is the...

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