محاضرات الميكانيك الهندسي
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© 2008جميع الحقوق محفوظة اليجوز استنساخ او طبع او توزيع او نشر اي نسخة من المحاضرات بدون اذن من المعهد التقني في الناصرية
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 )
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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 ) ?
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 1st & 2nd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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.
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 3rd Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 4th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 4th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 4th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 4th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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 .
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 5th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 5th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 5th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
===
=+=
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 5th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 6th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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 .
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 7th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 7th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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.
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 7th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
28
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 7th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
29
Correct the Wrong or Incomplete F.B.D . in the right column of the table below :
Engineering Mechanics – Statics For 1st Year Students – Mechanic Departement 7th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
=
=
=
=
=
=
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
.)().(
)()(
=+=
+=
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 8th&9th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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 .
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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.
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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
Engineering Mechanics – Statics For 1st Year Students – Mechanic Department 10th Week 2008 - 2009
By : Eng . YOUNIS FAKHER
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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