Mechanics - Dynamics

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Dynamic 3rd secondary Chapter one Newtons laws of motion - 1 -

Equations of motion uniform acceleration

Dynamics

Dicplacement

Distance

Revision

Introduction : Dynamics is a branch of mechanics which deals with moving bodies under the

Action of some influences of forces .

Velocity Vector V : A velocity vector is a veloc ity with direction .

The velocity is measured by : Km / hr , m / sec Or cm / sec where :

1000 m 5 1 km / hr m / sec

60 60 sec 18

1000Or we can say : 1 km / hr

2 1

100cm 250cm / sec

60 60 sec 9

Displacement vector S :

The of a particale during an interval of time t t t

is the change of the position of the particle from its initial position A

displacement

1 2at t to its terminal position B at t .

----------------------------------------------------------------------------------------------------------------------

2

2 2

1 V u a t " we use it if displacement is not given "

12 S u t a t " we use it if velocity is not given "

2

3 V u 2 a s " we use it if time is not given "

These equations are used only when

i S here equals to the whole displacement of the whole trip

ii If the

a body is travelling in a straight line with

Constant uniform acceleration If positive Or deceleration If negative .

Notes

o body started from rest : Initial velocity u v Zero

iii If the body stopped after a time t : Final velocity V Zero

iv Uniform velocity means that : S V T as a 0

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x -3




Very Important Remarks : Cases of Displacements S

Important Notes : 1 g , v , u , s are all positive in this case .

2 The distance in this case measure of the displacement.

3 The exp

ression : Left to fall or fell means U 0 .

4 The expression : The body is projected by Veloctiy...

" This is the Init

ial velocity : "U"

1 V U gt

"This equation to calculate the velocity of a body for a time period of t seconds"

1 2 S U T

2

2 2

g t2

"This equation can be used to calculate the distance traveled by a body in a time t"

3 V U 2 gs

" the above equation gives a relation between

the final velocity v of the body and the distances

traveled by the body"

Second : Vertical Motion Upwards

2

2 2

1 V u g t

1 2 S ut g t

2

3 V u 2g s

----------------------------------------------------------------------------------------------------------------------

The Vertical motion Downwards




maxS maxt

2

max

Finding the Time and Distance of maximum height :

u t Time of Max. height

g

u S

2g

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Expressions could help you Left to fall initial velocity = zero u 0 Fall from rest initial velocity = zero u 0 Comes to rest terminal velocity = zero . V 0

Projected upwards Gravity will be negative 2g -9.8 m/sec Uniform velocity acceleration = zero And S = V t Maximum velocity acceleration = zero Maximum height velocity = zero V 0 Mass of drop (each ) drop when it reaches the ground

Retarded acceleration = -ve ----------------------------------------------------------------------------------------------------------------------

Using the algebraic measure of vectors

0 S r r

d S d r1

d t d t

d S

2 Vd t

dV

3 Ad t

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Mass = original mass of the drop + (Rate of increase time)

Differentiation vector




th

Very important remarks

If S is a distance covered during n second

Distance during a given interval of time

th

th

th th

s for example :

Ex 1 : The distance covered in the 5 seconds :

4 5Here 5 starts from 4 5 seconds Then t 4.5 s from the begining of time

2

Ex 2 : The distance covered in the 5 and 6 seconds :

Time

rd th th

4 6 starts from 4 5 6 seconds Then t 5 s from the begining of time

2

Ex 3 : The distance covered in the 3 , 4 and 5 seconds :

2 5 Time starts from 2 3 4 5 seconds Then t 3.5 seconds

2

We know that the motion is something relative to another and it changes in description by the

Change of an observer .

For Example : If you look from the train window to the moving cars in the same dire

A

ction of the

motion of the train , They seen to move slowly , While we feel the contrary if the cars are moving

in the opposite direction of the train .

So , If the velocity vector of A is V and th B

BA

B A B A B BA A

e velocity vector of B is V

Then , The relative velocity vector of B with respect to A is denoted V

Where : V V V Or we can say : V V V

Velocity of B Velocity of B w.r

. to A Velocity of A

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The Relative Velocity

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2oGiven : u v 10 cm/sec a 2.5 cm/sec

Note : The velocity is in the same direction of the acceleration means that the velocity of the

body will increase as time increase " i.e: Imagine a bicycle is

22

2 2

2 2

2

moving on a bridge"

1 1i S u t at S 10 8 2.5 8 160 cm

2 2

1 5ii 700 10t 2.5 t 700 10t t Multiply by 4

2 4

2800 40t 5t 5t 40t 2800 0 divide by 5

t 8t 560 0 t 28 t 20 0 t 20 sec

Thus the body is at 700 far from the initial point after 20 seconds

st

BA

A car

B cycle

BA B

1 step : let the positive direction in the direction of the relative velocity here V

First : both are in the same direction

let V V 82 x

V V 43 x

V V

A BA

BA B

V V 43 x 82 x -39 x

The Magnitude of the relative velocity of the cyclist with respect to the car 39 km / hr

Second : both are in opposite direction

V V

A

BA BA

- V

V 43 x 82 x V 125 x Its Magnitude is 125 km / hr .

AV 82

BV 43

A B C

A B

xAV

BV

x

Example (1) A car moves on a straight road with velocity 82 km / hr . If it meets a motor - cyclist moves on

the same road with velocity 43 km / hr , Find the relative velocity of the cyclist with respect to

the c

ar in two cases :

First : The cyclist and the car move in the same diection .

Second : The cyclist and the car move in the opposite direction .

Answer

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Example (2)

2A body moves in a straight line with a uniform acceleration of magnitude 2.5 cm/sec , and

Initial velocity 10 cm/sec in the same direction of the acceleration , find :

i The distance covered by the bod

y after 8 seconds

ii After how many seconds the body is at 700 cm far from the begining point.

Answer




2 2 2

The word "fell" means U 0

1 After 3 second here means t 3 sec .

V U gt 9.8 3 29.4 m / sec

2 S 78.4 m U 0 g 9.8 m / sec .

1 1 S u t g t 78.4 9.8 t t 16 t 4

2 2

sec.

So , We need 4 sec to make the body reach the ground .

3 To get the velocity when the body reach the ground :

We have to know either the time of the body to reach the ground , Or the t

2 2

otal distance

of the body to reach ground.

So V U g t V 9.8 4 39.2 m / sec .

Or V U 2 g s V 2 9.8 78.4 39.2 m / sec .

S 78.4

u 0

ve

g

29.8 m / sec

Example (3)

A body fell vertically downwards from height of 78.4 m above the ground , Find :

1 The velocity of the body after 3 seconds .

2 The time taken to reach the ground .

3 The velocity when it reaches the ground .

Answer

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Mass :

The mass of a body is a ve scalar quantity which is proportional with the weight of this body .

The mass of a body is denoted by m .

Units of mass : 1 Ton 1000 Kgm & 1 Kgm 100

0 gm & 1Gram 1000 milligram .

Definition :

The momentum vector of a particle , At a certain instant is defined as the product of the mass

of the particle and its velocity vector at this instant , Momentum is denoted by H .

Rule of Momentum :-

From this definition , It is clear that the momentum of a body at a certain instant is a vector in

the same direction of the velocity vector .

H m v

Units of Momentum : H mv

Where m Mass , And v Velocity .

Its units may be for an example : gm . cm / sec & kg . m / sec & kg . m / hr .

Note : I always prefer to use the unit kgm . m / sec .

2

Since there is only one body in the problem , Then we don't need to show a direction .

To find the momentum : H mv

1So we must find v : u 0 a 9 cm / sec t 60 30 sec .

2

v u at v 9 30 270 c

m / sec .

H mv 7.5 270 2025 gm . cm / sec .

-------------------------------------------------------------------------------------------------------

For example, a heavy truck moving fast has a large momentumit takes a large and prolonged force to get the truck up to this speed, and it takes a large and prolonged force to bring it to a

stop afterwards. If the truck was lighter, or moving slower, then it would have less momentum.

Momentum can be defined as "mass in motion." All objects have mass; so if an object is

moving, then it has momentum - it has its mass in motion

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------------------------------------------------------------------------------------------------------- Example (1)

2A body of mass 7.5 gm , Moves from rest in a straight line with acceleration 9 cm / sec in the

1direction of its motion , Find its momentum after minutes from the begining of the motion .

2

Answer

Newtons laws of motion

1st

: Momentum




Example (2)

Find the momentum of a stone of mass 500 gm , When it is let to fall 4.9 meters vertically

downwards .

Answer

2 2 2

H mv So we have to get velocity :

u 0 " fall" s 4.9 m .

So , v u 2 g s v 2 9.8 4.9 96.04

v 9.8 m / sec .

H 500 9.8 4900 gm . m / sec .

------------------------------------------------------------------------------------------------------- Example (3)

Find the height from which a body of mass 500 gm , falls such that the magnitude of the momentum

when it collides with the ground equals the magnitude of the momentum of a body of mass 70 gm

moving with velocity of magnitude 432 km / hr .

Answer

2 1 1 2 2

1 1

1

22 2

5v 432 120 m / sec . So m v m v

18

500 V 70 120 V 16.8 m / sec .

u 0 v 16.8 m / sec s ??

v u 2g S 16.8 2 9.8 S S 14.4

m .

------------------------------------------------------------------------------------------------------- Example (4)

1A gun fires 300 bullets per minute ,If the mass of each bullet is kg and its velocity at the opening

5

of the gun is 200 m / sec , Find the momentum of the bullets fired per second in gm.cm / sec

Answer In order to find the momentum of bullets fired per second .

300 Number of bullets fired 5 bullets per second .

60

1Each one of them has mass kg .

5

Then the mass of the bullets in one secon

7

ds is :

1 5 1000 gm 1000 gm .

5

Momemtum H mv 1000 200 100 2 10 gm .cm / sec .

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2 2 21 12

1 1

1

Let the positive direction be .

Before impact : u 0 s 16.9 m .

v u 2 g s v 2 9.8 16.9

v 331.24 v 18.2 x

And H is the momentum vector of the ball before

2 2 22 2 2

impact .

5460 5460 m 18.2 m 300 gm .

18.2 Momentum after impact : u ?? v 0 " as the ball will become finally at rest" s 4.9

v u 2 g s 0 u 2 9.8 4.9 u 9.8 m / sec . u

2 1

-9.8 x

The change of momentum before and after impact is : m v v 300 9.8 18.2 8400 x .

Its magnitude is 8400 gm .m / sec .

ve

16.9 m1v

v 0

4.9 m

u ??

1 2 2 1

2 1 2 1

It is the change of the velocity of an object from T to T H H

Rule : The change in momentum : H H m v v It is also called Impulse

ve

2v

1v

A rubber ball of mass 300 gm moves horizontally with uniform velocity 135 cm / sec . It collides

4by a vertical wall , And rebounds in a perpendicular direction to the wall after loosing of the

5

magnitude of its velocity before collision, Find the magnitude of the ball change in momentum due

to collision with the wall .

u Not the same as V

the direction changed

u 0

nd

1 1

2

let the positive direction be in the direction of the 2 velocity

v 135 m / sec . v 135 x

The velocity after , Collision is opposite to x

4 1 After loosing of its velocity v 135 27

5 5

2 1 2 1

2 1

x

Change of momentum H H m v v

H H 300 27 135 x 48600 x Its magnitude 48600 gm . cm / sec

------------------------------------------------------------------------------------------------------- Example (5)

Answer

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Example (6) A ball is left to fall from a height of 16.9 m , And its momentum when it impinges with the

ground is 5460 gm . m / sec , Find its mass . If the ball rebounds to a height of 4.9 m , Then

Find the change of its momentum just before and after impact .

Answer

The change in momentum




1

1

2

1

In this problem, we want to find the velocity of the bullet and the wood after impact

v be its velocity just before impact .

v 390 x

Let v be the velocity of the system after impact .

And m be

2 the mass of the bullet and m is the mass of the system .

As the momentum of the system does not change due to impact .

2v

1v

x

1m 120 gm

2m 3120 gm

PT P

Let x be the unit vector in direction of the projectile :

H m v 5 350 x 1750 kg . m / sec .

1 The tank is moving away of the canon means , That are in the same direction .

V V V

T

PT

PT P T

5350 x 45 x 337.5 x

18

Its momentum H m v 5 337.5 x 1687.5 x

The magnitude of the moment of the projectile 1687.5 kg . m / sec .

2 v v v

35

PT

50 x 45 x 362.5 x

18

Its momentum H m v 5 362.5 x 1812.5 x

The magnitude of its momentum is 1812.5 kg . m / sec .

x

x

TvPv

PvTv

1 2 1 1 2 2

2 2

H H m v m v

120 390 120 390 x 3000 120 v v 15 x

3120Thus the system will move after impact with velocity 15 m / sec in the same direction of th bullet .

Example (7) A bullet of mass 120 gm is fired with a velocity of 390 m / sec towards a wooden body of mass

3 kgm .which is at rest . If the bullet is imbedded in it and the system moves after that with a

certain velocity . Find their velocity , Given that the momentum of the system doesnot change due

to impact .

Answer

]

------------------------------------------------------------------------------------------------------- Example (8)

A fixed cannon fired a projectile of mass 5 kg with a velocity of 350 m / sec in a horizontal

direction towards a tank moving with a velocity of 45 km / hr and it hit it , Find the absolute

value of the

momentum of the projectile , Then Calculate the magnitude of the momentum

of the projectile relative to the tank if :-

1 The tank is moving away of the canon 2 The tank is moving towards the canon .

Answer




5.4 m

u 0

t 1.5 secx

a -2.7

First Find magnitude of the swimmer velocity directly before collision with water :

v u g t 0 9.8 1.5 14.7 m / sec

Let u be a unit vector directed vertically downwards .

The vel

1

2 2 2

22

ocity directly before collision v 14.7 x

Second Magnitude of swimmer velocity directly after collision with water :

v u 2a s 0 u 2 2.7 5.4

u 5.4

2

2 1 2 1

u 5.4 meter / sec .

The velocity directly after collision v 5.4 x

Change of mometum due to collision H H m v v

40 5.4 x 14.7 x 372 x

Its magnitude 372 kgm . m / sec .

V 0

V ??u ??

2 2

10.5 Water condineces at rate 0.0105 gm / sec .

1000

So , To know the total mass of the water drop in the 1000 m : We must get the time then .

1 1 100 s u t g t 1000 9.8 t t

2 2 7

sec .

100Then the time water drop travel from 1000 m till it reaches the ground is sec .

7

100Then the mass of the water vapour at this time 0.0105 0.15 gm .

7

The total mass of the drop 0.15 0.25 0.4 gm

.

100To get the velocity :- v u g t 9.8 140 m / sec .

7

The momentum H mv 0.4 140 56 gm . m / sec .

Example (9) Water vapour condinces on the surface of water drops , while it is falling at rate 10.5 milligram

/ sec , The mass of one falling water drop is 0.25 gm , Find the momentum in gm . m / sec of

one water drop when it reaches the surface of the ground from a height 1000 meters .

Answer

------------------------------------------------------------------------------------------------------- Example (10)

A swimmer of mass 40 kgm jumped vertically from rest to the water surface of a swimming

pool , He collided with the surface after 1.5 sec , Then he douse vertically into the water in a

retarded motion 2 with uniform acceleration of magnitude 2.7 m / sec and covered a distance is

5.4 meter before he starts the ascending , Calculate magnitude of his change of momentum due

to collision with water .Answer




Example (11)

A rocket is projected vertically upwards with velocity 180 km / hr , If its mass at any instant

is m 25 0.001t kgm , Find the rate of change in its momentum after 15 seconds .

Answer

Very important note : When the mass of the body comes variable in the problem , we must use

the rule of derivative only .

180u km / sec u 0.05 km / sec .

3600

v u g t v 0.05 9.8 t

H m v 25 0.001t 0.05 9.8 t x

2

d H-9.8 25 0.001t 0.001 0.05 9.8t At t 15 sec .

d t

d H-244 853 0.14695 -244.706 kg . m / sec .

d t

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Every body remains in its state of rest or uniform motion unless it is compelleted to change

that state by an external action called a force .

Discussion of the first law :

1 The law assumes that bodies which are at rest state or have uniform motion state is a

natural state of the body .

2 The law assumes that every body can not change any

of its natural state by itself , So this law

is called the law of inertia .

3 The law assumes that the existence of an external action called a force can only change the

state of the body .

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Newtons first law




A uniform motion on a horizontal plane under the

action of a horizontal force .

When a body of weight w moves horizontally by force F ,

You have

"Normal reaction of the road " " Wei

to know that it meets an opposite resistance called

" Resistance force " and it is denoted by R .

In this case F R Or F R 0

Also , We can say that : N W

ght "

A uniform motion of a body on a horizontal plane under

The action of an inclined force .

It a force is inclinedby angle to the horizontal .

Then : N F sin W

And R F cos

NF

R

W

F sin

F cos

Vertical uniform motion :

If a body of weight W moves uniformly vertically in a liquid ,

Then R W

R

W

Direction of

motion

Uniform motion on an inclined plane :

N F sin W Cos

F cos R W Sin

F

W

W cos

F cos F Sin

N

Motion

Case (1)

Case (2)

Case (4)

Case (3)

R

W Sin

FR

W

N

Motion

Motion

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Cases of uniform motion of a body




st

The train moves with a uniform velocity, then it will stop soon according to resistance

we have to use Newton's 1 law F R 2000

The resistance per ton 2000 250 8 kg .wt Case 1

F 2000 kg . wt R

The train moves with a uniform velocity, then it will stop soon according to resistance

The mass of the whole system is : Mass of the locomotive Mass of number of wagons

The mass 6 3n

It moves

st

with a uniform velocity :

So we have to use Newton's 1 law

810 F R 900 6 3n 15 900 90 45n 45n 810 n 18 wagons .

45

Motion900 kg . wtR

Notes :

1 The resistance of the plane to the moving body is always parallel to the plane and in the opposite

direction to the motion of the body .

2 The force generated by the motor of a car or a

train is always in the same direction of the motion .

3 When we say that the body is moving with maximum velocity , This means that it is moving

with a uniform motion Then a 0

4 If the resultant o

f forces , Acting on a body , Vanishes at any moment during its motion , Then

it moves from this moment with a uniform motion . " So , Sum of forces 0 when the body moves

with a uniform moti

1 1

2 2

22 2 1 1

2

2 2

on " .

5 In many times , The reaction R is variable as the velocity of the moving body .

R V If R V , Then : R A V where A is constant

R V

R V If R V , Then : R A V

R V

------------------------------------------------------------------------------------------------------- Example (1)

A train of mass 250 Ton moves with a uniform velocity along a horizontal plane . The force

of the engine is 2000 kg .wt ,Find the magnitude of the resistance for each ton of the mass .

Answer

------------------------------------------------------------------------------------------------------- Example (2)

A locomotive of mass 6 tons pulls a number of wagons the mass of each equals 3 tons

along a horizontal straight road with uniform velocity. If the magnitude of the driving force

of the locomotive equals 900 kgm.wt, and the resistance to the motion of the train equals

15 kgm.wt per each ton of its mass . Find the number of the pulled wagons .

Answer




The body moves in a uniform velocity : F R

3 4 31350 cos W 1350 W W 1440 kg . wt

4 5 4

3Also , N 1350 sin W N 1350 1440

5

N 630 kg.wt

N

W

1350 Sin

1350 cos

F 1350

3 w

4

4

35

3Sin

5

4Cos

5

R 4 3 12 kg . wt

The component of W in the plane direction downwards

1 1 1is W sin 3 Ton . wt 1000 50 kg . wt

60 20 20

The car moves uniformly With a uniform velocity

F R W sin F 12 50 62 kg . wt

N F

W

W sin W cos

R

A body is pulled along a horizontal straight road by a force of magnitude 1350 kgm.wt , and

3inclined at an angle of Sin to the horizontal, so the body moved with a uniform motion

5

against the road res3

istance which is equal to of its weight . Calculate the weight of the body 4

and the normal reaction of the road .

Example (3)

Answer

------------------------------------------------------------------------------------------------------- Example (4)

A car of weight 3 ton.wt moves with a uniform velocity up a plane inclines to the horizontal

1at an angle where Sin . If the resistance of the plane is 4 kg.wt per ton, find the

60

driving force of

the car in kg . wt .

Answer

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The engine stopped while moving downwards means

that the car is moving by its weight only No force exists

1 1 R 5Sin 5 ton.wt

100 20

1 R 1000 50 kg.wt

20

50So , the resistance per each ton is 10 kg.

5

wt 5

5Cos

RN

Motion

5 Sin

4.5 ton.wt 4.5 1000 4500 kg.wt

Case 1 : When the car engine is stopped and the car

moves downward

The car moves with its weight only

1 R 4500 Sin 4500

50

90 kg.wt

Case 2 : When the car engine is turned on and the car is moving upwards .

1 F 90 4500 180 kg . wt

50

4500Cos

Motion

4500 Sin

Motion

4500

N R

R 90

F

4500

4500 Sin 4500Cos

N

Example (5) The engine of a car of weight 5 ton.wt is stopped while it is moving downwards a road

1inclined to the horizontal at an angle where Sin with a uniform velocity ,

100

calculate the resistance for ea

ch ton of its mass in kg.wt.

Answer

------------------------------------------------------------------------------------------------------- Example (6)

A car of weight 4.5 ton.wt moves along the line of the greatest slope of a plane inclined

1at an angle of Sin , if the engine of this car is stopped when moving downwards with

50

uniform velocity . If the car engine is turned on, find the magnitude of the driving force of

this car such that it ascends this plane with uniform velocity given that the resistance of the

plane to the car is unchanged in the two cases of motion .

Answer




2

2 1

When the car moved in the inclined plane :

First : You have to know that : no force downward .

Then the car is moving with its weight only

1So , R 2500 sin 2500 78.125 kg . wt

32

3And R R

8

1

1 1

8 1 R 78.125 208 kg . wt

3 3

1From 1 F R 208 kg . wt

3

2500 sin

Motion

Motion

N

NW 2500

R

2500 cos

2500

Ramp

1F 1R

2 1

1 1

3Given : R R

8

When the car moves in the horizontal road with uniform velocity :

Then : F R 1

Example (7)

A car of weight 2.5 ton . wt moves on a straight horizontal road with a uniform velocity

1when it reached a ramp inclined to the horizontal at an angle whose sin is , The driver

32stopped the motor of the car . So , It moved down the ramp with a uniform velocity , Given

3 that the resistance of the ramp is equal to of the resistance of the horizontal road .

8Calculate the driving force of the motor ofthe car along the horizontal road measured in kg .wt

Answer

-------------------------------------------------------------------------------------------------------




From the figure : R 30 Sin 20 1

When the tension of the force reduced to 10 kg . wt ,

The body moved downwards and the 10 kg.wt

Became another resistance .

So , R 10 30 sin

o

R 30 Sin 10 2

Substitute 2 in 1 : 30 sin 10 3 sin 20

1 60 sin 30 Sin

2

30

Motion

Motion

Example (8) A body of mass 30 kg is placed on a plane inclined to the horizontal at an angle and is

pulled by a force of 20 kg.wt acting along the line of the greatest slope up the plane, it

moves uniformly up t

he plane against resistance of R kg.wt. When the tension is

reduced to 10 kg.wt, the body can move down the plane uniformly, find the measure of

the angle of inclination of the plane, given that the resistance of the plane doesn't change

in the two cases .

Answer

------------------------------------------------------------------------------------------------------- Example (9)

A body of weight 16 kg.wt is placed on a plane inclined to the horizontal at an angle of measure

30 ,the body is pulled by a rope of force F kg . wt upwards and this force inclined to the greatest

sl

ope at an angle 30 , If the body moves uniformly upwards the plane when the resistance of the

plane to the body is 4 kg . wt , Find F and the pressure of the body on the plane .

Answer

o o o

o

F cos 30 4 16 sin 30 F cos 30 12

12 F 8 3 kg . wt

cos 30

N F sin 30 16 Cos 60

1 3 N 8 3 16

2 2

N 8 3 4 3 4 3 kg . wt

-------------------------------------------------------------------------------------------------------

30 sin

N

R30cos

30

30 sin30

30cos

N

20

10 R

o16 sin30

N F cos 30

30

30

30 16

F

4

o16 cos30

F sin 30




2 2 2

2 22 2 2 o

2

Let F be the resultant of the two tensions

R T T 2 T T Cos " Static"

R 2T 2T Cos 2 200 2 200 Cos120

R 40000 R 200 gm . wt

2

2

1

1 1

2 2

11

R W R 30 gm.wt

R VAnd R V 1

R V

R 25 gm.wt when V 12cm / sec

30 V 30 12from 1 : V 14.4 cm/sec

25 12 25

30

Example (10 important) A body of mass 5 kg is placed on a horizontal plane and is attached to two horizontal ropes ,

The measure of the angle between them is 120 , When the two ropes are pulled by a force of

200 gm . wt ea

ch then the body moves on the plane uniformly , Find the magitude and the

direction of the force of resistance of the plane against the motion of the body .

Answer

------------------------------------------------------------------------------------------------------- Example (11)

A mettalic ball of weight 30 gm.wt is left to fall down in long vertical tube full of a viscial liquid.

If the resistance of the liquid to the motion of the ball varies directly as the magnitude of the

velocity of its motion in the tube , And given that the magnitude of the resistance of the liquid to

the ball equals 25 gm . wt , When the magnitude of the velocity of the ball equals 12 cm / sec ,

Calculate the magnitude of its velocity when it is uniform .

Answer

-------------------------------------------------------------------------------------------------------

N

FR

T

T5

R




1

22 1 1

2

2 2

2 2

2

1

2

When the train moves with a uniform velocity :

R 6250 kg . wt

R V R V 1

R V

R 25 45 1125 kg .wt when V 15 :

6250 Vfrom 1 : V 25 2 km / hr .

1125 15

33 1 1

3

2 2

1

331

3

Let the weight of the man and the parachute is W kg . wt

The man moves with uniform velocity

R V R W and R V 1

R V

8And R W when V 20 km / hr

125

W Vfrom 1 : V

8 20W125

125000 V 50 km / hr .

Example (12) A train of mass 45 tons moves on a straight horizontal road and the driving force of the engine

equals 6.25 tons . wt , Find the uniform velocity with which it moves , Gives that the resistance

to its motion is proportional to the square of its velocity and the resistance equals 25 kg . wt for

each ton of its mass when its velocity equals 15 km / hr .

Answer

------------------------------------------------------------------------------------------------------- Example (13)

An Aviator is tied to a parachute descends vertically downwards . Given that the air resistance

8is directly proportional to the cube of its velocity at any time , This resistance is equal to the

125

weight of the man and the parachute when its velocity is 20 km / hr . Find the maximum velocity the

man descends with . Answer

-------------------------------------------------------------------------------------------------------

R

45

R

W

6250




nd

Two rules are formed :

1 Vector form of the 2 law when the mass is Constant .

m a F Where a is the acceleration vector .

2 Algebraic f

nd

nd

orm of the 2 law when the mass is Constant .

ma F

: In our problems, Newton 2 law is the basic rule which will we use always

Note

F R ma Or F R ma

Rate of change of momentum with respect to the time is proportional to the impressed

force and takes place in the direction in which the force acts .

1 The sympolic form of the second law Equation of motion :

Suppose a force F acts on a body of mass m for a time t , and causes its velocity to change from u to v

This changing in velocities lead the appeare

nce of acceleration.

dThe second law states that : m v F ma K F

d t

And if the unit of the force magnitude is that which produces one unit of acceleration magnitude

When Acts on a body

with one unit vector of mass 1 1 K 1 K 1

-------------------------------------------------------------------------------------------------------

Very important note : The force vector F Or its algebraic measure F and the acceleration a

Must have the same direction

So from the opposite figure :

We can say :

-------------------------------------------------------------------------------------------------------

Newtons first law Newtons second law (1) Uniform velocity (Motion) Or

Maximum velocity exists without changing

Special case of Newtons 2nd law

(1) Uniform Acceleration Or changing the

uniform velocity due to force or resistance

The original rule which will be always used

(2) This means that a = 0 (2) This means that a 0

(3)

F R Or F R 0

(3)

OrF R ma F ma R

Newtons Second law

FR

a

Differences between Newtons first law and Newtons second law

FR FR




G W

M

2

2

2

a Newton kg.m /sec

It is the magnitude of the force which while acting on a body of mass one kilogram it

produces an acceleration of magnitude 1 m / sec .

b Dyne gm.cm / sec

It is the magnit

2

5

-5

ude of the force which while acting on a body of mass one gram it

produces an acceleration of magnitude 1 cm / sec .

Note 1 Newton 10 Dyne .

Dyne 10 Newton .

So

2

when applying our Rule : F ma

Dyne gm Cm / sec

F M a

Newton K

2g m / sec

BY

If a body of mass m is left to fall , It desends vertically by a uniform acceleration g because

the earth attracts it by a force W called weight .

We replace F ma w m g

If the mass of a body m 15Kg Its weight : W mg 15 9.8 147 N

If the mass of a body m 3 Ton Its weight : W mg 3 1000 9.8 29400 N

If the mass of a body m 200 gm Its weight : W mg

200 980 196000 Dyne.

w 117.6 The body whose weight 117.6 N Its mass : m 12 Kg

g 9.8

-------------------------------------------------------------------------------------------------------

------------------------------------------------------------------------------------------------------- Example (1)

-------------------------------------------------------------------------------------------------------

Fundamental units of forces

The relation between the weight of a body and its mass




6 6Ton . wt 1000 Kgm . wt 10 gm . wt 1000 9.8 N 980 10 Dyne

Note :

If the mass of a body X gm

9801000 Also , Kg gm dyne

1000 980

Its weight X gm.wt

If the mass of a body X Kgm Its weight X Kgm.wt

Proof :

7 9.8If m 7 Kgm W 7 9.8 Newton W 7Kgm . wt

9.8

1 Kgm .w t 9.8 Newton 1

Newton Kgm . wt9.8

2 gm .w t 980 Dyne 1

Dyne gm . wt980

53 Kgm .w t 9.8 10 Dyne 5

1Dyne Kgm.wt

9.8 10

-------------------------------------------------------------------------------------------------------

Example (2)

2

A body of mass 8.4 Kgm is moving in a straight line with uniform acceleration of magnitude

350 cm / sec . Find the magnitude of the force acting on that body .

First In Newton Second In Kgm .wt Third In dyne Answer

2

5

m 8.4 Kgm a 350 100 3.5 m / sec

F ma F 8.4 3.5 29.4 Newton

F 29.4 10 2940000 Dyne F 29.4 9.8 3 Kgm .wt

-------------------------------------------------------------------------------------------------------

From now on:

1 If the resistance is not mentioned in the problem, then F ma only

2 If the force is not mentioned in the problem, then -R ma only

3 If the engine is stopped, then F 0

-------------------------------------------------------------------------------------------------------

Summary of units

Very important note




22 2 2

5u 27 7.5 m / s s 9 m R ?? V 0

18-25

V u 2a s 0 7.5 2 9 a a m / sec8

And F R ma But there is no driving force mentioned :

-25 -6125 6125 1- R 245 Newton 9.8 R 9.8 78 kg.wt

8 8 8 8

Example (1)

A car of mass 245 kg moves with uniform velocity 27 km / hr , the brakes are used to stop the car

after covering a distance 9 meters , then find the magnitude of the force of the brakes in kg.wt

Answer

final

the uniform velocity changed when the brakes are used, then F 0 There is no driving force

and resistance R , acceleration appear retardation and V 0

------------------------------------------------------------------------------------------------------- Example (2)

7

2

A force of magnitude 4.2 10 dyne acts on a body in the state of rest to move it in a straight

line with uniform acceleration of magnitude 6 m / sec , find the magnitude of momentum of

this body after

1 a minute from the instant of start in kg .m / sec .2

Answer

7 5

from now on, if the resistance is not mentioned in the problem, then F ma only

F 4.2 10 10 420 Newton F ma 420 m 6

m 70 kgm

A

1nd u 0 And t 60 30 sec

2

V u at 0 6 30 180 m / sec

Magnitude of momentum mv 70 180 12600 kgm . m / sec

-------------------------------------------------------------------------------------------------------

We have three cases in this chapter

Case I : Motion under a single force

FR

a




4

2

2

2 2

So F R ma - R ma -5 10 1.6 1000 a

-125 a m / sec

4

5 25And u 45 m / sec , V 0

18 2

25 -125So v u 2a s 0 2 S S 2.5 m

2 4

R F

xa

final

When the brakes are used , then F 0 There is no driving force And resistance R appears

acceleration appear retardation and V 0

F

F F

F

When we are talking about rough planes use either resistance force F

the body is stopped after 2 seconds F 0

F R where R m g F m g

u 560 t 2 V 0

So, V u at 0 560 2a a -280 cm / sec

So F F ma

or

F- F ma - m g -280 m

-280 2

-980 7

FF F

a

m g

R

Example (3) A car of mass 1.6 tons moves along a horizontal straight road , And when it is moving with

velocity of magnitude 45 km / hr , The driving force of the car is stopped , And the brackes are

used so the ca

4

r stopped after a small interval of time , find the distance which the car travelled

in this interval if the total resistance to the car is constant and its magnitude 5 10 N .

Answer

------------------------------------------------------------------------------------------------------- Example (4)

A body is projected with velocity of 560 cm/sec on a rough horizontal plane so that it stopped

after 2 seconds from the begining of the motion, calculate the coefficient of friction between the

body and the plane.

Answer

-------------------------------------------------------------------------------------------------------




2

5u 63 17.5 m / s V 0 R 500 kg.wt 9.8 4900 Newton

18And F R ma But there is no driving force mentioned :

-7- R ma - 4900 7000a a m / sec

107

And V u at 0 17.5 t t 25 sec10

2

2 2 2

S 0.9 m V 0 Comes to rest u ??

And F R ma - R ma

- 2940 1200a a -2.45 m / sec

And V u 2a s 0 u 2 2.45 0.9 u 2.1m / sec

Retreats a distance here means that : R 300 kgm.wt 9.8 2940 Newton

nd

2

there is change in velocities acceleration appears use Newton's 2 law

1R 600 150 kg.wt R 150 9.8 1470 Newton

4

-49And - R ma -1470 600 a a m / sec

20

5u 864 240 m / sec t 40 sec

18

-49V u at V 240

2

40 V 142 m / sec.0

Example (5)

When a train was moving along a straight road with velocity 63 km / hr , its last wagon whose

mass is 7 ton is separated so it stopped due to a constant resistance of magnitude 500 kg.wt ,

find the time taken until it came to rest

Answer

let x be a unit vector in the direction of F F and a have the same direction

-------------------------------------------------------------------------------------------------------

Example (6)

When a cannon of mass 1.2 tons fires a projectile , it retreats on a horizontal ground a distance

of 90 cm , if the resistance of the ground to its motion is 300 kgm . wt , Then find the velocity

with which the cannon starts retreating.

Answer

-------------------------------------------------------------------------------------------------------

Example (7) A body of mass 600 kg moves horizontally in the space with uniform velocity of magnitude

864 km / hr . It intered a dusty cloud which acted upon by a resistance , Its magnitude equals

1 kgm . wt per each

4 kilogram of the body mass . Find the velocity of the body at the instant of

getting out of the cloud , If it remained 40 sec through it .

Answer

FR

a

FR

a




we have two kinds of bullet problems:

1 If the bullet penetrated the body with given thickness , the we will solve the problem from

the begining of the body .

2 If the bullet penetrated the body wi

nd

thout given thickness , the we will solve the problem far

from the body .

In our problem , we will use the 2 option

1 1u 40 , V 40 8 m / sec , t sec

5 1001

So V u at 8 40 a a -3200 m / se100

2

3

22 2

c

F R ma - R 35 10 -3200 R 112N

V u 2a s 0 8 2 -3200 S S 0.01m

u 40

v 81

t100

v 0

22 2 2

3

In this problem, thickness of wood is given , then we will solve the problem from the begining

of the wood u 100 , V 0 , S 0.16 m

So V u 2a S 0 100 2 0.16 a a -31250 m / sec

F R ma - R 32 10 -31

2

22 2 2

250

5000 R 1000 Newton 9.8 kg.wt

49When the target is of thickness 7 cm

u 100 , V ?? , S 0.07 m , a -31250 m / sec

V u 2a s V 100 2 -31250 0.07 V 75 m / s

u 100 v 0

S 0.16 m

Example (8) A bullet of mass 35 gm is fired horizontally with velocity of magnitude 40 meter / sec towards a

4fixed vertical target of wood ,it penetrated the target , and lost of the magnitude of its velocity

5

a1

fter sec , calculate the magnitude of the target resistance assuming that it is constant , 100

And find the distance which the bullet moves through the target before it comes to rest .

Answer

------------------------------------------------------------------------------------------------------- Example (9)

A bullet of mass 32 gm is fired horizontally with velocity of magnitude 100 meter / sec towards a

fixed target of wood to embed through it a distance 16 cm before coming to rest, find the magnitude

of the wood resistance in kg.wt assuming that it is constant , and if the same bullet is fired with the

same velocity at another fixed target made of the same wood of the first target and its thickness is

7 cm only, find the magnitude of the velocity with which the bullet gets out of this target.

Answer

-------------------------------------------------------------------------------------------------------




a

A B CAU 74 BV ?? CV 0

f .g wood

Example (10) A bullet of mass 49 gm is fired horizontally with velocity of magnitude 74 meter / sec at a body

which is at rest and formed of two adjacent layers , one of which is fiberglass andthe other is

wooden , if the bullet passes through the layer of fiberglass which is of thickness 7 cm , then it

is embedded in the wooden layer at a distance of 10 cm before it stops , if the resistance of the

wood is three times that of the fiberglass , then find the resistance of each in kgm.wt .

Answer

-------------------------------------------------------------------------------------------------------

0.07 0.1

A B

the problem mentioned the thickness which the bullet penetrated , then we will start our

problem from the begining of the wood

* Motion from A B Inside the fiberglass :

u 74 m / s V ?? S 0.0

22 2 2

B A B

1 1 1

22

B 1 1

2 1 1

7 m

V u 2a s V 74 2 0.07 a 1

49 -1000 - R ma - R a a R 2

1000 49

-1000 20V 74 2 0.07 R 5476 R 3

49 7

* Motion from B C Inside the wood :

49 -3000 - R ma' -3R a' a' R 4

1000 49

Also

B

B B C

2 2 2

C B B 1

1 1 1 1

1

V here is the initial velocity between B C

V u ?? V 0 S 0.1 m

20 V V 2a' s 0 V 2 0.1 a' 0 5476 R 0.2a'

7

20 -3000 20 600 0 5476 R 0.2 R 0 5476 R R

7 49 7 49

740R 5476 R

49

1

2

5476 49 1813Newton 9.8 37 kg.wt

740 5

R 3 37 111 kg.wt




2

2

2

B B

B

Body A moves with F 600 dyne

F ma 600 60a' a' 10 cm/sec

u 30 cm / s a' 10 cm/sec t 5 sec

1 1So S ut a' t S 30 5 + 10 25

2 2

S 275 cm

Then the distance between the two bodies after 5 seconds is 275 150 125 cm

Example (11)

A body of mass 12 kg is at rest , a force of magnitude 3 kg.wt acts on the body for 12 seconds ,

and after this time , the driver force stopped and the body moved with uniform velocity , then

find the distance the body covered after 30 seconds from the beginning .

Answer

------------------------------------------------------------------------------------------------------- Example (12)

A force of 600 Dynes acts on two bodies (A) and (B) placed at rest on a horizontal smooth

plane , the masses of them are 60 gm , 40 gm respectively and attached together by a string .

After 5 seconds from the action of the force , the body (B) is separated moves with a uniform

velocity and the body (A) only remains under the action of the force , Calculate the distance

between the two bodies 5 seconds after the instant of braking the string .

Answer

F

A B C

Au 0

F

a

F

Bu ??

AB AB

Au ??S

t 12sec t 18sec

u 0

uniform motion

No acceleration

2

2

* When the two bodies are connected :

F ma 600 60 40 a

600100 a 600 a 6 cm / sec

100

Thus the bodies move with acceleration 6 cm / sec

At t 5 sec V u a t 6 5 30 cm/sec

After 5 seconds the string cut and t

B

he two bodies separated when their velocities u 30cm / s :

Body B move without force F a 0

Then B moves with uniform motion :

S vt 30 5 150 cm

st

If the resistance is not mentioned in the problem, then F ma only

* For the 1 12 seconds " during the action force "

Motion from A B

t 12sec u 0 F 3 kg.wt 9.8 29.4 Newton

And F R ma F ma

2

22

B A B

29.4 29.4 12a a 2.45 m/sec

121 1

So S u t at S 2.45 12 176.4 m2 2

So V u at V 2.45 12 29.4 m / sec

Motion from B C : When the driver force stopped , it means that F 0 and there i

s no

acceleration body moves with uniform velocity for the next 18 seconds

S V t 29.4 18 529.2 m the total distance after 30 seconds 176.4 529.2 705.6 m




If the engine force stopped steam shutt off .

If the resistance are neglected .

If the body moved with uniform velocity or maximum V .

F

F Cos

F SinN

R

m g

u

F Sin

F Cos

m g

N

F

x

m g

Ground surface

R

x

Rm g

F

x

x

F

R

m g

mg R F ma

R Opposite to direction of motion m g Always Downwards

F According the given direction

Inside

the ground

acceleration appears

inside the ground

acceleration appears in the

air when there is another

force act on the body

Equations of motion of famous applications :

Application Drawing Equation

Motion on a horizontal plane

under a horizontal force F

and a and resistance R

F R m a


under a force F inclined at an

angle With the horizontal and

a resistance R .

F cos R ma

F sin N m g 0


under a force F inclined at an

angle with the vertical

F sin ma

N F cos m g 0

Motion vertically downward

through a Ground under

the weight mg and the earth

resistance R

No driver force except

Its weight

m g R m a

Motion vertically upwards

under a driving force F

and weight

F R m g m a

Motion vertically downwards

under the weight mg and a

resistance R and a lifting

force F

Very important Notes

F 0

R 0

a 0

Case 2 : Motion under more than two forces

R Fx




2

Let u be a unit vector in the direction of motion .

u 0 V 39.2 m / sec t 2 60 120 sec.

49V U at 39.2 0 120a a m / sec

150

Equation of motion of the car is resul

tant of the forces acting on the car ma

49F R ma F R ma F 50 9.8 3 1000

150

F 490 980 1470 Newton 9.8 150 kgm . wt

R

x

2

1260 2520 0 30V 2520 m / min V 42 m / sec and t 15 sec

0.5 60 2

42 V u at Then : 42 15a a 2.8 m / sec

15

And the equation of motion of train is : F R ma

R F ma R 3000 9.8 5.6 1000 2.8 13720 Newton 9.8 1400 kg.

wt

Magnitude of the resistance per ton 1400 5.6 250 Kg .wt .

x

R F

a

a

Example (13) A car of mass 3tons starting from rest along a horizontal straight road with uniform

acceleration and against constant resistance of magnitude 50 kg . wt , If the magnitude

of the car velocity is 39.2 m / sec after 2 minutes from the starting, calculate the magnitude

of the driving force of the car in kg . wt

Answer

------------------------------------------------------------------------------------------------------- Example (14)

A train of mass 5.6 ton moves from rest with a uniform acceleration along a horizontal straight

1road, it travelled a distance 1260 meter during a minute from the starting, find the magnitude

2

of the resistance per each ton of its mass if the magnitude of the driving force 3000kgm . wt .

Answer

-------------------------------------------------------------------------------------------------------




F

o

F

o

F

When we are talking about rough planes use resistance force F

the body is stopped after 2 seconds F 0

F R where R 360 980 420 980 Sin30 147000 dyne

3F F ma 420 980Cos30 147000 360 a

3

a 75

or

2

F

4.4 cm / sec Then u 0 a 754.4 t 9

So, V u at V 754.4 9 6789.64 cm / sec

When the action of the force stopped:

Then the body is moving under the action of resistance F only

R 360 980 352800 dyne and

2

F

3

3

3So, - F ma' - R ma' - 352800 360 a' a' -565.8 cm/sec

3 Then u 6789.64 a' -565.8 V 0

So, V u at 0 6789.64 565.8 t t 12 sec

FF

420

360 g

R

F

F

F

2

When we are talking about rough planes use either resistance force F

F R where R 3.5 g 3.5 9.8 34.3 Newton

And F F ma 3.15 9.8 0.3 34.3 3.5a

a 5.88 m / sec

u 0 a 5.88 t 5

So

or

, V u at V 5.88 5 29.4 m / s

FF

F 3.15 9.8

a

3.5 g

R

o420 Cos30

o420 Sin30

o30

FF

360 g

R

o420 Sin30

Example (15)

A piece of iron of mass 3.5 kg starts motion from rest on a rough horizontal plane when a

horizontal force of magnitude 3.15 kg.wt acts on it, if the coefficient of friction between the

body and the plane equals 0.3, find the velocity of the piece of iron after 5 seconds from the

begining of motion.

Answer

------------------------------------------------------------------------------------------------------- Example (16)

A body of mass 360 gm is placed on a rough horizontal plane whose coefficient of friction with

3the body equals . A force of magnitude 420 gm.wt acted on the body in a direction inclined to

3

the horizo ontal at an angle of measure 30 , so that it moved from rest with a uniform acceleration.

Find the velocity of the body after 9 seconds from the begining of motion, and if the action of the

force is stopped after that, determine the time that the body makes until it comes to rest.

Answer




The train moved from O to A under the driving force F and the resistance R and from

A to B only under R .

If U is a unit vector in the direction of motion .

Equation of the train d

2

a

uring OA is : F R ma

6750 9.8 25 9.8 30 30 1000a

9.8 6750 750 3000a

9.8 6000 30000a a 1.96 m / sec

Magnitude of the velocity at A : V U at 0 1.96 5 60 588 m / sec.

Equation of motion during AB

2

2 2

B A

:

when engine stopped and u 588 changed to V 0 F 0 and acceleration must change

-R ma' , where a' is the algebraic measure of the acceleration during AB

-25 30 9.8 30 1000 a' a' -0.245 m / sec

And V V 2

2

2

a s 0 588 2 0.245S

0.49 S 588 S 705600 m .

u

R FO B

R

a'a

A

1 1 1U niform velocity means a 0 F R F R w m g 0.5 9.8 0.98 N

5 5 5

0.98 1F R kg.wt 1000 100 gm.wt, And when force increased with 15 gm.wt :

9.8 10

Then F R ma 115 980 100 980 0.5 1000a

2 14700 500a a 29.4 cm/sec

Example (17) A train of mass 30 ton started from rest and moved with uniform acceleration along a straight

horizontal road under the action of a driving force of magnitude 6750 kg.wt and a resistance

its magnitude is 25 kg.wt per each ton of its mass, and after 5 minutes of the start, the driving

force is stopped and the resistance still constant , find the distance which the train moved after

that instant before it comes to rest .

Answer

------------------------------------------------------------------------------------------------------- Example(18)

1A car of mass kg moves along a horizontal st. road with uniform velocity and the magnitude of

2

1the resistance to its motion equals of its weight, find magnitude of the driving force of the car mot

5or

in gm.wt. And if the force increases with magnitude 15 gm.wt , find magnitude of this acceleration .

Answer




Before separation : The train is moving with uniform velocity this means that :

F R 5 9.8 160 7840 Netwon driving force for all train .

After separation : The wagon moves

with no force Resistance only and the train .

goes force Resistance

The initial veocity of the locomotive and the wagon at the instant of sparation was u 45 km / h

So , Aft

er separation regard to wagon :

Its initial velocity directly after separation was:

5 u 45 12.5 m / s

18

offcourse its acceleration is decreasing

So F R ma - R ma

-5 9.8 16 16 1000a

2

2

22

1 1 1

a - 0.049 m/s

u 12.5 m/sec a -0.049 m/s t 2 60 120 sec

So , lets find the distance of the wagon after 2 minutes :

1 1S ut at S 12.5 120 -0.049 120 S 1147.2 meters

2 2

After separatio

n regard to the locomotive train :

Hence there is a driving force still and resistance , the train has initial velocity

u 12.5 m/s after separation directly , off course , Its acceleration will increas

2

22

2

e

as it becomes lighter F R ma' 7840 5 9.8 160 16 160 16 1000 a'

784 49 784 144000 a' a' m / sec

144000 9000

49 1 49u 12.5m/s a' m / s t 120 sec S 12.5 120 120 1539.2 m

9000 2 9000

Then the di

2 1

stance between them after 2 min. covered by each of them :

S S 1539.2 1147.2 392 meters

R F

FR R

a

1S2S

u 12.5 m/secu 12.5 m/sec

Example (19) A locomotive pulls a train moves horizontally in a straight road with a uniform velocity

45 km / h, the mass of the locomotive and the train together is 160 ton and the assistance force

is 5 kg.wt for each ton of the mass ,calculate the driving force of the locomotive. If the last

wagon is separated from the train, given that the mass of the wagon is 16 ton , find the

distance between it and the rest of the train 2 minutes after the instant of separating .

Answer




Note : the acceleration appears in the air beside the gravity as the helicopter changes its velocity

F R m g ma

F 750 9.8 2 1000 9.8 2 1000 4.9

F driving force 36750 Newton 9.8 3750 kg.wt

If the h

elicopter rises with uniform velocity :

F R m g 0 F R m g

F 750 9.8 2 1000 9.8 26950 Newton 2750 kg.wt

The train is uniform F R

The driving force F 10 9.8 60

F 5880 Newton is the driving force of the train

after sepration with respect to the cap

- R ma -10 9.8 1 1 1000a a -0.0

298m/s

5u 44.1 12.25 m/s a -0.098 V 0

18

V u at 0 12.25 0.098t t 125 seconds

After seperation with respect to the train : Its acceleration increased

F R ma' 58

` 2

2

4980 10 9.8 60 1 60 1 1000 a' a' m/s

29500

49When u 12.25 m/s & a' m/s t 125 sec

29500

49 27 V u a' t 12.25 125 12 m/s

29500 59

FR R

a'

u 12.25 m/secu 12.25 m/sec

FR

R

m g

a

F

Example (20) A train of mass 60 tons moves with a uniform velocity 44.1 km / h against a resistance of

magnitude 10 kg.wt . for each ton of the mass of the train . If the last cab is separated from

the train , given that the mass of the cab is one ton , find the time taken by the separated

cab till it comes to rest and the velocity of the train at this moment .

Answer

------------------------------------------------------------------------------------------------------- Example (21)

2A helicopter of mass 2 tons rises vertically with acceleration of magnitude 4.9 m / s against

resistances of air of magnitude 750 kg.wt, find the driving force of the engine of the helicopter,

also find the force of the engine if the helicopter rises with a uniform velocity assuming that the

resistance does not change .

Answer




2 2 2

B B

B

2

2 2

c B

u 0 S 22.5 m

V u +2 g s V 2 9.8 22.5 441

V 21 m/s

From B C "Inside sand acceleration appears"

m g R m a 0.5 9.8 372.4 0.5 a

a -735 m/s

V V

2a s 0 441 2 735 S

S 0.3m 30 cm

22.5

B

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Mechanics - Dynamics