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Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009

By : Eng . YOUNIS FAKHER

1

Resolution & Composition of a force Engineering Mechanics : may be defined as a science which describes and predicts the condition of rest or motion of bodies under the action of forces.

Vector & Scalar quantities : Vector quantities : are the quantities which have magnitude and direction .such as:

Force , weight , distance , speed , displacement , acceleration ,velocity . Scalar quantities : are the quantities which have only magnitude , such as :

Time , size , sound , density , light , volume . Force : A "force" is an action that changes, or tends to change, the state of motion of the body upon which it acts. It is a vector quantity that can be represented either mathematically or graphically A complete description of a force MUST include its:

1. MAGNITUDE 2. DIRECTION and SENSE 3. POINT OF ACTION

Resolution & Composition of a force : Let the force ( F) shown in fig.(1) with the direction (θ )

We can resolve this force into two components :

1‐ horizontal component ( Fx ) which lies on x‐ axis

2‐ vertical component ( Fy ) which lies on y‐ axis

as shown in fig.(2) fig.(1 )



2

from fig.(2) : The horizontal component may be determined as : Fx = F . cos θ The vertical component may be determined as : Fy = F . sin θ

EX (1 ) : Find the two components of the force ( 100 N ) if : θ = 30o , 120o , 270o fig. ( 2 ) Solution : θ = 30o : Fx = F . cos θ = 100 * cos 30

= 100 * 23 = 50 3 N

Fy = F . sin θ = 100 * sin 30 = 100 * 0.5 = 50 N θ = 120o : Fx = F . cos θ = 100 * cos 120 = 100 * (‐ 0.5 ) = ‐ 50 N Fy = F . sin θ = 100 * sin 120

= 100 *23 = 50 3 N

θ = 270o : Fx = F . cos θ = 100 * cos 270 = 100 * ( 0 ) = 0 Fy = F . sin θ = 100 * sin 270 = 100 * ( ‐ 1 ) = ‐ 100 N



3

EX ( 2 ) : The direction of the force ( P ) is ( 30O ) , Find the horizontal component if the vertical component is ( 30 N ) ? Solution : From the diagram shown : Fy = 30 N Fy = F . sin θ 30 = P * sin 30 30 = P * 0.5 P = 30 / 0.5 = 60 N Fx = F . cos θ

= 60 * cos 30 = 60 *23 = 30 3 N

Composition of a force : Let we have ( Fx ) is the horizontal component and ( Fy ) is the vertical component for the force ( F ) shown in fig. From the shape ABC we get : AC 2 = AB2 + BC2 F2 = Fx 2 + Fy 2

2+2= )()( FyFxF

Determination of the direction of a force : The direction of a force can be determined by :

)FxFy( 1- tan =θ

EX ( 3 ) : Determine the magnitude and direction of a force ( P ) , if the horizontal and vertical components are ( 20 N ) , ( 40 N ) respectively ? Solution :

We have : Fx = 20 N , Fy = 40 N , 2+2= )()( FyFxF

o43.63)2040(tan)(tan

N 72.44200001600400)40()20(

1-1-

22

===

==+=+=

FxFy

F

θ



4

EX ( 4 ) : The line of action of the ( 300 N ) force runs through the points ( A ) and ( B ) as shown in fig . Determine the ( X ) and ( Y ) scalar components of ( F ) . Solution :

NFyF

NFxF

BCL

ACL

ABL

2.141178*300sin.

7.2641715*300cos.

82)26(

152)87(

172)62(2)87(

===

===

=+=

=−−=

=−−+−−=

θ

θ

PROBLEMS 1 ‐ Determine a pair of horizontal and vertical components of

the ( 340 N ) force ? 2 ‐ Determine the horizontal & vertical components of the

force ( 200 N ) ?



5

3 ‐ Resolve the ( 100 N ) force into horizontal and vertical components for each of the following values of ( θ ): a‐ 20o b‐ 80o c‐ 240o d‐ 210o 4 ‐ The horizontal component of the force ( F ) is ( 60 N ) to

the right through the original point . Determine the vertical component and the magnitude of ( F ) ?

5 ‐ The body on the ( 30o ) incline is acted upon by a force

( P ) inclined at ( 20o ) with the horizontal . if ( P ) is resolved into components parallel and perpendicular to the incline and the value of the parallel component is ( 300 N ) , Compute the value of the perpendicular , and of ( P ) ?

6 – The Y – component of the force ( F )

which a person exerts on the handle of the box wrench is known to be ( 70 N ) .Determine the ( X ) component , and the magnitude of ( F ) .

Ans : Fx = 29.5 N , F= 75.8 N



6

7 – Determine the components of the ( 2 KN ) force

along the oblique axes ( a ) and ( b ) . Determine the projections of ( F ) onto the a – and b‐ axes .

8 ‐

9 ‐

10 : Write whether the following quantities are vectors or scalars : Force , sound , density , velocity , weight , time , mass , acceleration , light , area 11 : Determine the angle ( θ ) and locate the force on the coordinates when : Fx = Fy , Fx = ‐ Fy , ‐ Fx = ‐ Fy , ‐ Fx = Fy

Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009


7

Resultant of forces system The resultant is a representative force which has the same effect on

the body as the group of forces it replaces. A simplest force which can replace the original forces system without changing its

external effect on a rigid body. The symbol of resultant force is: The unit of resultant force is : Newton (N)

Types of forces system 1‐ coplanar forces system :

a‐ concurrent coplanar forces system b‐ non‐concurrent coplanar forces system

2‐ Non coplanar forces system : a‐ concurrent non‐coplanar forces system b‐ non‐concurrent non‐coplanar forces system

Resultant of concurrent coplanar forces system We will find out the resultant force for many forces acting on a rigid body by using

the following equations :

( ) ( )22

332211

332211

+=

==

yx

nny

nnx

RRR

FFFFRFFFFR

θθθθθθθθ

sin....sin.sin.sin.cos....cos.cos.cos.

mmmm

mmmm

The direction of resultant force may be determined as :

⎟⎠⎞

⎜⎝⎛

= −

x

y

RR1tanθ

Ex ( 1 ) : Find the resultant force for the concurrent coplanar forces system, shown in fig. Solution

( ) ( )( ) ( ) N

RRR

NFFFR

NFFFR

yx

y

x

2028.604.192

8.6045sin9060sin100200sin.sin.sin.

4.19245cos9060cos100200cos.cos.cos.

22

22

51

332211

52

332211

=+=

+=

−=−−•==

+=+−•==

θθθ

θθθ

mm

mm



8

Ex ( 2 ) : The 200 N force is a resultant of two forces, one of the forces "P" has the direction along the tine AB and the other force "Q" is on the horizontal direction, determine them. Solution

( )

( )

NQNP

QPFFR

FFRQP

QPFFR

FFR

R

y

R

x

252295138

060200 -

50160 060200

53

2211

2211

54

2211

2211

..

sinsinsin.sin.sin

sin.sin..

coscoscos.cos.cos

cos.cos.

=∴=

−∗=∗=∗

=+=

+∗=∗=∗

=

θθθθθ

θθθθθ

m

m

m

m

Ex ( 3 ) : Determine the resultant force for the forces system shown in fig. Solution

( )

( )

( ) ( )N 820178605192

N 60.78 45200025090100

N 519245200025090100

22

22

332211

332211

.).().(

sinsinsinsin.sin.sin.

.coscoscos

cos.cos.cos.

=−+=

+=

−=−+=

==

−+==

yx

y

x

RRR

FFFR

FFFR

θθθ

θθθ

mm

mm



9

Ex ( 4 ) : The 1000 N force is a resultant of two forces, one of which is 600 N, Determine the other force? Solution

( )

( )( ) ( )( )

N

FFF

FN θFF

FFRFFR

FN- θFF

FFFR

FFR

yx

y

R

y

x

R

x

6294202386860

2________ 02386 600601000

1________ 860 360501000

600601000

22

22

22

2254

2211

2211

22

22

2253

2211

2211

.).(

.sinsinsin

sin.sin.sinsin.sin.

coscos.

coscoscos.cos.cos

cos.cos.

=

+=

+=

⇒=

+∗=∗=∗

=⇒=

+=∗−

+∗=∗−=∗

=

θθθθ

θθ

θθ

θθθθθ

m

m

m

m

Ex ( 5 ) : Determine the resultant force of the concurrent coplanar forces system shown in fig. Solution

( ) ( ) ( ) ( )N

RRR

FFFRN

FFFR

yx

y

x

455417073170726

7073190304510025

7072690304510025

2222

332211

332211

...

.sinsinsin

sin.sin.sin..

coscoscoscos.cos.cos.

=

+=+=

=++=

==

++==

θθθ

θθθ

mm

mm



10

Ex ( 6 ) : The 624 N force is a resultant of two force P & Q shown in fig. determine these forces . Solution

( )

( )( )2________ 300

42001246

sin.sin.sin1________ 609

246cos.cos.cos

54

54

135

2211

53

53

1312

2211

QPQP

QPFFR

QPQP

FFR

R

R

−=−=

∗−∗=∗=∗+=

∗+∗=∗=∗

θθθ

θθθ

m

m

( ) ( )

NQ NP

: in Subst

330630

21

==

Ex ( 7 ) : Determine the forces P and Q when the 125 N force is the resultant force of these forces Solution

NQQ

PQFFR

PQP

FFR

R

R

251312581100

125

258190125

135

135

54

2211

1312

53

2211

..

sin.sin.sin.

coscos.cos.cos

=

∗−=

−=∗=∗

=∴

+∗=∗=∗

θθθ

θθθ

m

m



11

Ex ( 8 ) : Determine the resultant of concurrent coplanar forces system shown in fig. Solution

( ) ( )N 201.8

) 60.78 - ( 192.5) (

N 60.78 -

4590601005

1200

N 5192

4590601005

2200

2222

332211

332211

=

+=+=

=

−−=

==

+−=

=

yx

y

x

RRR

FFFR

FFFR

sinsin*

sin.sin.sin..

coscos*

cos.cos.cos.

θθθ

θθθ

mm

mm

Ex ( 9 ) : The screw eye in Fig. is subjected to two

forces, and Determine the magnitude and direction of the resultant force.

Solution



12

Ex ( 10 ) : Determine the resultant ( R ) of the two forces system shown in fig. Solution h = 6 sin 60 = 5.2 m b = 6 cos 60 = 3 m

( ) ( )

N 523.8 ) 392.84 ( 346.58) (

N 392.84 9.40sin600

sin.sin.N 5.346

9.40cos600800cos.cos.

9.40)866.0(tan

866.033

2.53

tan

22

22

2211

2211

1

=

+=

+=

====

−==

==

=+

=+

=

−

yx

y

x

RRR

FFR

FFR

bh

θθ

θθα

α

m

m

Ex ( 11 ) : Determine the angle ( θ ) so that the resultant ( R ) of the two forces system shown in fig. is 100 N Solution

( ) ( )( ) ( )

o

FFFFR

3.51625.0cos

cos*40*70*24070100

cos222

212

22

1

=

=

++=

++=

θ

θθ

θ



13

Ex ( 12 ) : Find out the magnitude of the force ( P ) and the resultant ( R ) of the two forces to rise the cylinder up without touch the vertical wall . Solution

NRRR

PRFFR

NPP

PRFFR

R

R

4.30647.11893.187

60sin8.136sin20090sinsinsin20090sin

sin.sin.sin8.136

5.04.68060cos70cos20090cos*

cos.cos.cos

2211

2211

=+=

+=+=

=∗=∴

−=−=

=∗

θθθθθθ

θθθ

m

m

PROBLEMS 1 ‐ Two forces are applied at the end of a screw eye

in order to remove the post. Determine the angle and the magnitude of force F so that the resultant force acting on the post is directed vertically upward and has a magnitude of 750 N.

2 – Determine the resultant ( R ) of the two forces

shown in fig.



14

3 ‐ The ratio of the lift force ( L ) to the drag force ( D ) for the simple airfoil is ( L / D = 10 ) . If the left force on a short section of the airfoil is ( 50 N ) , Compute the magnitude of the resultant force ( R ) and the angle ( θ ) which it makes with the horizontal . Ans. R = 50.2 N , θ = 84.3o

4 ‐ 5 – 6 ‐ Determine the magnitude ( Fs ) of the tensile spring

force in order that the resultant of ( Fs ) and ( F ) is a vertical force . Determine the magnitude ( R ) of this vertical resultant force .

Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 4th Week 2008 - 2009


15

Moment of a force

The moment of a force : is the ability of the force to produce turning or twisting about an axis or point or line. Mathematically: The moment of a force = the applied force X perpendicular distance M = F * d M = the moment of a force (N.m) F = applied force ( N ) d = perpendicular distance between the point of action of the force and moment center. Ex ( 1 ) : Determine the moment of the force 225 N about the Points A , B , and C .

Solution

MA=|F|dA=225*0.6 = 135 Nm

MB=|F| dB=225*0.4 = 90 Nm

MC=|F| dC=225*0.8 = 180 Nm

Ex ( 2 ) : Determine the moment of the force 500 N about the point A and B .

Solution Cos (60) = 200/L Cos (60) = dac/(L-160) L = 200/ Cos (60) L=160+dac/Cos (60) dac=200-160 Cos (60)=120 mm dac=120 mm MA=|F|dAC=500*0.12 = 60 Nm MB=|F| dB=500*0.2 = 100 Nm



16

Ex ( 3 ) : Find the moment of the force 200 N About the point ( A ) shown in fig. Solution Fx = F . cos θ = 200 cos 45 = 200 * 0.707 = 141.42 N Fy = F . sin θ = 200 sin 45 = 200 * 0.707 = 141.42 N M1 = Fx * d = 141.42 * 10 = 1414.2 N . cm M2 = Fy * d = 141.42 * 20 = 2828.4 N .cm M ( A ) = M1- M2 = - 1414.2 N .cm Ex ( 4 ) : Determine the moment of the force ( 70 N ) shown in fig. about the Point ( A ) . Solution Fx = F . cos θ = 70 cos 30 = 70 * 0.866 = 60.62 N Fy = F . sin θ = 70 sin 30 = 70 * 0.5 = 35 N M1 = Fx * d = 60.62 * 8 = 484.97 N . m M2 = Fy * d = 35 * 4 = 140 N . m M ( A ) = M1 + M2 = 484.97 N + 140 = 624.97 N . m Ex ( 5 ) : Find the distance ( Xn ) , if the moment of the force ( F ) about the point ( A ) is equal to zero . Solution Fx = F . cos θ = 20 cos 30 = 20 * 0.866 = 17.32 N Fy = F . sin θ = 20 sin 30 = 20 * 0.5 = 10 N M1 = Fx * d = 17.32 * 120 = - 2078.46 N . cm M2 = Fy * d = 10 * Xn = 10 Xn N .cm M ( A ) = - M1+ M2 0 = - 2078.46 + 10 Xn Xn = 2078.46 / 10 = 207.84 cm



17

) : 6Ex ( The moment of the force ( F ) about ( A ) is ( 3100 ) N . m ), the moment of this force about ( B ) is ( ‐ 350 N.m ) , the distance between ( A ) and ( B ) is ( 30 m ) , Find the magnitude and the location of the force ( F ) if its component along the line ( AB ) is ( 5 N ). Solution Given Fx = 5 N M ( A ) = 3100 , M ( B ) = - 350 M ( A ) = F . sin θ * x F . sin θ = 3100 / x M ( B ) = F . sin θ * ( 30 – x ) F . sin θ = - 350 / ( 30 – x )

xx −=

303503100

X= 20 m

Fy = F . sin θ = 3100 / x = 3100 / 20 = 5 3 N

2222 355 )()()()( +=+= FyFxF = 10 N Ex ( 7 ) : A ( 150 N ) force acts on the end of the ( 900 mm ) lever as shown in fig. Determine the moment of the force about ( O ) . Solution Q = 150 sin 20 = 51.3 N Mo = ‐ Q ( 0.9 ) = ‐ 51.3 * 0.9 = ‐ 46.2 N.m = 64.2 N.m



18

Ex ( 8 ) : Knowing that the distance AB is 250 mm determine the maximum moment about ( B ) which can be caused by the ( 150 N ) force . In what direction should the force act ? Solution MB = 150 * 0.25 = 37.5 N.m θ = 90 – 70 = 20 OR : θ = 180 – ( 90 + 70 ) = 20 Ex ( 9 ) : A ( 150 N ) force is applied to the control lever at ( A ) , knowing that the distance AB is ( 250 mm ) Determine the moment of the force about ( B ) when θ is 50 o . Solution BC = 250 cos 70 = 85.5 mm AC = 250 sin 70 = 234.92 mm P = 150 cos 50 = 96.41 N Q = 150 sin 50 = 114.9 N M B = P * AC + Q * BC = 96.41 * 234.92 + 114.9 * 85.5 = 3247 N.mm = 32.47 N . m



19

PROBLEMS 1 ‐ Find the distance ( Y ) if the moment of the

forces system shown in fig. is equal to ( 12750 N.m ) about the point ( O )

2 – The moment of a certain force ( F ) is ( 180 N.m )

clockwise about ( O ) and ( 90 N . m ) counterclockwise about ( B ) , if it's moment about ( A ) is zero , Determine the force ?

3 ‐ Forces of ( 4 N & 6 N ) act on the lamina ABCD as

shown in fig . Find the total moments of these forces about the point ( O ) .

4 ‐ Calculate the moment of the

( 300 N ) force on the handle of the monkey wrench about the center of the bolt .

5 ‐ A force F of magnitude 60 N is applied to the gear

Determine the moment of F about point O .



20

Moment of Couple Couples : A special case of moments is a couple. A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. It does not produce any translation, only rotation. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment.

Ex ( 1 ) Compute the magnitude and direction of the resultant couples action on the body shown Solution : Mc = 60 * 4 – 40 * 4 = 240 – 160 = 80 N .m Ex ( 2 ) A lug wrench is used to tighten a hex‐head bolt , Determine the magnitude ( F ) of the equal forces exerted on the six contact points as shown in fig. Solution On the lug wrench : Mc = F * d = 250 * 350 *2 = 175000 N . mm On the nut : Mc = 3 F * d 175000 = 3 F * 25 F = 175000 / 75 = 2333.33 N



21

Ex ( 3 ) Determine the moment associated with the forces shown in fig. Solution Mc = F * d = 1.5 * 16 = 24 Ib .in Ex ( 4 ) In the design of lifting hook , the action of the applied force ( F ) at the critical section of the hook is a direct pull at ( B ) and a couple . if the magnitude of the couple is ( 4000 N.m ) , Determine the magnitude of the force ( F ) . Solution Mc = F * d F = Mc / d = 4000* 12 / 4 = 12000 N

Ex ( 5) : A die is being used to cut threads on a rod . if the ( 150 N ) forces are applied as shown , Determine the magnitude of ( F ) of the equal forces exerted on the ( 1/4 " ) rod by each of the four cutting surfaces so that their external effect on the rod is equivalent to that of the two ( 15 N ) forces . Solution Mc = F * d = 15 * 10 = 2 * F * 0.25 F = 300 N



22

Ex ( 6 ) : A multiple – drilling is used to drill simultaneously six holes in the steel plate shown in fig. . Each drill exerts clockwise couple of magnitude (5 N.m ) on the plate . Determine an equivalent couple formed by the smallest possible forces acting : a ‐ at A and C b ‐ at A and D c – on the plate Solution a –

Mc = 5 N.m * 6 holes = 30N.m Mc = F * d 30 = F * 0.3 F = 100 N

b ‐

NFF

dFMcmmAD

80375.0*30

* 375)300()225( 22

===

=+=

c –

NFF

dFMcmmBC

605.0*30

* 500)400()300( 22

===

=+=



23

PROBLEMS 1 – Replace the couple and force

shown by a single force ( F ) applied at a point ( D ) , locate ( D ) by determining the distance ( b ) .

2 – The indicated force – couple system is

applied to a small shaft at the center of rectangular plate , Replace this system by a single force and specify the coordinate of the point on the y – axis through which the line of action of this resultant force passes .

3 – The wrench is subjected to the

( 200 N ) force and the force ( P ) as shown in fig. if the equivalent of the two forces is a force ( R ) at ( O ) and a couple of ( 20 N . m ) , Determine ( P ) and ( R ) .

Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 6th Week 2008 - 2009


24

Resultant of forces system Resultant of non-concurrent coplanar forces system

We will find out the resultant force for many non – concurrent forces acting on a rigid body by using the following equations :

( ) ( )22

332211

332211

+=

==

yx

nny

nnx

RRR

FFFFRFFFFR

θθθθθθθθ

sin....sin.sin.sin.cos....cos.cos.cos.

mmmm

mmmm

The direction of resultant force may be determined as :

⎟⎠⎞

⎜⎝⎛

= −

x

y

RR1tanθ

Ex ( 1 ) : Determine the force ( P ) shown in fig. knowing that the resultant of the two forces pass through the point ( A ). Solution

NP

PPP

PP

MPR

PR

FFRPR

PR

FFR

A

y

x

68.6747.266

1800007.266180000

029125.24180000

0300*174*100*

171*300*600

0)2.(..........2425.0600sin

171*600sin

sin.sin. )1...(..........*9701.0cos

174*cos

cos.cos.

2211

2211

−==

==−+

=−+

=

+−=∗

+−=∗

=−=∗

−=∗

=

∑θ

θ

θθθ

θ

θθ

m

m



25

Ex ( 2 ) : Determine the distance from point ( A ) to the line of action of the resultant of the three forces shown in fig. Solution R = 2 – 8 – 4 = ‐ 10 KN MA = ‐ 4 * 1 – 8 * 1 + 2 * 3 = ‐ 6 KN.m MA = R * d ‐6 = ‐ 10 * d d = 6 / 10 = 0.6 m PROBLEMS 1 ‐ Find the resultant force for the

non ‐ concurrent coplanar forces system, shown in fig

2 – Two integral pulleys are subjected

to the belt tension shown . if the resultant of these forces passes through the center ( O ) , Determine ( T ) and the magnitude of ( R ) and the counterclockwise angle ( θ ) it makes with the x – axis .

3 – Replace the three forces acting on

the bent pipe by a single equivalent force ( R ) . specify the distance ( x ) from point ( O ) th the point on x – axis through which the line of action of ( R ) passes .



26

FREE BODY DIAGRAM Free body diagram : is a sketch to show all the forces and reactions acting on the body For example : The free body diagram includes external forces applied to the body and external reaction forces resulting from the method of supporting the body.

Free – body diagram and the mechanical effects

The name of the body The effect of the body Free-body diagram

Earth

W

Flixible cables

And ropes

Cantilever beam

Smooth surface

Rollers , balls ,

cylinders

Smooth pins



27

Ex ( 1 ) : Draw Free – body diagram for Solution the 50 N sphare shown in fig.

Ex ( 2 ) : Draw Free – body diagram for Solution the 50 N sphare shown in fig.

----------------------------------------------------------------------------------------------------- Ex ( 3 ) : Draw Free – body diagram for Solution the ropes system shown in fig.



28



29

Correct the Wrong or Incomplete F.B.D . in the right column of the table below :



30

Draw the FREE- BODY DIAGRAM ( F . B . D ) for the following :

Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009


31

Equilibruim 1- For Coplanar forces system :

a- concurrent coplanar forces system

Rx = 0 , R y = 0 , R = 0

b - non-concurrent coplanar forces system

Rx = 0 , R y = 0 , R = 0 , Σ M = 0

2 - Non coplanar forces system :

a. concurrent non-coplanar forces system

Rx = 0 , R y = 0 , R = 0 , Σ M = 0

b. non-concurrent non-coplanar forces system

Rx = 0 , R y = 0 , R = 0 , Σ M = 0

) : 1Ex ( Determine the tension in the cord and the reaction of inclined plane

acting on the sphere of ( 50 N ) weight shown in fig.

Solution we draw F.B.D for the sphere , then :

Σ Fx = 0

T cos 75 – R1 cos 35 = 0 ……….. ( 1 )

Σ Fy = 0

T sin 75 + R1 sin 35 – 50 = 0 ……....( 2 )

Subst. ( 2 ) in ( 1 ) we get :

T = 4361 N

R1 = 137.7 N



32

Ex ( 2 ) : Determine the reactions at the points

( A ) and ( B ) , the angle beam was in

equilibrium state as shown in fig .

Solution Σ M ( A ) = 0

500* 10 – N * 15 = 0

N = 5000 / 15 = 333.34 N

Σ Fy = 0

Ay + N – 500 = 0

Ay + 333.34 – 500 = 0

Ay = 166.67 N

Σ Fx = 0 , Ax = 0

RA = Ay = 166.67 N

Ex ( 3 ) : Determine the tension forces ( T1 )

and ( T2 ) in the equilibrium system

shown in fig. .

Solution Σ Fx = 0 T1 . cos ( 0 ) + T2 .cos ( 60 ) – 2000 cos (15 ) = 0 T1 + 0.5 T2 – 1931.85 = 0 …….. ( 1 )

Σ Fy = 0

T1 . sin ( 0 ) + T2 . sin ( 60 ) – 2000 . sin ( 15 ) – 100 = 0 0.866 T2 – 617.63 = 0 T2 = 713.2 N Subs. in ( 1 ) T1 = 1575 N



33

Ex ( 4 ) : Determine the tension in each cord shown in fig. ( TA , TB , TC , TD ) . Solution By using Lami' s rule :

NT

T

TNT

T

T

C

C

C

D

D

D

64341

408660

9040

120

201

4050

9040

150

..

sinsin

.

sinsin

=

=

=

=

=

=

NT

T

TTNT

T

TT

A

A

CA

B

B

CB

64347070

64347070

135135

98487070

643450

13590

..

..

sinsin

..

..

sinsin

=

=

=

=

=

=



34

Ex ( 5 ) : Find out the reaction on the cylinder ( A )

and the total force acting on the pin ( O )

Solution

Σ M ( O ) = 0

2 * 250 - RA * 400 = 0

400 RA = 500

RA = 1.25 KN

Σ Fy = 0

Oy – 2 = 0

Oy = 2 KN

Σ Fx = 0

RA – Ox = 0

Ox = RA = 1.25 KN

NF

OyOxF

3522251 22

22

.)().(

)()(

=+=

+=



35

Ex ( 6 ) : Find out the reactions at the points ( A ) & ( B ) .

Solution

Σ M ( A ) = 0

- RA * 600 + 300 * 9.8 * 1000 = 0

- 600 RA + 2940 000 = 0

RA = 4900 N = 4.9 KN

Σ Fy = 0

RA – RB - 300 * 9.8 = 0

4900 - RB - 2943 = 0

RB = 1950 N = 1.95 KN



36

PROBLEMS 1 - A ( 200 N ) weight of the block shown in fig. is

supported by a pin and bracket at ( A ) and by a cable ( BC ) , Determine the reaction at ( A ) and the tension in the cable .

2 - The 200-N force produces a torque (moment) of

40 N.m about the axis of the bolt in order to tighten the hexagonal nut . Find the forces between the smooth jaws of the wrench and the nut if contact occurs at the corners A and B of the hexagon .

3 - The pin at A can support a maximum force of 3.2 KN .

What is the corresponding maximum load L which can be supported by the bracket ?

4 - A block placed under the head , of the claw hammer as

shown greatly facilitates the extraction of the nail . If the 200-N pull on the handle is required to pull the nail calculate the tension T in the nail and the magnitude A of the force exerted by the hammer head on the block . The contacting surfaces at A are sufficiently rough to prevent slipping



37

5 - The ring supports the 1000-N load and is held in position by the two cables attached to vertical walls . Find the tensions T1 and T2 by at least two different ways .

6 - The cable from A to B is 6 m long and supports

the 100-kg crate from the small pulley . Calculate the tension T in the cable .



38

Friction

Friction force : is the force resisting the relative motion of two surfaces in contact There are two types of friction : Static friction Static friction is friction between two objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The coefficient of static friction, typically denoted as μs, is usually higher than the coefficient of kinetic friction. Kinetic friction Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). The coefficient of kinetic friction is typically denoted as μk, and is usually less than the coefficient of static friction. The coefficient of friction: (also known as the frictional coefficient) is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction (the two materials slide past each other easily), while rubber on pavement has a high coefficient of friction (the materials do not slide past each other easily). Coefficients of friction range from near zero to greater than one – under good conditions The coefficient of friction is a dimensionless quantity symbolized by the Greek letter( μ ) and is used to approximate the force of friction. Friction can be viewed, again as an approximation, as being of two primary types, static or kinetic. The coefficient of static friction is defined as the ratio of the maximum static friction force (F) between the surfaces in contact to the normal (N) force. The coefficient of kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces in contact to the normal force:

NFf

=µ

µ : coefficient of friction fF : Friction force

N : Normal force Both static and kinetic coefficients of friction depend on the pair of surfaces in contact; their values are usually approximately determined experimentally. For a given pair of surfaces, the coefficient of static friction is usually larger than that of kinetic friction; in some sets the two coefficients are equal, such as Teflon‐on‐Teflon.



39

Angle of friction ( α ) The angle of friction for two surfaces in contact is defined as the angle that the maximum contact force makes with the direction of normal force

)(tan)tan(1 µα

αµ−=∴

=

EX ( 1 ) : In fig. shown , a = 12 cm , b = 18 cm , h = 15 cm ,

W = 100 N , µ = 0.24 , P = 24 N Is the body sliding or turning over or stay at rest ? Solution :

cmPaWh 5.2

)24(2)12(100

*2.

max ==

When bh >max then the force doesn't effect on the body , therefore , the body is not turnover .

PFWF ==== maxmax , 24)100(24.0.µ Then the body tends to slide and doesn't turnover EX (2 ) : A block with ( 200 N weight rests on a rough horizontal plane , is subjected to the force ( P = 40 N ) which inclined ( 25 o ) . Determine the coefficient of friction. Solution :

17.0217

25.36)217(25.36

.21720025sin40020025sin40

0

25cos400

==

=

==+==−−

=

=

=

∑

∑

µ

µ

µ NFNN

N

F

FF

f

y

f

x



40

EX (2 ) : A wooden block ( 3000 N ) weight , the coefficient of friction between the block and the floor is ( 0.35 ) , Determine whether pushing or pulling process by the force ( P ) is suitable to make the block tend to move to the right with a least force ( P ) , ( i . e , which is easy pushing or pulling ) Solution : In case of Pushing

NP

PPP

PP

NFPNPN

F

PF

FPF

f

y

f

f

x

4.1519691.0

10501050691.0

1050175.0866.0)30005.0(35.0866.0

.300030sin

0300030sin

0

866.0

030cos0

==

=+=

+=

=+=

=−−

=

=

=−

=

∑

∑

µ

In case of Pulling

NP

PPP

PP

NFPN

PNPN

F

PF

FPF

f

y

f

f

x

6.1008041.1

10501050041.1

175.01050866.0)5.03000(35.0866.0

.5.03000300030sin

0300030sin

0

866.0

30cos0

==

=−=

−=

=−+=

+−==−+

=

=

=

=

∑

∑

µ



41

EX ( 3 ) : The ( 85 N ) force ( P ) is applied to the ( 200 N ) crate . Determine the magnitude and

direction of the friction force ( fF ) exerted

by the horizontal surface on the crate . Solution :

N 85F , volidassumption,

100)200(5.0.200

0

850

mequilibriu Assume

max

=<

====

=

==

=

∑

∑

f

f

y

f

x

FF

NNFNN

F

NPFF

µ

EX ( 4 ) : The tongs are designed to handle hot steel tubes which are being heat‐ treated in an oil bath , for a ( 20o ) jaw opening . What is the maximum coefficient of friction between the jaws and the tube which will enable the tongs to grip the tube without slipping. Solution :

176.0 10tan

tanmin

=== φµ



42

EX ( 5 ) : The magnitude of the force ( P ) is slowly increased .Does the homogeneous box of mass ( m ) slip or tip first ? State the value of ( P ) which would cause each occurrence . Neglect any effect of the size of the small feet . Solution :

( )

( )

( ) ( )occur will,

53.030sin230cos

0*2*30sin*30cos0)(

44.05.0*5.0866.0

*5.0

30sin.30cos.

.)30sin.30(cos.30sin.30cos

30sin..30cos.30sin...30cos

.)30sin(30cos: ) 1 (in ) 2 ( subst.

) 2 .........( 30sin030sin

0) 1 .(.......... ..30cos

030cos... , .

030cos0

Slips

SlippingthentipPslipP

tipPmgmgP

dmgdPdPCM

Tips

slipPmgmg

mgP

mgPmgPP

PmgPNPNmgP

NPNmgP

PmgNNPmgNN

FNNP

PNNNFNF

PFFF

CC

CC

CB

CB

y

CB

CB

CCBB

CB

x

<

==+

=

=−+

=

==+

=

+=

=+=+

−=+−−=

+−−=

−+−==+−+

=

+==+−−

==++−−

=

∑

∑

∑

Q

µµ

µµµµ

µµµµµµ

µµ

µµµµ

µµ



43

EX ( 6 ) : The body ( A ) is to be tend to move to the right by the inclined force ( P ) which is ( 25 N ) in case of pushing and it is ( 10 N ) in case of pulling , Determine the weight of the body and the coefficient of friction between the body and the floor . Solution :

ans 74.0565.16

66.85

66.8ans 65.16

)5(65.21)5.12(66.85

66.85.12

65.215

66.8

)5.(66.8.530sin10

030sin100

66.830cos100Pulling of caseIn

5.1265.21

)5.12.(65.21.5.1230sin25

030sin25

0

65.2130cos250Pushing of caseIn

=−

=−

=

=−=+

−=

+

−=

−=⇒=−=+−=

=−+⇒=

==⇒=

+=

+=⇒=+=+=

=−−

=

==⇒=

∑∑

∑∑

W

WWW

WW

W

WNFWWN

WNF

NFF

W

WNFWWN

WN

F

NFF

f

y

fx

f

y

fx

µ

µ

µµ

µ

µµ



44

PROBLEMS 1 ‐ The system of two blocks , cable and

fixed pulley is initially at rest . Determine the horizontal force ( P ) necessary to cause motion , then determine the corresponding tension ( T ) in the cable .

2 -Determine the magnitude ( P ) of the

horizontal force required to initiate motion of the block of mass ( m o ) for the cases : a – P is applied to the right . b – P is applied to the lift .

3 – The sliding glass door rolls on the

two small lower wheels ( A ) and ( B ) . Under normal conditions the upper wheels don't touch their horizontal guide . a – compute the force ( P ) required

to slide the door at a steady speed if the wheel ( A ) becomes " frozen " and does not turn in its bearing .

b – Rework the problem if wheel ( B ) becomes " frozen " instead of wheel ( A ) . if the coefficient of friction between a frozen wheel and supporting surface is ( 0.3 ) , and the center of mass of the ( 140 N ) door is at its geometric center . Neglect the small diameters of the wheel

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