Pawel Keblinski

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ENGR-1100 Introduction to Engineering Analysis

• Pawel Keblinski
• Materials Science and Engineering MRC115
• Office hours Tuesday 1-3
• phone [phone]
• email keblip_at_rpi.edu

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ENGR-1100 Introduction to Engineering Analysis

• TA Igor Bolotnov, Mechanical, Aerospace and
  Nuclear Engineering JEC5204
• Office hours Tuesday 3-5
• phone [phone] emailboloti_at_rpi.edu

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ENGR-1100 Introduction to Engineering Analysis

• SI (Supplemental Instruction)
• (Begins Wed. Sept. 8)
• Sun. 8pm-10pm - DCC 330
• Wed 8pm-10pm - Sage 5510

Drop-in Tutoring - DCC 345 (Begins Tue. Sept.
7th) Sundays - 3-5pm M-Th - 7-9pm
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ENGR-1100 Introduction to Engineering Analysis

• Lecture 3

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Lecture Outline

• Rectangular components of a force
• Resultant force by rectangular components

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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.
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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
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Example

• Determine the x and y scalar component of the
  force shown in figure 2-47.
• Express the force in Cartesian vector form.

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Solution
x
Fxcos(570) 275149.8 lb
Fysin(570) 275230.6lb
F Fxi Fyj149.8 i 230.6 j
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Class Assignment Exercise set 2-49 please submit
to TA at the end of the lecture

• Determine the x and y scalar components of the
  force shown in figure 2-49.
• Express the force in Cartesian vector form.

y
Solution a) Fx-440lb Fy-177.9lb b) F-440 i
- 177.9 j lb
x
220
F475 lb
Figure 2-49
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Three Dimension Rectangular Components of a Force
F Fx i Fy j Fz k F cos qx i F cos
qy j F cos qz k FeF
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The Scalar Components of a Force
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Azimuth Angle
q- azimuth angle f- elevation angle
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Finding the direction of a force by two points
along its line of action
A(xA, yA, zA) B(xB, yB, zB)
yB-yA (xB-xA)2 (yB-yA)2
(zB-zA)2
cosqy
zB-zA (xB-xA)2 (yB-yA)2
(zB-zA)2
cosqz
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Example

• For the force shown in the figure
• Determine the x, y, and z scalar components of
  the force.
• Express the force in Cartesian form.

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Solution
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Fx Fxy cos q-237.5sin(370)-142.9 N Fy Fxy
sin q-237.5cos(370)-189.7 N
Or if we follow the obtained formula Fx Fxy cos
q237.5cos(2330)-142.9 N Fy Fxy sin
q237.5sin(2330)-189.7 N
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Class Assignment Exercise set 2-55 please submit
to TA at the end of the lecture

• For the force shown in Fig. P2-58
• Determine the x,y, and z scalar components of the
  force.
• Express the force in Cartesian form.

z
Solution a) Fx-583 lb Fy694lb Fz423
lb b) F-583 i 694 j 423 k lb
F1000 lb
250
y
1300
x
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Resultant Force by Rectangular Components
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Rx (Ax Bx )
Ry (Ay By )
Rz (Az Bz )
The magnitude
The direction
qxcos-1(Rx/R) qycos-1(Ry/R) qzcos-1(Rz/R)
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Example
Determine the magnitude and direction of the
resultant force of the following three forces.
Solution
RF1F2F3-383 i 269.6 j
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RF1F2F3-383 i 269.6 j
The force magnitude
R
R 468.4 N
qxcos-1(Rx/R)
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Class Assignment Exercise set 2-71 please submit
to TA at the end of the lecture
Determine the magnitude R of the resultant of the
forces and the angle qx between the line of
action of the resultant and the x-axis, using the
rectangular component method
y
F1600 lb
Solution R1696 lb 10.220
F2700 lb
450
150
x
300
F3800 lb
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The scalar (or dot) product
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
ABBAAB cos(q)
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• The scalar product of the two vectors written in
  Cartesian form are
• AB (Ax i Ay j Az k) (Bx i By j Bz
  k)
• Ax Bx (ii) Ax By (ij) Ax Bz (ik)
• Ay Bx (ji) Ay By (jj) Ay Bz (jk)
• Az Bx (ki) Az By (kj) Az Bz (kk)

Therefore AB Ax Bx Ay By Az Bz
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Example
Determine the angle q between the following
vectors A3i 0j 4k and B2i -2j
5k
AB320(-2)4526
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Orthogonal (perpendicular) vectors
(w1 ,w2 ,w3)
z
w
(v1 ,v2 ,v3)
v
y
x
w and v are orthogonal if and only if wv0
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Properties of the dot product
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Using dot product for a force system

• Determine
• The magnitude and direction (qx, qy, qz)
• 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 a between force F1 and F2.

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Solution

• F1 F1 e1

e1 1.5/(1.52624.52)1/2 i 6/(1.52624.52)1/2
j 4.5/(1.52624.52)1/2 k e1 0.196 i
0.784 j 0.588 k
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L1(221.52)1/22.5 ft
L22.5 tan(600)4.33 ft

• F2 F2 e2

e2 1.5/(1.52(-2)24.332)1/2 i
-2/(1.52(-2)24.332)1/2 j 4.33
/(1.52(-2)24.332)1/2 k e2 0.3 i - 0.4 j
0.866 k
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R F1 F2 130.8 i 139.35 j 384.3 k lb
R 429 lb
qxcos-1(Rx/R) qycos-1(Ry/R) qzcos-1(Rz/R)
qxcos-1(130.8/429)
qx72.20
qycos-1(139.35/429)
qy71.10
qzcos-1(384.3/429)
qz26.40
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b) The magnitude of the rectangular component of
the force F1 along the line of action of force
F2.
F1e2(58.8 i 235.3 j 176.5 k)(0.3 i - 0.4 j
0.866 k) 58.80.3235.3(-0.4)176.50.86676 lb
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c) The angle a between force F1 and F2.
F1F2 F1 F2 cos(a)
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Class assignment Exercise set 2-63

• Two forces are applied to an eye bolt as shown in
  Fig. P2-63.
• Determine the x,y, and z scalar components of
  vector F1.
• Express vector F1 in Cartesian vector form.
• Determine the angle a between vectors F1 and F2.

Fig. P2-63
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Solution
9.7 ft
F1x F1 cos(qx) 900 (6)/9.7-557 lb
F1y F1 cos(qy) 900 3/9.7278.5 lb
F1z F1 cos(qz) 900 7/9.7649.8 lb
b) Express vector F1 in Cartesian vector form.
F1 -557 i 278.5j 649.8 k lb
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c) Determine the angle a between vectors F1 and
F2.
cos(a) F1e2 / F1 e2
F1 -557 i 278.5j 649.8 k lb
e2
6/9j
3/9 k
-0.67 i 0.67 j 0.33 k
-6/9 i
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Class Assignment Exercise set 2-78 please submit
to TA at the end of the lecture
Determine the magnitude R of the resultant of the
forces and the angles qx, qy, qz between the line
of action of the resultant and the x-, y-, and
z-coordinate axes, using the rectangular
component method.
Solution R28.6 kN qx 82.20 qy 69.60 qz 22.00