me270 reviewproblems sol

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ME 270 - Fall 2004 SOLUTION

Examination No. 1

PROBLEM NO. 1

Given: Force-couple system I shown below left is made up of four forces F 1 , F 2 ,

F 3 and F acting at points A, B, C and D, respectively, in addition to acouple M. Force F and couple M are known to have magnitudes of 200lb and 900 ft-lb, respectively.

Force-couple system II shown below right is made up of a force R at B

and a couple MB , with R and MB , having known magnitudes of 300 lband 600 ft-lb, respectively.

Find: If System II is to be the equivalent force-couple of System I, what are themagnitudes of the forces F 1, F 2 and F 3?

For System I:

F x!( ) I = F " F 1 = 200 " F 1F y!( )

I = F

3" F

2

M B!( ) I = 2F

1+ 4F

3+ 2F " M

= 2F 1+ 4F

3+ 2 200( )" 900 = 2F 1 + 4F 3 " 500

2ft

C

D

B

A

4ft 2ft

F3

F2

F1

F = 200lb

M = 900 ft-lb

System I

C

D

B

A

53.13°

R

MB

System II

equivalent

x

y


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For System II:

F x!( ) II

= Rcos53.13° = 300( ) 0.6( ) =180

F y!( ) II

= Rsin53.13° = 300( ) 0.8( ) = 240

M B!( ) II

= M B = 600

Making System I equivalent to System II

F x!( ) I = F x!( )

II " 200# F

1=180 " F

1= 20lb

M B!( ) I = M B!( )

II " 2F

1+ 4F

3#500 = 600 "

F 3=

1100 #2F 1

4=

1100 #2 20( )

4= 265

lb

F y!( ) I = F y!( )

II " F

3# F

2= 240 " F

2= F

3# 240 = 25 lb


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ME 270 - Fall 2004 SOLUTION

Examination No. 1

PROBLEM NO. 2

Given: Pole OE is supported by a ball-and-socket joint at O and cables AC and

BE. A downward vertical force F acts at end E. Consider the weight of the pole to be negligible and the ball-and-socket joint at D to befrictionless.

Find:

a) Complete the free body diagram of the pole given below right. b) Resolve the forces of cables AC and BE on the pole in terms of

their unknown magnitudes and known unit vectors.c) If F = 1000 lb, find the tension in cables AC and BE.

Forces on pole due to cables:

F AC =

F AC

r A /C

r A /C

=

F AC

3i !1.5 j + 4k

32+1.5

2+ 4

2

# $ & ' =

F AC 0.5747

i !0.2873

j + 0.7663

k ( )

F BE

= F BE r B / E

r B / E

= F BE !3i ! 3 j + 2k

32+ 3

2+ 2

2

# $

& ' = F BE !0.6396i !0.6396 j + 0.4264k ( )

Using FBD above:

2 ft

2 ft

3 ft

3 ft

1.5 ft

A

B

C E

O

F

1.5 ft

x

y

z

A

B

C E

O

F

x

y

z

O z

O y

O x

F AC F BE


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M O! = rC /O " F AC + r E /O " F BE + r E /O " #Fk ( ) = 0

where

rC / O ! F AC =

i j k

0 1.5 0

0.5747F AC "0.2873F AC 0.7663F AC

=1.1495F AC i " 0.8621F AC k

r E / O ! F BE =

i j k

0 3 0

"0.6396F BE "0.6396F BE 0.4264 F BE

=1.2792F BE i +1.9188F BE k

r E / O ! F BE =

i j k

0 3 0

0 0 "1000

= "3000i

Therefore,

1.1495F AC

+1.2792F BE

! 3000( )i + !0.8621F AC +1.9188F BE ( )k = 0

Balancing coefficients:

i : 1.1495F AC

+1.2792F BE ! 3000 = 0

k : !0.8621F AC

+1.9188F BE

= 0 " F BE

=0.8621

1.9188F AC

= 0.4493F AC

Therefore,

1.1495F AC

+1.2792 0.4493F AC ( )! 3000 = 0 "

F AC

=3000

1.1495 + 1.2792( ) 0.4493( )=1740 lb

F BE = 0.4493F AC = 0.4493( ) 1740( ) = 772.8 lb


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