chapter 4 matriculation stpm

7/28/2019 Chapter 4 matriculation STPM

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PHYSICS CHAPTER 4

1

CHAPTER 4:

FORCES

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PHYSICS CHAPTER 4

4. FORCES

4.1 Basic of Forces and Free Body Diagram

4.2 Newtons Laws of Motion

2

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PHYSICS CHAPTER 4

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At the end of this chapter, students should be able to:

Identify the forces acting on a body in different situations.

Weight

Tension

Normal force

Friction

Determine weight, static friction and kinetic friction

Draw free body diagram

Determine the resultant force

Learning Outcome:


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PHYSICS CHAPTER 4


Weight, is defined as the force exerted on a body under gravitational

field.

It is a vector quantity.

It is dependant on where it is measured, because the value ofgvaries at different localities on the earths surface.

It always directed toward the centre of the earth or in the same

direction of acceleration due to gravity, g.

The S.I. unit is kg m s-2 orNewton (N).

Equation:

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PHYSICS CHAPTER 4

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Tension, T

The tension force is the force that is transmitted through a

string, rope, cable or wire when it is pulled tight by forcesacting from opposite ends. The tension force is directed

along the length of the wire and pulls equally on the objects

on the opposite ends of the wire.

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Figure 4.1


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Normal (reaction) force,

is defined as a reaction force that exerted by the surface to

an object interact with it and the direction always

perpendicular to the surface.

An object lies at rest on a flat horizontal surface as shown in

Figure 4.2.

RorN

N

gmW

0mgNFyThereforeFigure 4.2

Action: weight of an object is exerted on the

horizontal surface

Reaction: surface is exerted a force, Non theobject

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A free body diagram is defined as a diagramshowing the chosen body by itself, with vectors

drawn to show the magnitude and directions of allthe forces applied to the body by the other bodies

that interact with it.


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PHYSICS CHAPTER 4

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Friction

is defined as a force that resists the motion of one surface

relative to another with which it is in contact. is independent of the area of contact between the two surfaces..

is directly proportional to the reaction force.

OR

Coefficient of friction, is defined as the ratio between frictional force to reactionforce.

OR

is dimensionless and depends on the nature of the surfaces.

Nf

where

N

f

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PHYSICS CHAPTER 4

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There are three types of frictional force :

Static,fs

(frictional force act on the object before its move)

Kinetic,fk (frictional force act on the object when its move)

Rolling,fr (frictional force act on the object when its rolling)

Caution:

The direction of the frictional force exerted by a surfaceon an object is always in the opposite direction of the

motion.

The frictional and the reaction forces are always

perpendicular.

Nf kk

Nf ss

Nf rr

skr fff where

thus skr

Can be ignored

Simulation 4.1

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http://localhost/var/www/apps/conversion/tmp/scratch_10/AF_0516.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_10/AF_0516.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_10/AF_0516.html


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PHYSICS CHAPTER 4

Example 4.1:

A mass is resting on a flat surface which has a normal force of98N, with a coefficient of static friction of 0.35. What force

would it take to move the object?

9

Solution: N = 98N, s = 0.35

Nf ss

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PHYSICS CHAPTER 4

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Example 4.2:

A 15 kg piece of wood is placed on top of another piece ofwood. There is 35N of static friction measured between them.

Determine the coefficient of static friction between the two

pieces of wood.

Solution: N = mg = 15(9.81) = 147.15 N, Fs

= 35 N

N

fss

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PHYSICS CHAPTER 4

Example 4.3

A dock worker loading crates on a ship finds that a 15 kg crate,initially at rest on a horizontal surface, requires a 50 N

horizontal force to set it in motion. However, after the crate is in

motion, a horizontal force of 30 N is required to keep it moving

with a constant speed. The acceleration of gravity is 9.8 ms-2.

Find the coefficient of kinetic friction.

11

Solution:

Mass of crate = m = 15 kg

Force required to set the crate in motion = F1 = 50 N

Force required to keep the crate in moving at constant speed =

fk = 30 N

Acceleration of gravity = g = 9.81 ms-2

Normal force, N = mg = =

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PHYSICS CHAPTER 4

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Solution:

From

N

fkk

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PHYSICS CHAPTER 4

Resultant force Is defined as a single force that represents the combined

effect of two or more forces

1313

The figure above shows three forces F1, F2 and F3 acted on a

particle O. Calculate the magnitude and direction of the

resultant force on particle O.

Example 4.4:y

30o

O

)N30(2F

)N10(1

F

30ox

)N40(3F

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PHYSICS CHAPTER 4

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30o

Solution :

O

y

x

3F

30o

y3F

321 FFFFFr

yxr FFF

xxxx FFFF 321

yyyy FFFF 321

xF2

1F2F

60o

yF2

x3F

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PHYSICS CHAPTER 4

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Vector x-component y-component

1F

3F

2F

N01 xF 11FFy N011 yF

60cos302 xF N152 xF

60sin302

yF N622 yF30cos403 xF

N34.63 xF

30sin403 yFN203 yF

Vector

sum

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PHYSICS CHAPTER 4

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y

xO

Solution :

The magnitude of the resultant force is

and

Its direction is 162 from positive x-axis OR 18 above negative x-

axis.

22 yxr FFF

x

y

F

F 1tan

yF

xF

162

rF

18

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PHYSICS CHAPTER 4

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1. Given three vectorsP, Q andR as shown in Figure 4.3.

Calculate the resultant vector ofP, Q andR.

ANS. : 49.4 m s2; 70.1 above + x-axis

Exercise 4.1:

Figure 4.3

y

x0

502sm10 R

2sm35 P

2sm24 Q

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PHYSICS CHAPTER 4

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At the end of this chapter, students should be able to:

State Newtons First Law

Define mass as a measure of inertia.

Define the equilibrium of a particle.

Apply Newtons First Law in equilibrium of forces

State and apply Newtons Second Law

State and apply Newtons Third Law.

Learning Outcome:

4.2 Newtons laws of motion

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PHYSICS CHAPTER 4

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4.2 Newtons laws of motion

4.2.1 Newtons first law of motion states an object at rest will remain at rest, or continues to

move with uniform velocity in a straight line unless it isacted upon by a external forces

OR

The first law gives the idea of inertia.

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PHYSICS CHAPTER 4

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4.2.2 InertiaInertia

is defined as the tendency of an object to resist any changein its state of rest or motion.

is a scalar quantity.

Mass, m

is defined as a measure of a bodys inertia. is a scalar quantity.

The S.I. unit of mass is kilogram (kg).

The value of mass is independent of location.

If the mass of a body increases then its inertia will increase.

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PHYSICS CHAPTER 4

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Figures 4.4a and 4.4b show the examples of real experience ofinertia.

Figure 4.4

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PHYSICS CHAPTER 4

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4.2.3 Equilibrium of a particle

is defined as the vector sum of all forces acting on a particle(point) must be zero

.

The equilibrium of a particle ensures the body in translationalequilibrium and its condition is given by

This is equivalent to the three independent scalar equationsalong the direction of the coordinate axes,

There are two types of equilibrium of a particle. It is

Static equilibrium (v=0) body remains at rest (stationary).

Dynamic equilibrium (a=0) body moving at a uniform(constant) velocity.

Newtons first

law of motion

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PHYSICS CHAPTER 4

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Problem solving strategies for equilibrium of a

particle

The following procedure is recommended when dealing with

problems involving the equilibrium of a particle:

Sketch a simple diagram of the system to help

conceptualize the problem.

Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and

construct a table to resolve the forces into their

components.

Apply the condition for equilibrium of a particle in

component form :

Solve the component equations for the unknowns. 0xF 0yFand

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PHYSICS CHAPTER 4

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A load of 250 kg is hung by a cranes cable. The load is pulled by a

horizontal force such that the cable makes a 30 angle to thevertical plane. 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 s2)

Solution :

Example 4.5:

30

F F

Free body diagram of the load :

gm

T

yT3060

xT

kg250m

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PHYSICS CHAPTER 4

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Solution :

1st method :

a.

Since the load is in the equilibrium, then

Thus

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

0xF 060cos

TF

kg250m

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

gm

0 9.81250mg

F

F 0

T

60cosT 60sinT

2453

0F

(1)

(2) 0yF 0245360sin T

060cos2833

F

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PHYSICS CHAPTER 4

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30

Solution :

2nd method :

a. Since the load is in the equilibrium, then a closed triangle of

forces can be sketched as shown below.

b.

30sinT

F

kg250m

30cosT

mg

30sin2833 F

F

gm

T

From the closed triangle of forces, hence

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PHYSICS CHAPTER 4

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Calculate the magnitude and direction of a force that balance the

three forces acted at point A as shown in Figure 4.5.

Example 4.6:

N121

FN202F

N303F

30.055.0

45.0A

Figure 4.5

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PHYSICS CHAPTER 4

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Solution :

To find a force to balance the three forces means the system must

be in equilibrium hence

N30N;20N;12 321 FFF


1F 55.0cos12

F

xF yF

6.88

55.0sin129.83

2F 30.0cos20

17.3

30.0sin2010.0

3F 45.0cos30

21.2

45.0sin3021.2

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Solution :

The magnitude of the force,

and its direction,

0 yF021.210.09.83 yF

222y

2

x

FFF 1.3731.6

x

y1

F

F tan

31.6

1.37tan 1

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PHYSICS CHAPTER 4

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30

A window washer pushes his scrub brush up a vertical window at

constant speed by applying a force Fas shown in Figure 4.6.The brush weighs 10.0 N and the coefficient of kinetic friction is

k= 0.125. Calculate

a. the magnitude of the force F,

b. the normal force exerted by the window on the brush.

Example 4.7:

F

50.0

Figure 4.6

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PHYSICS CHAPTER 4

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Solution :

a. The free body diagram of the brush :

The brush moves up at constant speed (a=0) so that

Thus

0.125;N10.0 kW

W

F

N

kf

constant

speed


F

50.0cosF

kf

0Nk

50.0sinF

W

0 10.0

N

N 0

N0.125

0amF

50.0cosFN0 xF (1)

(2)10.00.12550.0sin NF

0 yF

50.0

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PHYSICS CHAPTER 4

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Solution :

a. By substituting eq. (1) into eq. (2), thus

b. Therefore the normal force exerted by the window on the brush

is given by

10.050.0cos0.12550.0sin FF

50.0cosFN

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PHYSICS CHAPTER 4

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Exercise 4.2:

Use gravitational acceleration, g= 9.81 m s2

1.

The system in Figure 5.8 is in equilibrium, with the string at the

centre exactly horizontal. Calculate

a. the tensions T1, T2 and T3.

b. the angle .

ANS. : 49 N, 28 N, 57 N; 29

Figure 4.7

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PHYSICS CHAPTER 4

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Exercise 4.2:

2.

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 tothe vertical is 20 and if B makes an angle of 50 to the vertical

as shown in Figure 4.8, Determine the tension in ropes A and B.

ANS. : 134 N; 300 N

Figure 4.8

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PHYSICS CHAPTER 4

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Exercise 4.2:

3.

A block of mass 3.00 kg is pushed up against a wall by a force

P that makes a 50.0 angle with the horizontal as show in

Figure 4.9. The coefficient of static friction between the blockand the wall is 0.250. Determine the possible values for the

magnitude ofP that allow the block to remain stationary.

ANS. : 31.8 N; 48.6 N

Figure 4.9

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PHYSICS CHAPTER 4


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Newtons second law of motion

states the rate of change of linear momentum of a moving

body is proportional to the resultant force and is in thesame direction as the force acting on it

OR

its can be represented by

where

momentumlinearinchange:pd

intervaltime:dt

forceresultant:F

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PHYSICS CHAPTER 4


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

Case 1: Object at rest or in motion with constant velocity but with

changing mass. For example : Rocket

and

0dt

vd

and

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PHYSICS CHAPTER 4


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PHYSICS CHAPTER 4

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Case 2:

Object at rest or in motion with constant velocity and constant

mass.

Thus

dt

vdm

dt

dmvF

Newtons 1st

law of motion

0dt

pdF

where and

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PHYSICS CHAPTER 4


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Case 3:

Object with constant mass but changing velocity.

The direction of the resultant force always in the same

direction of the motion oracceleration.

dt

vdm

dt

dmvF

0dt

dman

d

and

where

objectanofmass:m

onaccelerati:a

forceresultant:F

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PHYSICS CHAPTER 4


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Newtons 2nd law of motion restates that The acceleration of

an object is directly proportional to the nett force acting on

it and inversely proportional to its mass.OR

One newton(1 N) is defined as the amount of nett force that

gives an acceleration of one metre per second squared to abody with a mass of one kilogramme.

OR 1 N = 1 kg m s-2

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

uniform acceleration (not at rest or not in the

equilibrium) hence

amFFnett

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When a book is placed on the table. (refer to Figure 4.11)

If a car is accelerating forward, it is because its tyres arepushing backward on the road and the road is pushingforward on the tyres.

A rocket moves forward as a result of the push exerted on it

by the exhaust gases which the rocket has pushed out.

In all cases when two bodies interact, the action and reactionforces act on different bodies.

Figure 4.11

Force by the book on the table (action)

Force by the table on the book (reaction)

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PHYSICS CHAPTER 4


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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 equation above to solveproblems in mechanics:

Identify the object whose motion is considered.

Determine the forces exerted on the object.

Draw a free body diagram for each object. is defined as a diagram showing the chosen body by

itself, with vectors drawn to show the magnitude anddirections of all the forces applied to the body by theother bodies that interact with it.

Choose a system of coordinates so that calculations may besimplified.

Apply the equation above,

Along x-axis:

Along y-axis:

maFFnett

xxmaF

yy maF

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PHYSICS CHAPTER 4


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PHYSICS CHAPTER 4

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Three wooden blocks connected by a rope of negligible mass are

being dragged by a horizontal force, Fin Figure 4.12.

Suppose thatF= 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.

Example 4.8:

Figure 4.12

1T

m1 m2 m32T

F

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Solution :

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

(2)

1T

m1

m2

m3

2T

F

aTF 1x 31000

1T

a

aTTF 21x 15

2T

a

amTF 32x(3)

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Solution :

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

045 aT1 (4)

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PHYSICS CHAPTER 4


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Two objects of masses m1 = 10 kg and m2 = 15 kg are connected

by a light string which passes over a smooth pulley as shown inFigure 4.13. Calculate

a. the acceleration of the object of mass 10 kg.

b. the tension in the each string.

(Giveng= 9.81 m s2

)Solution :

a. For the object m1= 10 kg,

Example 4.9:

Figure 4.13

m1

m2

1T

gmW 11

amgmTF 111y

(1)agT 1010 a where TTT 21

Simulation 4.2

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Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side

and in contact with each another. They are pushed along a smoothfloor under the action of a constant forceFof magnitude 200 Napplied to A as shown in Figure 4.14. Determine

a. the acceleration of the blocks,

b. the force exerted by A on B.

Solution :

a. Let the acceleration of the blocks is a. Therefore

Example 4.10:

ammF BAx

N200kg;30kg;10 Fmm BA

Figure 4.14

A BF

ammF BA

Simulation 4.3

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Solution :

b. For the object A,

From the Newtons 3rd law, thus

OR

For the object B,

amFFFABAx

5.010200 BAFF

a

BAF

A

BABF

a

amFFBABx

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1. A block is dragged by forces,F1 andF2 of the magnitude

20 N and 30 N respectively as shown in Figure 4.15. The

frictional forcefexerted on the block is 5 N. If the weight ofthe block is 200 N and it is move horizontally, determine the

acceleration of the block.

(Giveng

= 9.81 m s2)

ANS. : 1.77 m s2

Exercise 4.3:

50a

1F

2F

f

20

Figure 4.15

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PHYSICS CHAPTER 4


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2. One 3.5 kg paint bucket is hanging by a massless cord from

another 3.5 kg paint bucket, also hanging by a massless cordas shown in Figure 4.16. If the two buckets are pulled upward

with an acceleration of 1.60 m s2 by the upper cord, calculate

the tension in each cord.

(Giveng= 9.81 m s2)

ANS. : 39.9 N; 79.8 N

Exercise 4.3:

Figure 4.16

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PHYSICS CHAPTER 4


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PHYSICS CHAPTER 4

THE END

Next ChapterCHAPTER 5 :

Work, Energy and Power

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