ENGR-1100 Introduction to Engineering Analysis Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu

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:boloti@rpi.edu

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

ENGR-1100 Introduction to Engineering Analysis Lecture 3

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. Fy Fx x y F

F Fy Fx F = Fx + Fy= Fx i + Fy j q 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 q F= Fx2+ Fy2 Fy = F sin q q=tan-1(Fy/Fx)

Example Determine the x and y scalar component of the force shown in figure 2-47. Express the force in Cartesian vector form. F=275 lb x y Figure 2-47 570

Solution F=275 lb Fy 570 Fx Fx=cos(570) *275=149.8 lb Fy=sin(570) *275=230.6lb F= Fxi+ Fyj=149.8 i + 230.6 j

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 F=475 lb Figure 2-49

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 Where eF= cos qx i + cos qy j + cos qz k is a unit vector along the line of action of the force. x y z F Fx=Fxi Fz=Fzk Fy=Fyj qx qy qz

The Scalar Components of a Force Fx = F cos qx ; Fy = F cos qy ; Fz = F cos qz ; qx=cos-1(Fx/F); qy=cos-1(Fy/F); qz=cos-1(Fz/F); F= Fx2 + Fy2 + Fz2 cos2 qx +cos2 qy +cos2 qz=1 0< q <1800 x y z F Fz qx qy qz Fy Fx

Azimuth Angle x y z F x y z Fz = F sin f ; Fy= Fxy sin q=F cos f sin q Fxy = F cos f; Fx Fx= Fxy cos q=F cos f cos q q- azimuth angle f- elevation angle

Finding the direction of a force by two points along its line of action x y z F B A xB xA yB yA zB zA A(xA, yA, zA) ; B(xB, yB, zB) cosqx= xB-xA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 yB-yA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 cosqy= zB-zA (xB-xA)2+ (yB-yA)2+ (zB-zA)2 cosqz=

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. x y z F=745 N 370 300

Solution z F=745 N Fz Fxy y x Fz = F sin f = 745 sin(600)=411.4 N 370 300 Fz Fz = F sin f = 745 sin(600)=411.4 N Fxy = F cos f = 745 cos(600)=237.5 N f Fxy

x y z Fz Fxy=237.5 N Fx Fy F=-142.9 i – 189.7 j + 411.4 k b) 370 Fxy=237.5 N q Fx Fy Fx= Fxy cos q=-237.5*sin(370)=-142.9 N Fy= Fxy sin q=-237.5*cos(370)=-189.7 N Or if we follow the obtained formula: Fx= Fxy cos q=237.5*cos(2330)=-142.9 N Fy= Fxy sin q=237.5*sin(2330)=-189.7 N F=-142.9 i – 189.7 j + 411.4 k b)

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 Fy=694lb Fz=423 lb b) F=-583 i +694 j + 423 k lb F=1000 lb 250 y 1300 x

Resultant Force by Rectangular Components z x y A B 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) = (Ax +Bx )i + (Ay +By ) j + (Az +Bz) k Rx= (Ax +Bx ); Ry= (Ay +By ); Rz= (Az +Bz )

Rx= (Ax +Bx ); Ry= (Ay +By ); Rz= (Az +Bz ) R= Rx2 + Ry2 + Rz2 The magnitude: The direction: qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R);

Example Determine the magnitude and direction of the resultant force of the following three forces. F3= 600*cos(1200) i + 600*sin(1200)j+0 k Solution: F2=500*cos(2100) i +500*sin(2100)j+0 k F1=350 i+ 0 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: R The direction: qx R= Fx2 + Fy2 + Fz2 = 3832 + 269.62+02 R= 468.4 N qx=cos-1(Rx/R) qx=cos-1(Rx/R)= cos-1(-383/468.4)=144.80 qx=144.80

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 F1=600 lb Solution: R=1696 lb 10.220 F2=700 lb 450 150 x 300 F3=800 lb

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 q A B A•B=B•A=AB cos(q) 0< q <1800 Finding the rectangular scalar component of vector A along the x-axis Ax=A•i=A cos(qx) Along any direction An=A•en=A cos(qn) A n y q 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) 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 Therefore: A•B = Ax Bx + Ay By + Az Bz

Determine the angle q between the following vectors: Example Determine the angle q between the following vectors: A=3i +0j +4k and B=2i -2j +5k A•B=AB cos(q) cos(q)= A•B/ AB A = 32+42 = 5 B = 22 +22 +52 = 5.74 AB= 28.7 A•B=3*2+0*(-2)+4*5=26 cos(q)= 26/28.7 q= 25.10

Orthogonal (perpendicular) vectors (w1 ,w2 ,w3) z w (v1 ,v2 ,v3) v y x w and v are orthogonal if and only if w·v=0

Properties of the dot product If u , v, and w are vectors in 2- or 3- space and k is a scalar, then: u·v= v·u u·(v+w)= u ·v + u·w k(u·v)= (ku) ·v = u·(kv) v·v> 0 if v=0, and v·v= 0 if v=0

Using dot product for a force system x y z F1 =300 lb 600 1.5 ft 6 ft 2 ft 4.5 ft F2 =240 lb 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.

x y z F1 =300 lb 600 1.5 ft 6 ft 2 ft 4.5 ft F2 =240 lb 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

F2= F2 e2 z L1=(22+1.52)1/2=2.5 ft L2 L2=2.5 tan(600)=4.33 ft y L1 x F1 =300 lb 600 1.5 ft 6 ft 2 ft 4.5 ft F2 =240 lb L1=(22+1.52)1/2=2.5 ft L2 L2=2.5 tan(600)=4.33 ft L1 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 + Ry2 + Rz2 = 130.82 + 139.352+384.32 = 429 lb R= 429 lb qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R); qx=cos-1(130.8/429) qx=72.20 qy=cos-1(139.35/429) qy=71.10 qz=cos-1(384.3/429) qz=26.40

b) The magnitude of the rectangular component of the force F1 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 a between force F1 and F2. F1•F2= F1 F2 cos(a) cos(a)= 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(a)=18321/72000 a = 75.250

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. x y z 3 ft 4 ft 6 ft F2=700 lb F1=900 lb Fig. P2-63

Solution +72 d1 = d1 = x12+ y12+ z12 (-6)2 +32 =9.7 ft F2=700 lb F1=900 lb +72 d1 = d1 = x12+ y12+ z12 (-6)2 +32 =9.7 ft F1x = F1 cos(qx) =900 *{(–6)/9.7}=-557 lb F1y = F1 cos(qy) =900 *{3/9.7}=278.5 lb F1z = F1 cos(qz) =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 a between vectors F1 and F2. cos(a)= F1•e2 / F1 e2 x y z 3 ft 4 ft 6 ft F2=700 lb F1=900 lb F1 = -557 i + 278.5j+ 649.8 k lb d2 = x22+ y22+ z22 d2 = (-6)2 +62 +32 =9 ft F1•e2 = (-557)*(-2/3)+278.5*2/3+649.8*0.33=771.4lb e2 = -6/9 i + 6/9j + 3/9 k = -0.67 i + 0.67 j + 0.33 k cos(a)=771.4/900 a=310

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: R=28.6 kN qx= 82.20 qy= 69.60 qz= 22.00