Lecture 3 - Rensselaer Polytechnic Institute · Lecture outline • Forces as vectors • Resultant...
Transcript of Lecture 3 - Rensselaer Polytechnic Institute · Lecture outline • Forces as vectors • Resultant...
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Lecture 3Notes courtesy of: Prof. Yoav Peles
ENGR-1100 Introduction toEngineering Analysis
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Lecture outline
• Forces as vectors• Resultant of two concurrent forces
• Resultant of three or more concurrentforces
• Resolution of force into components
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Characteristics of a force
• A force is characterized by the following:
(1) Magnitude
(2) Direction
(3) Point of application
FB
FA
F
B
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x
y
F
θ
Force representation
x
y
F6m
5m
x
y
z
θx
θy
θz
F
x
y
z
F
7 m
5 m
4 m
Two dimensions
Three dimensions
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Principle of Transmissibility
• The external effect of a force on a rigidbody is the same for all points ofapplication of the force along its line ofaction (force is a sliding vector!)
Push PullLine of action
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Forces classification
• Contact or surface forces• Example: push or pull
• Body forces• Example: gravitational forces, magnetic forces
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Concurrent forces
• A force system is said to be concurrent ifthe action lines of all forces intersect at acommon point
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Example(a) Determine the x and y scalar components of the force
shown in the figure.
(b) Express the force in Cartesian vector form.
F=275 lb
x
y
570
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Solution
Fx = 275 * cos(570) = 149.8 lb
Fy = 275 * sin(570) = 230.6 lb
x
F=275 lby
570
Fy
Fx
F= Fxi+ Fyj=(149.8 i + 230.6 j) (lb)
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3-D rectangular components of a force
x
y
z
F
Fx=Fxi
Fz=Fzk
Fy=Fyjθx θy
θz
F = Fx + Fy + Fz= Fx i + Fy j + Fz k
= F cos θx i + F cos θy j + F cos θz k = FlF
WherelF= cos θx i + cos θy j + cos θz k is a unitvector along the line of action of the force.
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The scalar components of a forceFx = F cos θx ; Fy = F cos θy ; Fz = F cos θz ;
θx=cos-1(Fx/F); θy=cos-1(Fy/F); θz=cos-1(Fz/F);
F= Fx2 + Fy
2 + Fz2
x
y
z
F
Fx
Fyθx θy
θz
Fz cos2 θx +cos2 θy +cos2 θz=10< θ <1800
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Azimuth angle
x
y
z
F
θ φ
θ- azimuth angleφ- elevation angle
θ
x
y
z
Fxy
Fz
Fxy
Fxy = F cos φ ;
Fz
Fz = F sin φ
Fx
Fx= Fxy cos θ =F cos φ cos θ
Fy
Fy= Fxy sin θ =F cos φ sin θ
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Finding the direction of a force bytwo points along its line of action
x
y
z
FΒΑ
xB
xA
yB
yA
zBzA
cosθx=xB-xA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
cosθy=yB-yA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
zB-zA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2cosθz=
Α (xA, yA, zA) ; Β (xB, yB, zB)
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Example
For the force showna) Determine the x, y, and z scalar components of the force.b) Express the force in Rectangular form.
x
y
zF=475 N
370
300
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Solution
x
y
zF=475 N
370
300
Fz
Fxy
Fz = F sin φ = F sin(600) =411.4 N
Fxy = F cos φ = F cos(600) =237.5 N
φ
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F = (-142.9 i – 189.7 j + 411.4 k) Nb)
Fx
Or if we follow the obtained formula:Fx= Fxy cos θ =237.5*cos(2330 )= -142.9 NFy= Fxy sin θ =237.5*sin(2330) = -189.7 N
x
y
z
Fz
370
Fxy=237.5 N
θ
Fx= Fxy cos θ =−237.5*sin(370 )= -142.9 N
Fy= Fxy sin θ =−237.5*cos(370) = -189.7 N
Fy
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Resultant by rectangularcomponents
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
The sum of the two forces are:
R=A+B=(Ax i + Ay j + Az k) + (Bx i + By j + Bz k) =
Rx= (Ax +Bx )i; Ry= (Ay +By )j; Rz= (Az +Bz )k
z
x
y
A
B
(Ax +Bx )i+ (Ay +By ) j + (Az +Bz) k
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The magnitude: R= Rx2 + Ry
2 + Rz2
The direction:
x=cos-1(Rx/R); y=cos-1(Ry/R); z=cos-1(Rz/R);
Rx= (Ax +Bx )i; Ry= (Ay +By )j; Rz= (Az +Bz )k
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ExampleDetermine the magnitude and direction of theresultant of the following three forces.
Solution:
F1=350 i
F2=500*cos(2100) i +500*sin(2100)j
F3= 600*cos(1200) i + 600*sin(1200)j
F1=350 i
F2= -433 i -250 j
F3= -300 i + 519.6 j
R=F1+F2+F3=-383 i + 269.6 j
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R=F1+F2+F3=-383 i + 269.6 j
The force magnitude:
x=cos-1(Rx/R)
R= Fx2 + Fy
2 =
R= 468.4 N
x=cos-1(Rx/R)= cos-1(-383/468.4)=144.80
x=144.80
R
The direction:
x3832 + 269.62
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Example
x
y
z
F1 =300 lb
600
1.5 ft6 ft2 ft
4.5 ft
F2 =240 lb
Determine:
a) The magnitude and direction (x, y, z)
of the resultant force.
b) The magnitude of the rectangular
component of the force F1 along
the line of action of force F2.
c) The angle between force F1 and F2.
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Solution
• F1= F1l1
l1 = 1.5/(1.52+62+4.52)1/2 i + 6/(1.52+62+4.52)1/2 j+4.5/(1.52+62+4.52)1/2 k
l1 = 0.196 i + 0.784 j+ 0.588 k
F1 = (58.8 i + 235.3 j+ 176.5 k) lb
x
y
z
F1 =300 lb
60
1.5 ft6 ft2 ft
4.5 ft
F2 =240 lb
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L1=(22+1.52)1/2=2.5 ft
x
y
z
F1 =300 lb
600
1.5 ft6 ft2 ft
4.5 ft
F2 =240 lb
L1
L2
L2=2.5 tan(600)=4.33 ft
• F2= F2 e2
e2 = 1.5/(1.52+(-2)2+4.332)1/2 i -2/(1.52+(-2)2+4.332)1/2 j+4.33 /(1.52+(-2)2+4.332)1/2 k
e2 = 0.3 i - 0.4 j+ 0.866 k
F2= (72 i - 96 j+ 207.8 k) lb
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R= F1 + F2 = 130.8 i + 139.35 j+ 384.3 k lb
R= Rx2 + Ry
2 + Rz2 = 130.82 + 139.352+384.32 = 429 lb
x=cos-1(Rx/R); y=cos-1(Ry/R);z=cos-1(Rz/R);
R= 429 lb
x=cos-1(130.8/429)
y=cos-1(139.35/429)
z=cos-1(384.3/429)
x=72.20
y=71.10
x=26.40