ch4 forces mh

7/31/2019 Ch4 Forces MH

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4.1.1 Basic of forces

4.1.2 Free body diagram

4.1.3 forces acting on a body

i) Weight

ii) Tension

iii) Normal forceiv) Friction

4.1.4 Static friction and kinetic friction

4.2.1 Newtons First Law and inertia4.2.2 The equilibrium of a particle4.2.3 Newtons First Law in equilibrium

of forces

4.2.4 Newtons Second Law4.2.5 Newtons Third Law


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i) Weight

ii) Tension



4.2.1 Newtons First Law and inertia4.2.2 the equilibrium of a particle4.2.3 Newtons First Law in equilibrium

of forces

4.2.4 Newtons Second Law4.2.5 Newtons Third Law


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Remarks


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5


Force is defined as something capable of changingstate of motion or size or dimension of a body.

Figure 4.1

Figure 4.2

There are fourtypes offundamental forces innature:

(refer to figures 4.1

and 4.2)

The forces involveattraction between

massive body. is a long-range forces.

the weakest forces innature.


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6

(refer to figures 4.3) The attractive and

repulsive forcesbetween electriccharges.

is a long-range forces.

(refer to figures 4.4)

The attractive forces

bonding neutron andproton in atomicnucleus.

is a short-range forces

and the strongestforces in nature.

Figure 4.3

Figure 4.4


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(refer to figures 4.5)

cause the unstablecondition for atomicnucleus and is responsiblefor the radioactive decay.

is a short-range forces and12 times weak comparewith electromagneticforces.

7

Force- Is a vector quantity.

- The S.I. unit of force, Fis or


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- A graphical tool which is a

- Is extremely useful analyzingforces and motion

- Drawn only on an object

- Picture that show the size and

direction of all forces acting on

an object in a given situation

is defined as

.


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Basic Forces are:


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Steps to drawing a free body diagram

M

Which one would you like to select todraw FBD?

What do you think are the forces actingon this object?Gravitational forcegMFG

The force supporting the

object exerted by the floor

Me

Which one would you like to select todraw FBD?

What do you think are the forces actingon this elevator?

N

gMFG

Gravitational force The force pulling theelevator (Tension)m

What about the box in the elevator?

Gravitational

force

Normal force

N

T

gMFG gmFGB

T

gMFG

N

gmFBG

N


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N

f Fapplied

W

W = mgN = W

f = Fapplied


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12

- is defined as

- It is a .

- It is .

- because the .

- It always

or in the

.

- The S.I. unit is or .


W

W mg


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13Weight

Weight

Weight


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14

- Is a force which is transmitted through a string,rope, cable or wire when it is pulled tight by

forces acting from opposite ends.

- The tension force is directed along the length of

the wire and pulls equally on the objects on theopposite ends of the wire.

- It is a .

- The S.I. unit is or


T


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Rope attached to a wall

PullTension Tension

PullTension Tension

Pulled object


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16

- is defined as

.

- It is a .

- The S.I. unit is or


N


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Horizontal surface

N

gmW

Weight is exerted on the

horizontal surface

Surface is exerted aforce, N on the object

Inclined plane

N

xW

xy

Weight is exerted on the

horizontal surface

: y-component of the

objects weight is exertedon the inclined surface.

yW


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- Is defined as

.

- The exerted by a

surface on an object is always in the

.

- The are always.

- is independent of the area of contact between the

two surfaces.

- is directly .

f


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A box of mass mis pulled along a horizontalsurface by a horizontal force, F

F

a

gm

N

f

horizontal force


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Inclined plane

Consider a box of mass mis pulled up along aninclined plane by a force,F

aN

gmW

xy

yW

x

W

F

f

Consider a box of mass m

is pulled down along aninclined plane by a force,

F

a

N

gmW

xy

yW

x

W

F

f


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Is defined as the ratio between frictional

force to normal force

OR

Depends on the nature of the surfaces. Without unit

N

f

Nf where

-

- caused by irregularities in surfaces.

- Always acts against the direction of

motion.


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Static Frictionno motion (v = 0 m s-1)

Static Friction,fs frictional force act on theobject

Fapplied = fs and

If static equilibrium,

W

N

Fapplied

fs

Nf ss


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Static Frictionno motion

W

N

Fapplied

fs

If increaseFapplied

,fs

also increases, up tofs(max)

If increasem,fs(max) also increases

On horizontal surface,N = mg

fs(max) m

fs(max) N


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Static Frictionno motion

W

N

Fapplied

fs

fs(max) is proportional to N,f

s(max)N

= constant = s

s = coefficient of static friction, fs

sN


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W

N

Fapplied

fk

Kinetic Friction,fk frictional force act on theobject

Kinetic Frictionmoving


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W

N

Fapplied

fk

fk is proportional to N, fkN

= constant = k

k = coefficient of kinetic friction, fk

= k

N



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W

N

Fapplied

fk


IfFapplied= fk , then speed is constant ( a = 0 m s-2)

kinetic friction < static friction

IfFapplied fk , then speed changes ( a 0 m s-2)


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29

k sk sf f

Static Friction Kinetic Friction

Action

Bigger SmallerMagnitude

Equation

Bigger Smaller

on the object on the object

Nf ss k kf N


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Frictional Forces


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Frictional Forces


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Types of Friction

I better besafe Ump!!

To initiate motion of the boxthe man must overcome theForce ofStatic Friction

Upon sliding, the baseball

player will come to acomplete stop due to theForce ofKinetic Friction


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Static & Kinetic Friction Coefficients


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Static Friction VS Kinetic Friction


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x-component :

y-component :

35

Consider a box of mass mis pulled along a

horizontal surface by a horizontal force,F asshown in figure.

maFF nettx

F

a

gm

N

f mafF

0yF

mgN


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Consider a box of mass mis pulled along aninclined plane by a force, Fas shown in figure.

x-component

(parallel to the

inclined plane):

y-component

(perpendicular to the

inclined plane):

a

N

gmW

xy

yW

xW

F

f

0yF

0yWN

mgN cos

maFx

mafWF x

fmgmaF sin


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37

A box of mass 20 kg is on a rough horizontal plane.

The box is pulled by a force, Fwhich is applied at an

angle of 30 above horizontal as shown in figure. If

the coefficient of static friction between the box and

the plane is 0.3 and the box moves at a constantspeed, calculate

(a) the normal reaction force,

(b) the applied force F,

(c) the static friction force. (Given g = 9.81 m s-2)

30

F


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a) Since the box moves at constant speed thus

x-component :

0.3kg;20 sm

30

F

constant speed

N

gm

s

f 30cosF

30sinF

0x

F

030cos sfF

0a

30cos

0.3NF

030cos NF s

(1)


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y-component :

By substituting eq. (1) into eq. (2), hence

b) Therefore the applied force is given by

c) The static friction force is

0yF

030sin mgFN

N167N

(2)

9.812030sin FN

N57.930cos

1670.3

F

19630sinFN

19630sin30cos

0.3

NN

Nfss N50.11670.3


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(a) the normal force,

(b) the nett force,

(c) the acceleration of

the block,

(d) the time taken for

the block to travel

30 m from rest.

(Given g = 9.81 m s-2)

F

20

A block of mass 200 kg is

pulled along an inclinedplane of 30 by a force, F

= 2 kN as shown in

figure. The coefficient of

kinetic friction of the

plane is 0.4. Determine:


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41

30

0.4N;2000kg;200 kFm

0yF030cos20sin mgFN

N1015N

F

x

y

20cosF

N

20sinF20

gm

kf 30cosmg

30sinmg

a

030cos9.8120020sin2000 N

a) y-component :


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42

b) The nett force is directed along the inclined

plane surface.

x-component :

c)

d) Given

xnett FF

knett fmgFF 30sin20cos

NmgFF knett 30sin20cos

N492nettF

10150.430sin9.8120020cos2000 nettF

maFnett

a200492 2sm2.46a0m;30 us

2

2

1atuts 22.46

2

130 t0

s4.94t


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A 5.00 kg object placed on a

frictionless, horizontal table isconnected to a string that

passes over a pulley and then

is fastened to a hanging 9.00

kg object as in the figure.

(a) Sketch free body diagrams

of both objects,

(b) Calculate the acceleration

of the two objects and thetension in the string.

(Given g= 9.81 m s 2)


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Two object are connected by a light string that

passes over a frictionless pulley as in figure.

The coefficient of kinetic friction of the plane is

0.3 and m1= 2.00 kg, m2= 6.00 kg and = 55 .

a) Sketch free body diagrams of both objects.

b) Determine

(i) the accelerations of

the objects,

(ii) the tension in the string(iii) the speed of each object

2.00 s after being released

from rest. (g= 9.81 m s 2)


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45

A 5.00 g bullet is fired horizontally into a 1.20 kg

wooden block resting on a horizontal surface. Thecoefficient of kinetic friction between block and

surface is 0.20. The bullet remains embedded in

the block, which is observed to slide 0.230 m

along the surface before stopping. Calculate theinitial speed of the bullet. (Given g= 9.81 m s 2)

Tips : Use

Newtons second law of motion involving

acceleration. Principle of conservation of linear

momentum.

Equation of motion for linear motion.


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46

The block shown in figure, has mass, m=7.0 kg

and lies on a smooth frictionless plane tilted at anangle, = 22.0 to the horizontal.

a) Determine the acceleration of the block as it

slides down the plane.

b) If the block starts from

rest 12.0 m up the plane

from its base,

calculate the blocks

speed when it reaches

the bottom of the incline

plane. (g= 9.81 m s 2)


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47

A block is dragged by forces, F1and F2of the

magnitude 20 N and 30 N respectively as shownin figure 3.12. The frictional force fexerted on the

block is 5 N. If the weight of the block is 200 N

and it is move horizontally, determine the

acceleration of the block.(Given g= 9.81 m s 2)

50

a

1F

2F

f

20


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4.1.2 Free body diagram4.1.3 forces acting on a body

i) Weight

ii) Tension



4.2.1 Newtons First Law and inertia4.2.2 The equilibrium of a particle4.2.3 Newtons First Law in equilibrium

of forces

4.2.4 Newtons Second Law

4.2.5 Newtons Third Law


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50

Newtons First Law of Motion states

OR

Newtons first law is oftencalled the law of inertia


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51

- is defined as

- is a scalar quantity.

The examples of real experience of inertiaThe first law gives the idea of inertia


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52

is defined as

is a .

The S.I. unit of mass is .

The value of mass is .

If the mass of a body increases then its inertia

will increase.


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4.2.2 The equilibrium of a particle

is defined as the

. (whether

inside or outside of the

body).

The forces cause the

on

the body.

Figure 4.6 and Figure 4.7

show the examples of

concurrent forces.

2F

3F

1F

2F

3F

1

F

53


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The Equilibrium of a particle

is defined as the.

Newtons first lawof motion

The equilibrium of a

particle ensures thebody inand its

is given by

54


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This is equivalent to thethree independent

scalar equations along

the direction of the

coordinate axes,

0

,0

,0

z

y

x

F

F

F

There are two types ofequilibrium of a particle.

It is equilibrium

body remains at.

equilibriumbody moving at a

.

(v= 0)

(a= 0 )

55


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Polygon of forces

1F

2F

A particle in equilibrium

as a result of two forces

acting on it as shown inFigure.

They are equal in

magnitude but opposite

in the direction, thus

021 FFF

0xF

OR

0yF

56


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A particle in equilibrium

as a result of three forces

acting on it as shown in

Figure.2F

1F

3F

They are form a closed

triangle of forces, thus

2F

1F

3F

0321 FFFF

i.e. 0xF and 0yF

57


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A particle in equilibriumas a result of four

forces acting on it as

shown in Figure.

They will form a closed

polygon of forces, thus 1F

2F

3F

4F

1F

2F

3F

4F

04321

FFFFF

i.e. 0xF and0yF

58


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4.2.3 Newtons First Law in equilibrium

of forcesEquilibrium:The state of a body in which there is nochange in motion. At rest orwith constant velocity.

The net external force acting on abody in equilibrium must be equal tozero.

The force that brings an acceleratingobject into equilibrium must be equaland opposite to the force causing theobject to accelerate.

Fx = 0 and Fy = 0.


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Body is in when it is at or

moving with in anframe of reference

o Hanging lamp

o Suspension bridge

o Airplane flying at constant speed

o Other examples?


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61

F

gm

T

yT30

60

xT

Solution : kg250m

30

F

Free body diagram of

the load :

A load of 250 kg is hung

by a cranes cable. Theload is pulled by a

horizontal force such that

the cable makes a 30

angle to the verticalplane. If the load is in the

equilibrium, calculate

a) the magnitude of the

tension in the cable,b) the magnitude of the

horizontal force.

(Given g=9.81 m s 2)


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a)

Force x-comp

(N)

y-comp (N)

mg

F

T

0

F

60cosT

250 9.81

mg

2453

0

60sinT

Since the load is in the

equilibrium, then

0F

0xF

060cos

TF

N2833T

(1)

(2)

0yF

0245360sin T

Thus

b) By substituting eq. (2) into eq. (1), therefore

060cos2833 F N1417F


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63

30

a) Since the load is in the equilibrium, then a

closed triangle of forces can be sketched as

shown below.

b) sin30F

T

(250)(9.81)cos30

T

N2833T

cos30

mg

T

(2833)sin30F

N1417F

F

gm

T

From the closed triangle of

forces, hence

sin30F T


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64

Calculate the magnitude and direction of a

force that balance the three forces acted atpoint A as shown in Figure above.

Example 2 :

N121FN202F

N303F

30.055.0

45.0A


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Solution :

To find a force to balance the three forcesmeans the system must be in equilibrium

hence

N30N;20N;12 321 FFF

Force x-component (N) y-component (N)

1F 55.0cos12

F

xF yF

6.88

55.0sin129.83

2F 30.0cos20

17.3

30.0sin20

10.0

3F 45.0cos30

21.2

45.0sin3021.2

0xF

021.217.36.88 xF

N31.6x

F


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Solution :

The magnitude of the force,

and its direction,

0yF

021.210.09.83 yFN1.37yF

222

y

2

x FFF 1.3731.6

N31.6F

x

y1

F

F tan

31.6

1.37tan

1

2.48

from the +x-axis

anticlockwise


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W

F

N

kf

constantspeed

50.0

F

50.0

A window washer pushes his

scrub brush up a vertical

window at constant speed

by applying a force as

shown in Figure 5.7.

The brush weighs 10.0 N andthe coefficient of kinetic

friction is k= 0.125.

Calculate

a) the magnitude of theforce

b) the normal force exerted

by the window on the

brush.

Example 3 :

67


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Solution :

a) The free body diagram of

the brush :

0.125;N10.0 kW

Force x-comp (N) y-comp (N)

F

50.0cosF

kf

0Nk

50.0sinF

W 0 10.0

N

N 0

N0.125

0amF

50.0cosFN

0xF

(1)

(2)sin50.0 0.125F N

0yF

The brush moves up

at constant speed(a= 0) so that

Thus

10.0

F= ? N= ?


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Exercise :Use gravitational acceleration, g= 9.81 m s 2

1.The system in Figure 5.8

is in equilibrium, with the

string at the centre

exactly horizontal.Calculate

a) the tensions T1, T2and T3.

b) the angle .


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2.

Figure 5.9

70

A 20 kg ball is supported

from the ceiling by a

rope A. Rope B pulls

downward and to the

side on the ball. If the

angle of A to the verticalis 20 and if B makes an

angle of 50 to the

vertical as shown in

Figure 5.9, Determinethe tension in ropes A

and B.


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3.A block of mass 3.00 kg ispushed up against a wall by a

force that makes a 50.0

angle with the horizontal as

show in Figure 5.10. The

coefficient of static frictionbetween the block and the

wall is 0.250. Determine the

possible values for the

magnitude of that allow theblock to remain stationary.


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states

OR its can berepresented by

where

momentumlinearinchange:pd

intervaltime:dt

forceresultant:F

4.2.4 Newtons Second Law


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73

From the Newtons 2nd law of motion, italso can be written as

and


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74

Object at or in motion

with and

.

Thus

dt

vdm

dt

dmvF

0dt

pdF

where and

Object at or in

motion withbut with

. For

example : Rocket

dt

vdm

dt

dmvF

0dt

vd

and


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Object with but .

The of the always in the

or .

dt

vdm

dt

dmvF

0dt

dmand

and

where

objectanofmass:m

onaccelerati:a

forceresultant:F


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Newtons 2nd law ofmotion restates that

.

OR

One newton (1 N) is

defined as

.

OR


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77

Notes:

is a nett force or effective force or

resultant force.

The force which causes the motion of an

object.

If the forces act on an object and the object

moving at

hence

amFFnett

F


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78

Applications of Newtons 2nd law of motion From the Newtons second law of motion, we

arrived at equation

There are five steps in applying the equationabove to solve problems in mechanics:

Identify the object whose motion isconsidered.

Determine the forces exerted on the object.

Draw a for each object.

Choose a system of coordinates so thatcalculations may be simplified.

Apply the equation above,

Along x-axis:

Along y-axis:

maFF nett

xxmaF

yy maF


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79

Three wooden blocks connected by a rope of

negligible mass are being dragged by ahorizontal force, in figure.

Suppose that F= 1000 N, m1 = 3 kg, m2 = 15 kg

and m3 = 30 kg. Determine

a) the acceleration of blocks system.

b) the tension of the rope, T1 and T2.

Neglect the friction between the floor and the

wooden blocks.

1T

m1 m2 m32T

F


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80

a) For the block, m1 = 3 kg

For the block, m2 = 15 kg

For the block, m3 = 30 kg

a amTFF 11x

(1)

amTTF 221x

aTT 21 15 (2)

1T

m1

m2

m3

2T

F

aTF 1x 31000

10003aT1

1T

a

aTTF 21x 15

2T

a

amTF 32x

aT2 30 (3)


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81

a) By substituting eq. (3) into eq. (2) thus

Eq. (1) (4) :

b) By substituting the value of acceleration into

equations (4) and (3), therefore

045aT1 (4)

48

1000a

2sm20.8a

N9361T

N6242T


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82

Two objects of masses m1

= 10 kg and m2 = 15 kg areconnected by a light string

which passes over a

smooth pulley as shown in

figure 3.10. Calculatea) the acceleration of the

object of mass 10 kg.

b) the tension in each

string.(Given g= 9.81 m s 2)

m1

m2

a) For the object m1=

10 kg,

1T

gmW 11

a

amgmTF 111y

(1)agT 1010

where TTT 21


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a) For the object m2=

15 kg,2T

gmW 22

a

amTgmF 222y

(2)agT 1515aTgFy 1515

25

9.815

25

5ga

21.96 m sa

Eq. (1) + (2) :

b) Substitute the value of

acceleration into

equation (1) thus

118NT

1.96109.8110T

N118TTT 21

Therefore


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84

states

. For example :

When the student push on the wall it will pushback with the same force.

A (hand)

B (wall)

ABF

BAF

is a force by the hand on the wall

Where

is a force by the wall on the handABF

BAF


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When a book is placed on the table.

If a car is accelerating forward, it is because itstyres are pushing backward on the road and theroad is pushing forward on the tyres.

A rocket moves forward as a result of the pushexerted on it by the exhaust gases which therocket has pushed out.

In all cases when two bodies interact, the.

85

Force by the book on the table

Force by the table on the book


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Consider a person standing inside a lift as shown

in figures (a), (b) and (c).

(a) Lift moving upward at a uniform velocity

86gmW

N

Since the lift moving at a

uniform velocity, thus

0yaTherefore

0yF

0mgN


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(c) Lift moving downwards at a constant

acceleration, a

Caution : Nis also known as

and Wis . 88

a

gmW

N

By applying the newtons

2nd law of motion, thus

yy maF

maNmg


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89

Two blocks, A of mass

10 kg and B of mass 30kg, are side by side and

in contact with each

another. They are

pushed along a smooth

floor under the action ofa constant force Fof

magnitude 200 N applied

to A as shown in figure.

Determinea) the acceleration of

the blocks,

b) the force exerted by A

on B.

A BF

a)Let the acceleration of

the blocks is a. Therefore

N200kg;30kg;10 Fmm BA

ammF BAx

2

sm5.0a

ammF BA

a3010200


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90

b. For the object A,

From the Newtons 3rd

law, thus* : force acted on A by B

: force acted on B by A

OR

For the object B,

150 NAB

F

x AB AF F F m a

200 10 5.0ABF

150 NBA ABF F

F

a

ABF

A

BBAF

a x BA BF F m a

30 5.0BAF

150 NBAF

ABF

BAF


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91

One 3.5 kg paint bucket is hanging by

a massless cord from another 3.5 kgpaint bucket, also hanging by a

massless cord as shown in figure

3.13. If the two buckets are pulled

upward with an acceleration of 1.60m s 2 by the upper cord, calculate the

tension in each cord.

(Given g= 9.81 m s 2)


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THE END

CHAPTER 5 :

Work, Energy and Power

ch4 forces mh

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