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  RENSSELAER POLY TECHNIC INSTITUTE TROY, NY

EXAM NO. 2 INTRODUCTION TO ENGINEERING ANALYSIS(ENGR-1100) – Fall 10

NAME: ______________________________ Section: ___________ RIN: _______________________________ 

Wednesday, October 20, 20108:00 – 9:50

Please state clearly all assumptions made in order for full credit to be given. 

Problem Points Score

1 25

2 25

3 25

4 25

 Total 100

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Problem #1 (25 %)For the following matrices:

A =[4 5 6],

32

14 D ,

101

210

115

 C,

1

3

2

 B  

Determine the following expressions. If an expression is not valid just state the reason.a)  AB (4)

b)  BA (6)

c)  B +C (2)

d)  Determinant of C using the duplicate column method (4)

e)  BtC (4)

f)  Determinant of B (2)

g)  D-1 (3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

7/27/1

14/114/3

14

42

13

D 

142122x)1(3x4321-4 D g)

matrixsquareanotisBasdUndefine)f 

7191-62032-1-010 

1x)1(2x31x20x)1(1x3)1(x21x)1(0x35x2

101

210

11-5

132CB )e

21300102-5 

1x0x)1(0x2x51x1x11x0x0(-1)x2x15x1x1

0

1

1-

 

1

0

5

 

101

210

11-5

 C d)

sizesamethehavenotdomatricesthesincedUndefine)c

654

181512

12108

6x15x14x1

6x35x34x3

6x25x24x2

654

1

3

2

 BA)b

1761586x(-1)5x34x2

1

3

2

654AB )a

1-

t

 

1

6

2

1

2

3

2

1

3

1

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Problem #2 (25%) Three cables (AB, AC and AD) are used to tether a balloon as shown. The force (tension) in thecable AC is 300 N.

1. 

Identify the particle to be analyzed (3 points)2.  Draw a complete and separate FBD showing the particle and all the forces acting on it. (5points)

3.  Write the equilibrium equations (8 points)4.  Determine the vertical force P exerted by the balloon. (3 point)5.  Determine the magnitude of the force in the cables AB and AD (6 points)

Particle to be analyzed: A

FBD

kP P

k) je T 781.0625.0( ABABABAB T T

k jk ji

e 781.0625.0)5()4(

54022

AB

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k) jie T 745.0298.0596.0( ACACACAC T T

k jik ji

e 745.0298.0596.0

)5()2()4(

524

222

AC

k) jie T 857.00515.0( ADADADAD T T

k jik ji

e 857.00515.0)5()3(

50322

AD

N607.223443.89885.178(

 N745.0298.0596.0(300

 k) ji

 k) ji T

AC

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Equilibrium

0P T T T ADACAB

0kk) ji

 k) jik) j

P T

 T

AD

AB

857.00515.0(

N607.223443.89885.178(781.0625.0(

0857.0(N607.223781.0(

0N443.89)625.0(

0)515.0(N885.178

P T T

 T

 T

ADAB

AB

AD

 ))

N054.633

N109.143

N350.347

P

 T

 T

AB

AD

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Problem #3 (25 %)

 The beam fixed in the wall is subjected to forces T andF in addition to the weightW. Assumethat the beam is weightless. The forces T andF are given in the figure.

(1) Calculate the moment of  T about O. (4)(2) Calculate the moment of F about O. (4)(3) Calculate the moment of G about O. (4)(4) Replace T, F, andG by an equivalent resultant forceR at point O and a coupleC.

ExpressR andC in Cartesian vector form. (9)

(5) Calculate the scalar component of the moment of  T aboutz-axis (note thatz-axis followsthe right-hand rule). (4)Important: mark the senses of 2D moment and couple in your result.

Solution:

(1) M TO = Ty ×rOD=150 ×4 =600 Nm;

or: k

k ji

 TrM OD TO 600

0150260

004

Nm;

M TO =600 Nm (eqn: 1; number: 1; unit: 1; direction: 1)

(2) MFO =Fy ×rOB=300 sin30° ×3 =150×3 =450 Nm;

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or: k

k ji

FrM OBFO 450

0150260

003 Nm

MFO =450 Nm (eqn: 1; number: 1; unit: 1; direction: 1)

(3) MGO =G×rOB=-500 ×6 =-3000 Nm;MGO =3000 Nm (eqn: 1; number: 1; unit: 1; direction: 1)

(4) R = T +F +G =( Tx+Fx+Gx)i +( Ty+Fy+Gy) j =(-260+260) i +(150+150-500) j ==-200 j N.

R =-200 j N (eqn: 1; number: 1; unit: 1; vector form: 1) 

C =M TO +MFO +MGO =(600+450-3000)k =-1950k Nm;

C =1950 Nm (eqn: 1; number: 1; unit: 1; vector form: 1; direction: 1)

(5)  6001600 kMM  TO TOzprojM Tz Nm

Or use triple scalar product:

600101504

0150260

004

100

k TrM OD TOzprojM Tz Nm

(eqn: 2; number 1; unit: 1)

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Problem #4 (25%)