Mechanics - Dynamics

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rd _-₁

Equations of motion “

uniform acceleration”

Dynamics

Dicplacement Distance

Revision

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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

ظ ف ح ي

60 60 sec 18 1000 Or we can say : 1 km / hr     

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2 1 100cm 250 cm / sec

ظ ف ح ي

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 2

at t to its terminal position B at t .

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      2 2 2

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

2 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

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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

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       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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Very Important Remarks : Cases of Displacements S

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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

 

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ression : Left to fall or fell means U 0 . 4 The expression : The body is projected by Veloctiy... " This is the Init

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ial velocity : "U"

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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    

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2 2 2 g t 2

" 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

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the final velocity v of the body and the distances traveled by the body"

Second : Vertical Motion Upwards

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      2 2 2 1 V u g t 1 2 S u t g t 2 3 V u 2g s ---

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01009988836 – 01009988826 Email : [email protected] rd _-₃ -max S t_max

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   _max  2

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

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u0

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 Fall from rest initial velocity = zero

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u0

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 Comes to rest terminal velocity = zero .

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V 0

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 Projected upwards Gravity will be negative

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2

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g -9.8 m/sec

 Uniform velocity acceleration = zero And S = V t  Maximum velocity acceleration = zero

 Maximum height velocity = zero

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V 0

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 Mass of drop (each ) drop when it reaches the ground

 Retarded acceleration = -ve

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Using the algebraic measure of vectors

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   0 S r r d S d r 1 d t d t

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d S 2 V d t 

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dV 3 A d t  ---

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



time)

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th

Very important remarks

If S is a distance covered during n second

Distance during a given interval of time

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     th th th th s for example : Ex 1 : The distance covered in the 5 seconds :

4 5

Here 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

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             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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- -  2

o

Given : 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

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                            2 2 2 2 2 2 2 moving on a bridge" 1 1 i S u t a t S 10 8 2.5 8 160 cm 2 2 1 5 ii 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

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       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 

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      A BA BA - V V 43 x 82 x V 125 x Its Magnitude is 125 km / hr . A V 82 B V 43 A _B _C A B x A V B V 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

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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

--- Example (2)

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2

A 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

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y after 8 seconds

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

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            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

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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

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     

    

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 2 9.8 m / sec Example (3)

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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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rd _-₇

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 .

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

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

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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

1

So we must find v : u 0 a 9 cm / sec t 60 30 sec . 2 v u a t v 9 30 270 c            m / sec . H mv 7.5 270 2025 gm . cm / sec .     

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For example, a heavy truck moving fast has a large momentum—it 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)

2

A body of mass 7.5 gm , Moves from rest in a straight line with acceleration 9 cm / sec in the 1

direction of its motion , Find its momentum after minutes from the begining of the motion . 2

Answer

Newton’s laws of motion

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-Example (2)

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

Answer

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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 .               

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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

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2 1 1 2 2 1 1 1 2 2 2 5 v 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 .

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Example (4) 1

A 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

1 Each 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 m v 1000 200 100 2 10 gm .cm / sec .          

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2 2 2 1 1 2 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

            

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2 2 2 2 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 m 1 v v 0 4.9 m u??

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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  2 v 1 v

A rubber ball of mass 300 gm moves horizontally with uniform velocity 135 cm / sec . It collides 4

by 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

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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

         

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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

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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

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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 .

2 v 1 v x 1 m 120 gm 2 m 3120 gm

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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       

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T PT PT P T 5 350 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 5 0 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 T v P v P v T v

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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 3120

Thus 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 ]

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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

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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 .

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01009988836 – 01009988826 Email : [email protected] rd _-₁₁ -5.4 m u 0 t 1.5 sec x a -2.7

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     

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

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              1 2 2 2 2 2

ocity directly before collision v 14.7 x

Second Magnitude of swimmer velocity directly after collision with water : v u 2 a s 0 u 2 2.7 5.4 u 5.4

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         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   

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40 5.4 x 14.7 x 372 x Its magnitude 372 kgm . m / sec .

V 0 V ?? u ??

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

7 100

Then 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

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   .

100

To get the velocity :- v u g t 9.8 140 m / sec . 7

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

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   

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

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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 with uniform acceleration of magnitude 2.7 m / sec and covered a distance is2 5.4 meter before he starts the ascending , Calculate magnitude of his change of momentum due to collision with water .

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-Example (11)

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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 .

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180 u km / sec u 0.05 km / sec . 3600 v u g t v 0.05 9.8 t H m v 25 0.001 t 0.05 9.8 t x            

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2 d H -9.8 25 0.001t 0.001 0.05 9.8 t 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 .

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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

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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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rd _-₁₃

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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



   N F R W F sin



F cos  

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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

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

      F W



W cos

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

F cos

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F Sin

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N Motion

Case (1)

Case (2)

Case (4)

Case (3)

R W Sin

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F R W N Motion Motion

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rd _-₁₄

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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 3 n

It moves

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              

st

with a uniform velocity : So we have to use Newton's 1 law

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

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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

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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

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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

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          1 1 2 2 2 2 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

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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

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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 .

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rd _-₁₅

The body moves in a uniform velocity : F R

3 4 3 1350 cos W 1350 W W 1440 kg . wt 4 5 4 3 Also , N 1350 sin W N 1350 1440 5 N 630 kg.wt



                 N W 1350 Sin



1350 cos   F 1350 3 w 4  4 3 5 Sin 3 5   4 Cos 5  





R 4 3 12 kg . wt

The component of W in the plane direction downwards

1 1 1

is 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 3

inclined at an angle of Sin to the horizontal, so the body moved with a uniform motion 5

against the road resistance which is equal to 3 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 1

at 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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rd _-₁₆

-



The engine stopped while moving downwards means that the car is moving by its weight only No force exists

1 1 R 5 Sin 5 ton.wt 100 20 1 R 1000 50 kg.wt 20 50

So , the resistance per each ton is 10 kg. 5



         wt ₅



5 Cos  R N 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    



4500 Cos Motion 4500 Sin



Motion 4500 N R R90 F 4500 4500 Sin _{4500 Cos}_ N Example (5)

The engine of a car of weight 5 ton.wt is stopped while it is moving downwards a road 1

inclined 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 1

at 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 .

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rd _-₁₇

-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

1 So , R 2500 sin 2500 78.125 kg . wt 32 3 And R R 8



      

 

1 1 1 8 1 R 78.125 208 kg . wt 3 3 1 From 1 F R 208 kg . wt 3      2500 sin



Motion Motion N N W 2500 R 2500 cos



2500 Ramp 1 F 1 R



 

2 1 1 1 3 Given : 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 1

when it reached a ramp inclined to the horizontal at an angle whose sin is , The driver 32

stopped 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 . 8

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

Answer

(18)

rd _-₁₈

- 

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 R 30 cos 30 30 sin 30 30 cos N 20 10 R o 16 sin30 N _{F cos 30}_ 30 30 30 16 F 4 o 16 cos 30 F sin 30

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- - 









2 2 2 2 2 2 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 1 1 R W R 30 gm.wt R V And R V 1 R V R 25 gm.wt when V 12 cm / sec 30 V 30 12 from 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 F R T T 5 R

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- 

 

1 2 2 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 V from 1 : V 25 2 km / hr . 1125 ₁₅



               

 

3 3 1 1 3 2 2 1 3 3 1 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 8 And R W when V 20 km / hr 125 W V from 1 : V 8 ₂₀ W 125



             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 8 is 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

(21)

rd _-₂₁

- 

 

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

orm of the 2 law when the mass is Constant . m a 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. d

The second law states that : m v F m a 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 :

---

Newton’s first law Newton’s second law

(1) Uniform velocity (Motion) Or

Maximum velocity exists without changing “Special case of Newton’s 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)

Or

F

 

R

ma

F



ma



R

Newton’s Second law

F R

a

Differences between Newton’s first law and Newton’s second law

F

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01009988836 – 01009988826 Email : [email protected] rd _-₂₂ -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 m a

Dyne gm Cm / sec F M a Newton K          2 g 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



m a



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

(23)

rd _-₂₃

-6 6

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

Note :

If the mass of a body X gm

980 1000

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 9.8 If m 7 Kgm W 7 9.8 Newton W 7 Kgm . wt 9.8               

 

1 Kgm .w t 9.8 Newton Newton 1 Kgm . wt 9.8 

 

2 gm .w t 980 Dyne Dyne 1 gm . wt 980 

 

5 3 Kgm .w t  9.8 10 Dyne Dyne 1 ₅ 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 m a 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 m a only 2 If the force is not mentioned in the problem, then -R m a only 3 If the engine is stopped, then F 0

  

---

Summary of units

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- 

 









2 2 2 2 5 u 27 7.5 m / s s 9 m R ?? V 0 18 -25 V u 2 a s 0 7.5 2 9 a a m / sec 8

And F R m a 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 m a only F 4.2 10 10 420 Newton F m a 420 m 6 m 70 kgm A              1 nd u 0 And t 60 30 sec 2 V u a t 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

F R

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01009988836 – 01009988826 Email : [email protected] rd _-₂₅ -4 2 2 2 2 So F R m a - R m a -5 10 1.6 1000 a -125 a m / sec 4 5 25 And u 45 m / sec , V 0 18 2 25 -125 So v u 2 a s 0 2 S S 2.5 m 2 4                    _ _        R F x a





 





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 a t 0 560 2 a a - 280 cm / sec So F F m a



                     or F - F m a -



m   

 

g -280 m -280 2 -980 7



    F F 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

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-



2 5 u 63 17.5 m / s V 0 R 500 kg.wt 9.8 4900 Newton 18

And F R m a But there is no driving force mentioned : -7 - R m a - 4900 7000 a a m / sec 10 7 And V u a t 0 17.5 t t 25 sec 10                       







 

2 2 2 2 S 0.9 m V 0 Comes to rest u ?? And F R m a - R m a - 2940 1200 a 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 1 R 600 150 kg.wt R 150 9.8 1470 Newton 4 - 49 And - R m a -1470 600 a a m / sec 20 5 u 864 240 m / sec t 40 sec 18 - 49 V u a t 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 F R a F R a