Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276...

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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

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

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.

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+

Fy = F sin tan-

1(Fy/Fx)

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

Figure 2-47

Solution

F=275 lby

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 2-49please submit to TA at the end of the lecture

F=475 lb

Figure 2-49

(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

Fx=Fxi

Fz=Fzk

Fy=Fyjx y

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

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

- azimuth angle- elevation angle

Fxy = F cos

Fz = F sin ;

Fx= Fxy cos Fcos cos

Fy= Fxy sin Fcos sin

Finding the direction of a force by two points along its line of action

cosx= xB-xA

(xB-xA)2+ (yB-yA)2+ (zB-zA)2

cosy= yB-yA

cosz= zB-zA

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.

zF=745 N

Solution

zF=745 N

Fz = F sin = 745 sin(600)=411.4 N

Fxy = F cos = 745 cos(600)=237.5 N

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

Fxy=237.5 N

Fx= Fxy cos 237.5*sin(370)=-142.9 N

Fy= Fxy sin 237.5*cos(370)=-189.7 N

F=-142.9 i – 189.7 j + 411.4 kb)

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.

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

Rx= Ax +Bx ); Ry= Ay +By ); Rz= Az +Bz )

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

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

3832 + 269.62+02

The direction:

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

F1=600 lb

450 F2=700 lb

F3=800 lb

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)

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)

(w1 ,w2 ,w3)

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:

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

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

F1 =300 lb

1.5 ft6 ft2 ft

4.5 ft

F2 =240 lb

F1 =300 lb

1.5 ft6 ft2 ft

4.5 ft

F2 =240 lb

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

3 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

3 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.

3 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

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

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

Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276...

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