# Solution manual vector mechanics engineers dynamics 8th beer chapter 03

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Chapter 3, Solution 1.

Resolve 90 N force into vector components P and Q
where Q = ( 90 N ) sin 40°
= 57.851 N
Then M B = − rA/BQ

= − (0.225 m )(57.851 N )
= −13.0165 N ⋅ m
M B = 13.02 N ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 2.

Fx = ( 90 N ) cos 25°

= 81.568 N
Fy = ( 90 N ) sin 25°
= 38.036 N
x = ( 0.225 m ) cos 65°
= 0.095089 m

y = (0.225 m ) sin 65°
= 0.20392 m

M B = xFy − yFx
= ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N )
= −13.0165 N ⋅ m

M B = 13.02 N ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 3.

Px = ( 3 lb ) sin 30°
= 1.5 lb

Py = ( 3 lb ) cos 30°
= 2.5981 lb

M A = xB/ A Py + yB/ A Px
= ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb )
= 16.0335 lb ⋅ in.

M A = 16.03 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 4.

For P to be a minimum, it must be perpendicular to the line joining points
A and B
with rAB =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.
 y
 

α = θ = tan −1  
x
 4.8 in. 
= tan −1 

 3.4 in. 

= 54.689°
Then

M A = rAB Pmin

or

Pmin =

M A 19.5 lb ⋅ in.
=
rAB
5.8822 in.

= 3.3151 lb
∴ Pmin = 3.32 lb

54.7°
or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Pmin = 3.32 lb

35.3°

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Chapter 3, Solution 5.

M A = rB/ A P sin θ

By definition
where

θ = φ + ( 90° − α )

and

φ = tan −1 

 4.8 in. 

 3.4 in. 

= 54.689°
Also

rB/ A =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.

Then

(17 lb ⋅ in.) = ( 5.8822 in.)( 2.9 lb ) sin ( 54.689° + 90° − α )

or

sin (144.689° − α ) = 0.99658

or

144.689° − α = 85.260°; 94.740°
∴ α = 49.9°, 59.4°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 6.

(a)

(a) M A = rB/ A × TBF

M A = xTBFy + yTBFx
= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°
= 386.41 N ⋅ m
or M A = 386 N ⋅ m
(b)

(b) For FC to be a minimum, it must be perpendicular to the line
joining A and C.

∴ M A = d ( FC )min
d =

with

( 2 m )2 + (1.35 m )2

= 2.4130 m
Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min

( FC )min
and

= 160.137 N
 1.35 m 
 = 34.019°
 2m 

φ = tan −1 

θ = 90 − φ = 90° − 34.019° = 55.981°
∴ ( FC )min = 160.1 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

56.0°

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Chapter 3, Solution 7.

(a)

M A = xTBFy + yTBFx

= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°
= 386.41 N ⋅ m
or M A = 386 N ⋅ m

(b)

Have
or

M A = xFC

FC =

MA
386.41 N ⋅ m
=
2m
x

= 193.205 N

(c)

∴ FC = 193.2 N
For FB to be minimum, it must be perpendicular to the line joining A
and B

∴ M A = d ( FB )min
with
Then

d =

= 2.0396 m

386.41 N ⋅ m = ( 2.0396 m ) ( FC )min

( FC )min
and

( 2 m )2 + ( 0.40 m )2

= 189.454 N

 2m 
 = 78.690°
 0.4 m 

θ = tan −1 

or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

( FC )min

= 189.5 N

78.7°

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Chapter 3, Solution 8.

(a)

(

)

M B = rA/B cos15° W

= (14 in.)( cos15° )( 5 lb )
= 67.615 lb ⋅ in.
or

M B = 67.6 lb ⋅ in.

(b)

M B = rD/B P sin 85°
67.615 lb ⋅ in. = ( 3.2 in.) P sin 85°
or

(c)

P = 21.2 lb

For ( F )min, F must be perpendicular to BC.
Then,

M B = rC/B F
67.615 lb ⋅ in. = (18 in.) F
or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

F = 3.76 lb

75.0°

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Chapter 3, Solution 9.

Slope of line EC =

(a)

Then

and
Then

TABx =

35 in.
5
=
76 in. + 8 in. 12

12
(TAB )
13

=

12
( 260 lb ) = 240 lb
13

TABy =

5
( 260 lb ) = 100 lb
13

M D = TABx ( 35 in.) − TABy ( 8 in.)

= ( 240 lb )( 35 in.) − (100 lb )( 8 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.
(b) Have

M D = TABx ( y ) + TABy ( x )
= ( 240 lb )( 0 ) + (100 lb )( 76 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 10.

Slope of line EC =

35 in.
7
=
112 in. + 8 in. 24

Then

TABx =

24
TAB
25

and

TABy =

7
TAB
25

M D = TABx ( y ) + TABy ( x )

Have

∴ 7840 lb ⋅ in. =

24
7
TAB ( 0 ) +
TAB (112 in.)
25
25
TAB = 250 lb
or TAB = 250 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 11.

The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y ( d )
M D = 1152 N ⋅ m

where

(TAB max ) y
Now

sin θ =

= TAB max sin θ = ( 2880 N ) sin θ

1.05 m

(d

∴ 1152 N ⋅ m = 2880 N 



+ 0.24 ) + (1.05 ) m
2

2

1.05

(d

+ 0.24 ) + (1.05 )
2

( d + 0.24 )2 + (1.05)2

or
or

2

 (d )



(d

or

= 2.625d

+ 0.24 ) + (1.05 ) = 6.8906d 2
2

2

5.8906d 2 − 0.48d − 1.1601 = 0

Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 12.

with d AB =

( 42 mm )2 + (144 mm )2

= 150 mm
sin θ =

42 mm
150 mm

cosθ =

144 mm
150 mm

and FAB = − FAB sin θ i − FAB cosθ j
=

2.5 kN
( − 42 mm ) i − (144 mm ) j
150 mm 

= − ( 700 N ) i − ( 2400 N ) j
Also rB/C = − ( 0.042 m ) i + ( 0.056 m ) j
Now M C = rB/C × FAB
= ( − 0.042 i + 0.056 j) × ( − 700 i − 2400 j) N ⋅ m
= (140.0 N ⋅ m ) k
or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

M C = 140.0 N ⋅ m

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Chapter 3, Solution 13.

( 42 mm )2 + (144 mm )2

with d AB =

= 150 mm
sin θ =

42 mm
150 mm

cosθ =

144 mm
150 mm

FAB = − FAB sin θ i − FAB cosθ j
=

2.5 kN
( − 42 mm ) i − (144 mm ) j
150 mm 

= − ( 700 N ) i − ( 2400 N ) j
Also rB/C = − ( 0.042 m ) i − ( 0.056 m ) j
Now M C = rB/C × FAB
= ( − 0.042 i − 0.056 j) × ( − 700i − 2400 j) N ⋅ m
= ( 61.6 N ⋅ m ) k
or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

M C = 61.6 N ⋅ m

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Chapter 3, Solution 14.

ΣM D :

 88

 105

M D = ( 0.090 m ) 
× 80 N  − ( 0.280 m ) 
× 80 N 
137
137

= −12.5431 N ⋅ m
or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

M D = 12.54 N ⋅ m

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Chapter 3, Solution 15.

Note: B = B ( cos β i + sin β j)

B′ = B ( cos β i − sin β j)
C = C ( cos α i + sin α j)
By definition:

B × C = BC sin (α − β )

(1)

B′ × C = BC sin (α + β )

(2)

Now ... B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α − sin β cos α ) k
and

(3)

B′ × C = B ( cos β i − sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α + sin β cos α ) k

(4)

Equating the magnitudes of B × C from equations (1) and (3) yields:
BC sin (α − β ) = BC ( cos β sin α − sin β cos α )

(5)

Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields:
BC sin (α + β ) = BC ( cos β sin α + sin β cos α )

(6)

Adding equations (5) and (6) gives:
sin (α − β ) + sin (α + β ) = 2cos β sin α
or

sin α cos β =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

1
1
sin (α + β ) + sin (α − β )
2
2

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Chapter 3, Solution 16.

Have d = λ AB × rO/ A
where λ AB =

rB/ A
rB/ A

and rB/ A = ( −210 mm − 630 mm ) i
+ ( 270 mm − ( −225 mm ) ) j

= − ( 840 mm ) i + ( 495 mm ) j
rB/ A =

( −840 mm )2 + ( 495 mm )2
= 975 mm

Then λ AB =
=

− ( 840 mm ) i + ( 495 mm ) j
975 mm

1
( −56i + 33j)
65

Also rO/ A = ( 0 − 630 ) i + ( 0 − (−225) ) j

= − ( 630 mm ) i + ( 225 mm ) j

∴d =

1
( −56i + 33j) × − ( 630 mm ) i + ( 225 mm ) j
65

= 126.0 mm
d = 126.0 mm W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 17.

(a)
where

λ =

A×B
A×B

A = 12i − 6 j + 9k
B = − 3i + 9 j − 7.5k

Then
i
j
k
A × B = 12 − 6 9
− 3 9 − 7.5

= ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 ) k
= 9 ( − 4i + 7 j + 10k )
And A × B = 9 (− 4) 2 + (7)2 + (10)2 = 9 165

∴λ =

9 ( − 4i + 7 j + 10k )
9 165
or λ =

(b)
where

λ =

A×B
A×B

A = −14i − 2 j + 8k
B = 3i + 1.5j − k

Then

j k
i
A × B = −14 − 2 8
3 1.5 −1
= ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k

= 5 ( −2i + 2 j − 3k )
and

A × B = 5 (−2)2 + (2)2 + (−3)2 = 5 17

∴λ =
or λ =

5 ( −2i + 2 j − 3k )
5 17

1
( − 2i + 2 j − 3k )
17

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

1
( − 4i + 7 j + 10k )
165

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Chapter 3, Solution 18.

(a)

Have A = P × Q
i j k
P × Q = 3 7 −2 in.2
−5 1 3
= [ (21 + 2)i + (10 − 9) j + (3 + 35)k ] in.2

(

) (

) (

)

= 23 in.2 i + 1 in.2 j + 38 in.2 k
∴A=

(23)2 + (1)2 + (38) 2 = 44.430 in.2
or A = 44.4 in.2 W

(b)

A = P×Q
i j k
P × Q = 2 − 4 3 in.2
6 −1 5
= [ (−20 − 3)i + (−18 − 10) j + (−2 + 24)k ] in.2

(

) (

) (

)

= − 23 in.2 i − 28 in.2 j + 22 in.2 k
∴A=

(− 23)2 + (−28)2 + (22) 2 = 42.391 in.2
or A = 42.4 in.2 W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 19.

(a)

Have

MO = r × F

i

j

k

= − 6 3 1.5 N ⋅ m
7.5 3 − 4.5
= [ (−13.5 − 4.5)i + (11.25 − 27) j + (−18 − 22.5)k ] N ⋅ m
= ( −18.00i − 15.75 j − 40.5k ) N ⋅ m
or M O = − (18.00 N ⋅ m ) i − (15.75 Ν ⋅ m ) j − ( 40.5 N ⋅ m ) k W
(b)

Have

MO = r × F
i
j
k
= 2 − 0.75 −1 N ⋅ m
7.5
3
− 4.5
= [ (3.375 + 3)i + (−7.5 + 9) j + (6 + 5.625)k ] N ⋅ m
= ( 6.375i + 1.500 j + 11.625k ) N ⋅ m
or M O = ( 6.38 N ⋅ m ) i + (1.500 Ν ⋅ m ) j + (11.63 Ν ⋅ m ) k W

(c)

Have

MO = r × F
i
j k
= − 2.5 −1 1.5 N ⋅ m
7.5 3 4.5
= [ (4.5 − 4.5)i + (11.25 − 11.25) j + (−7.5 + 7.5)k ] N ⋅ m
or M O = 0 W

This answer is expected since r and F are proportional ( F = −3r ) . Therefore, vector F has a line of action
passing through the origin at O.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 20.

(a)

Have

MO = r × F
i
j k
= − 7.5 3 − 6 lb ⋅ ft
3 −6 4
= [ (12 − 36)i + (−18 + 30) j + (45 − 9)k ] lb ⋅ ft
or M O = − ( 24.0 lb ⋅ ft ) i + (12.00 lb ⋅ ft ) j + ( 36.0 lb ⋅ ft ) k W

(b)

Have

MO = r × F
i
j k
= − 7.5 1.5 −1 lb ⋅ ft
3 −6 4
= [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5)k ] lb ⋅ ft
or M O = 0 W

(c)

Have

MO = r × F
i
j
k
= − 8 2 −14 lb ⋅ ft
3 −6 4
= [ (8 − 84)i + (−42 + 32) j + (48 − 6)k ] lb ⋅ ft
or M O = − ( 76.0 lb ⋅ ft ) i − (10.00 lb ⋅ ft ) j + ( 42.0 lb ⋅ ft ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 21.

With

TAB = − ( 369 N ) j

JJJG
= ( 369 N )

( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k
( 2.4 m )2 + ( −3.1 m )2 + ( −1.2 m )2

TAD = ( 216 N ) i − ( 279 N ) j − (108 N ) k
Then

R A = 2 TAB + TAD
= ( 216 N ) i − (1017 N ) j − (108 N ) k

Also

rA/C = ( 3.1 m ) i + (1.2 m ) k

Have

M C = rA/C × R A
i
j
k
= 0
3.1
1.2 N ⋅ m
216 −1017 −108
= ( 885.6 N ⋅ m ) i + ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k
M C = ( 886 N ⋅ m ) i + ( 259 N ⋅ m ) j − ( 670 N ⋅ m ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 22.

Have

M A = rC/ A × F

where

rC/ A = ( 215 mm ) i − ( 50 mm ) j + (140 mm ) k
Fx = − ( 36 N ) cos 45° sin12°

Fy = − ( 36 N ) sin 45°
Fz = − ( 36 N ) cos 45° cos12°

∴ F = − ( 5.2926 N ) i − ( 25.456 N ) j − ( 24.900 N ) k
and

i
j
k
M A = 0.215
− 0.050
0.140 N ⋅ m
− 5.2926 − 25.456 − 24.900
= ( 4.8088 N ⋅ m ) i + ( 4.6125 N ⋅ m ) j − ( 5.7377 N ⋅ m ) k
M A = ( 4.81 N ⋅ m ) i + ( 4.61 N ⋅ m ) j − ( 5.74 N ⋅ m ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 23.

Have

M O = rA/O × R

where

rA/D = ( 30 ft ) j + ( 3 ft ) k
T1 = − ( 62 lb ) cos10°  i − ( 62 lb ) sin10°  j

= − ( 61.058 lb ) i − (10.766 lb ) j
JJJG
AB
T2 = T2
AB
= ( 62 lb )

( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k
( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2

= (10 lb ) i − ( 60 lb ) j + (12 lb ) k
∴ R = − ( 51.058 lb ) i − ( 70.766 lb ) j + (12 lb ) k
MO

i
j
k
=
0
30
3 lb ⋅ ft
− 51.058 −70.766 12
= ( 572.30 lb ⋅ ft ) i − (153.17 lb ⋅ ft ) j + (1531.74 lb ⋅ ft ) k
M O = ( 572 lb ⋅ ft ) i − (153.2 lb ⋅ ft ) j + (1532 lb ⋅ ft ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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Chapter 3, Solution 24.

(a)

Have

M O = rB/O × TBD

where

rB/O = ( 2.5 m ) i + ( 2 m ) j
TBD = TBD

JJJG
BD
BD

 − (1 m ) i − ( 2 m ) j + ( 2 m ) k 
= ( 900 N ) 
( −1 m ) 2 + ( − 2 m ) 2 + ( 2 m ) 2

= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k
Then

MO

i
j
k
= 2.5
2
0 N⋅m
− 300 − 600 600
M O = (1200 N ⋅ m ) i − (1500 N ⋅ m ) j − ( 900 N ⋅ m ) k W
continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

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(b)

Have

M O = rB/O × TBE

where

rB/O = ( 2.5 m ) i + ( 2 m ) j
TBE = TBE

JJJG
BE
BE

 − ( 0.5 m ) i − ( 2 m ) j − ( 4 m ) k 
= ( 675 N ) 
( 0.5 m )2 + ( −2 m )2 + ( − 4 m )2

= − ( 75 N ) i − ( 300 N ) j − ( 600 N ) k
Then

MO

i
j
k
= 2.5
2
0 N⋅m
− 75 − 300 − 600
M O = − (1200 N ⋅ m ) i + (1500 N ⋅ m ) j − ( 600 N ⋅ m ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell