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Pawel Keblinski
Materials Science and Engineering MRC115
Office hours: Tuesday 1-3
phone: (518) 276 6858
email: [email protected]
ENGR-1100 Introduction to Engineering Analysis
TA: Igor Bolotnov, Mechanical, Aerospace and Nuclear Engineering
JEC5204
Office hours: Tuesday 3-5
phone: (518) 276 812 email:[email protected]
ENGR-1100 Introduction to Engineering Analysis
SI (Supplemental Instruction)
(Begins Wed. Sept. 8)
Sun. 8pm-10pm - DCC 330
Wed 8pm-10pm - Sage 5510
ENGR-1100 Introduction to Engineering Analysis
Drop-in Tutoring - DCC 345
(Begins Tue. Sept. 7th)
Sundays - 3-5pm
M-Th - 7-9pm
Lecture 3
ENGR-1100 Introduction to Engineering Analysis
Lecture Outline
• Rectangular components of a force
• Resultant force by rectangular components
Rectangular Components of a Force
A force F can be resolved into a rectangular component Fx along the x-axis and a rectangular component Fy along the y-axis . The forces Fx and Fy are the vector components of the force F.
FFy
Fx
x
y
The force F and its two dimensional vector components Fx
and Fy can be written in Cartesian vector form by using unit
vectors i and j.
F = Fx + Fy= Fx i + Fy j
F=| F |
Fx = F cos F= Fx2+
Fy2
Fy = F sin tan-
1(Fy/Fx)
FFy
Fx
x
y
Example
(a) Determine the x and y scalar component of the force shown in figure 2-47.
(b) Express the force in Cartesian vector form.
F=275 lb
x
y
Figure 2-47
570
Solution
x
F=275 lby
570
Fx=cos(570) *275=149.8 lb
Fy=sin(570) *275=230.6lb
Fy
Fx
F= Fxi+ Fyj=149.8 i + 230.6 j
Class Assignment: Exercise set 2-49please submit to TA at the end of the lecture
F=475 lb
x
y
Figure 2-49
220
(a) Determine the x and y scalar components of the force shown in figure 2-49.
(b) Express the force in Cartesian vector form.
Solution
a) Fx=-440lb
Fy=-177.9lb
b) F=-440 i - 177.9 j lb
Three Dimension Rectangular Components of a Force
x
y
z
F
Fx=Fxi
Fz=Fzk
Fy=Fyjx y
z
F = = Fx i + Fy j + Fz k
= F cos x i + F cos y j + F cos z k = FeF
Where eF= cos x i + cos y j + cos z k is a unit
vector along the line of action of the force.
The Scalar Components of a Force
x
y
z
F
Fx
Fyx y
z
Fz
Fx = F cos x; Fy = F cos y; Fz = F cos z;
xcos-1(Fx/F); ycos-1(Fy/F); zcos-1(Fz/F);
F= Fx2 + Fy
2 + Fz2
cos2 x+cos2 y+cos2 z=10< <1800
Azimuth Angle
x
y
z
F
- azimuth angle- elevation angle
x
y
z
Fxy
Fz
Fxy
Fxy = F cos
Fz
Fz = F sin ;
Fx
Fx= Fxy cos Fcos cos
Fy
Fy= Fxy sin Fcos sin
Finding the direction of a force by two points along its line of action
x
y
z
F
xB
xA
yB
yA
zBzA
cosx= xB-xA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
cosy= yB-yA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
cosz= zB-zA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
xA, yA, zAxB, yB, zB
Example
For the force shown in the figurea) Determine the x, y, and z scalar components of the force.b) Express the force in Cartesian form.
x
y
zF=745 N
Solution
x
y
zF=745 N
Fz
Fxy
Fz = F sin = 745 sin(600)=411.4 N
Fxy = F cos = 745 cos(600)=237.5 N
Fx
Or if we follow the obtained formula:Fx= Fxy cos 237.5*cos(2330)=-142.9 NFy= Fxy sin 237.5*sin(2330)=-189.7 N
x
y
z
Fz
Fxy=237.5 N
Fx= Fxy cos 237.5*sin(370)=-142.9 N
Fy= Fxy sin 237.5*cos(370)=-189.7 N
Fy
F=-142.9 i – 189.7 j + 411.4 kb)
Class Assignment: Exercise set 2-55please submit to TA at the end of the lecture
For the force shown in Fig. P2-58a) Determine the x,y, and z scalar components of the force.b) Express the force in Cartesian form.
x
y
z
F=1000 lb
Solution
a) Fx=-583 lb
Fy=694lb
Fz=423 lb
b) F=-583 i +694 j + 423 k lb
Resultant Force by Rectangular Components
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 ); Ry= Ay +By ); Rz= Az +Bz )
z
x
y
A
B
Ax +Bx )i + Ay +By ) j + Az +Bz) k
The magnitude: R= Rx2 + Ry
2 + Rz2
The direction:
xcos-1(Rx/R); ycos-1(Ry/R); zcos-1(Rz/R);
Rx= Ax +Bx ); Ry= Ay +By ); Rz= Az +Bz )
ExampleDetermine the magnitude and direction of the resultant force of the following three forces.
Solution:F1=350 i+ 0 j + 0 k
F2=500*cos(2100) i +500*sin(2100)j+0 k
F3= 600*cos(1200) i + 600*sin(1200)j+0 k
F1=350 i
F2= -433 i -250 j
F3= -300 i + 519.6 j
R=F1+F2+F3=-383 i + 269.6 j
R=F1+F2+F3=-383 i + 269.6 j
The force magnitude:
xcos-1(Rx/R)
R= Fx2 + Fy
2 + Fz2 =
R= 468.4 N
xcos-1(Rx/R)= cos-1(-383/468.4)=144.80
x=144.80
R
3832 + 269.62+02
The direction:
x
Class Assignment: Exercise set 2-71please submit to TA at the end of the lecture
Determine the magnitude R of the resultant of the forces and the angle x between the line of action of the resultant and the x-axis, using the rectangular component method
x
F1=600 lb
y
450 F2=700 lb
F3=800 lb
150
300
Solution:
R=1696 lb 10.220
The scalar (or dot) product
A•B=B•A=AB cos()
0< <1800
Finding the rectangular scalar component of vector A along the x-axis
Ax=A•i=A cos(x)
The scalar product of two intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them
Along any direction
An=A•en=A cos(n)
A
n
y
en
et
At=A-An
The scalar product of the two vectors written in Cartesian form are:
A•B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k)
Ax Bx (i•i) + Ax By (i•j) + Ax Bz (i•k)+
Ay Bx (j•i) + Ay By (j•j) + Ay Bz (j•k)+
Az Bx (k•i) + Az By (k•j) + Az Bz (k•k)
Therefore: A•B = Ax Bx + Ay By + Az Bz
Since i, j, k are orthogonal:
i•j= j•k= k•j=(1)*(1)*cos(900)=0
i•i= j•j= k•k=(1)*(1)*cos(00)=1
Example
Determine the angle between the following vectors:
A=3i +0j +4k and B=2i -2j +5k
A•B=AB cos() cos()= A•B/ AB
A•B=3*2+0*(-2)+4*5=26
cos()= 26/28.7 = 25.10
Orthogonal (perpendicular) vectors
(v1 ,v2 ,v3)
x
z
v
(w1 ,w2 ,w3)
w
w and v are orthogonal if and only if w·v=0
y
Properties of the dot product
If u , v, and w are vectors in 2- or 3- space and k is a scalar, then:
a) u·v= v·u
b) u·(v+w)= u ·v + u·w
c) k(u·v)= (ku) ·v = u·(kv)
d) v·v> 0 if v=0, and v·v= 0 if v=0
x
y
z
F1 =300 lb
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.
Using dot product for a force system
Solution
• F1= F1 e1
e1 = 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
e1 = 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
1.5 ft6 ft2 ft
4.5 ft
F2 =240 lb
x
y
z
F1 =300 lb
1.5 ft6 ft2 ft
4.5 ft
F2 =240 lb
L1
L1=(22+1.52)1/2=2.5 ftL2
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
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
xcos-1(Rx/R); ycos-1(Ry/R); zcos-1(Rz/R);
R= 429 lb
xcos-1(130.8/429)
ycos-1(139.35/429)
zcos-1(384.3/429)
x72.20
y71.10
z26.40
b) The magnitude of the rectangular component of the force F1 along the line of action of force F2.
F1•e2=(58.8 i + 235.3 j+ 176.5 k)•(0.3 i - 0.4 j+ 0.866 k)=
58.8*0.3+235.3*(-0.4)+176.5*0.866=76 lb
c) The angle between force F1 and F2.
F1•F2= F1 F2 cos() cos()= F1•F2 / F1 F2
F1•F2= 58.8 i + 235.3 j+ 176.5 k)•(72 i - 96 j+ 207.8 k )=
58.8*72+235.3*(-96)+176.5*207.8=18321 lb
F1 F2=72000
cos()=18321/72000
Class assignment: Exercise set 2-63
x y
z
3 ft 3 ft
4 ft
3 ft
6 ftF2=700 lb
F1=900 lb
Two forces are applied to an eye bolt as shown in Fig. P2-63.
a) Determine the x,y, and z scalar components of vector F1.
b) Express vector F1 in Cartesian vector form.
c) Determine the angle between vectors F1 and F2.
Fig. P2-63
Solution
x y
z
3 ft 3 ft
4 ft
3 ft
6 ftF2=700 lb
F1=900 lb
d1 = (-6)2 +32 +72
F1x = F1 cos(x) =900 *{(–6)/9.7}=-557 lb
9.7 ft
d1 = x12+ y1
2+ z12
F1y = F1 cos(y) =900 *{3/9.7}=278.5 lb
F1z = F1 cos(z) =900 *{7/9.7}=649.8 lb
b) Express vector F1 in Cartesian vector form.
F1 = -557 i + 278.5j+ 649.8 k lb
c) Determine the angle between vectors F1 and F2.
x y
z
3 ft 3 ft
4 ft
3 ft
6 ftF2=700 lb
F1=900 lb
F1 = -557 i + 278.5j+ 649.8 k lb
d2 = (-6)2 +62 +32 9 ft
e2 =
cos()= F1•e2 / F1 e2
F1•e2 = (-557)*(-2/3)+278.5*2/3+649.8*0.33=771.4lb
cos()=771.4/900 =310
d2 = x22+ y2
2+ z22
+ 6/9j + 3/9 k = -0.67 i + 0.67 j + 0.33 k-6/9 i
Class Assignment: Exercise set 2-78please submit to TA at the end of the lecture
Determine the magnitude R of the resultant of the forces and the angles x, y, z between the line of action of the resultant and the x-, y-, and z-coordinate axes, using the rectangular component method.
Solution:
R=28.6 kN
x= 82.20
y= 69.60
z= 22.00