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COSMOS: Complete Online Solutions Manual Organization System
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 © 2007 The McGraw-Hill Companies.
Chapter 3, Solution 136.
First reduce the given force-couple system to an equivalent force-couple system ( ), BR M at point B.
( ) ( ) ( )2 2 2480 mm 560 mm 480 mmBDd = − + + −
880 mm=
( )132 N 480 560 480880BD BD BDF λ= = − + −F i j k
( )( )12 N 6 7 6= − + −i j k
( ) ( ) ( )2 2 2240 mm 220 mm 480 mmEBd = + − +
580 mm=
( )145 N 240 220 480580EB EB EBF λ= = − +F i j k
( )( )5 N 12 11 24= − +i j k
: BD EBΣ = +F R F F
( )( ) ( )12 N 6 7 6 5 N 12 11 24= − + − + − +i j k i j k
( ) ( ) ( )12 N 29 N 48 N= − + +i j k
( ) ( ) ( )2 2 2340 mm 240 mm 60 mmBFd = + + −
20 442 mm=
Then ( )20 N m 340 240 6020 442B
⋅= + −M i j k
( )20 N m 17 12 3442⋅
= + −i j k
COSMOS: Complete Online Solutions Manual Organization System
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 © 2007 The McGraw-Hill Companies.
Now determine whether R and BM are perpendicular
( ) ( )2012 29 48 17 12 3442B⋅ = − + + ⋅ + −R j j k i j kM
( )20 12 17 29 12 48 3442
= − × + × − ×
0=
∴R and BM are perpendicular so that ( ), BR M can be reduced to the single equivalent force
( ) ( ) ( )12.00 N 29.0 N 48.0 NR = − + +i j k
Now require /B B P= ×M r R
or ( ) ( )20 N m 17 12 3 0.480 0.480 N m442
12 29 48y z⋅
+ − = − − ⋅−
i j ki j k
( ) ( )20 12: 12 0.480 0.480 48442
z×= − − +j
or 1.449 mz =
( )20 3: 0.480 29 12442
y− ×= − +k
or 0.922 my =
∴ The point of intersection is defined by
0.922 my =
1.449 mz =
COSMOS: Complete Online Solutions Manual Organization System
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 © 2007 The McGraw-Hill Companies.
Chapter 3, Solution 137.
First, reduce the given force system to a force-couple at the origin.
Have : A GΣ + =F F F R
( ) ( ) ( ) ( ) ( ) ( ) ( )4 in. 6 in. 12 in. 10 lb 14 lb 4 lb 6 lb 2 lb
14 in. + −
∴ = + = + −
i j kR k i j k
and 56 lbR =
Have ( ): RO O C OΣ ∑ + ∑ =M r F M M×
( ) ( ) ( ) ( ) ( ) ( ){ }12 in. 10 lb 16 in. 4 lb 6 lb 12 lbRO = + + − M j k i i j k× ×
( ) ( ) ( ) ( ) ( ) ( ) ( )16 in. 12 in. 4 in. 12 in. 6 in.84 lb in. 120 lb in.
20 in. 14 in. − − +
+ ⋅ + ⋅
i j i j k
( ) ( ) ( )0 221.49 lb in. 38.743 lb in. 147.429 lb in.R∴ = ⋅ + ⋅ + ⋅M i j k
( ) ( ) ( )18.4572 lb ft 3.2286 lb ft 12.2858 lb ft= ⋅ + ⋅ + ⋅i j k
COSMOS: Complete Online Solutions Manual Organization System
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 © 2007 The McGraw-Hill Companies.
The force-couple at O can be replaced by a single force if the direction of R is perpendicular to .ROM
To be perpendicular 0RO =R M⋅
Have ( ) ( )4 6 2 18.4572 3.2286 12.2858 0?RO = + − + + =R M i j k i j k⋅ ⋅
73.829 19.3716 24.572= + −
0≠
∴ System cannot be reduced to a single equivalent force.
To reduce to an equivalent wrench, the moment component along the line of action of P is found.
1 RR O RM
R= =
RMλ ⋅ λ
( ) ( )4 6 218.4572 3.2286 12.2858
56 + −
= + +
i j ki j k⋅
9.1709 lb ft= ⋅
and ( )( )1 1 9.1709 lb ft 0.53452 0.80178 0.26726RM= = ⋅ + −M i j kλ
And pitch 1 9.1709 lb ft 1.22551 ft56 lb
MpR
⋅= = =
or 1.226 ftp =
Have
( ) ( )( )2 1 18.4572 3.2286 12.2858 9.1709 0.53452 0.80178 0.26726RO= − = + + − + −M M M i j k i j k
( ) ( ) ( )13.5552 lb ft 4.1244 lb ft 14.7368 lb ft= ⋅ − ⋅ + ⋅i j k
Require 2 /Q O=M r R×
( ) ( ) ( )13.5552 4.1244 14.7368 4 6 2y z− + = + + −i j k j k i j k×
( ) ( ) ( )2 6 4 4y z z y= − + + −i j k
From j: 4.1244 4z− = or 1.0311 ftz = −
From k: 14.7368 4 or 3.6842 fty y= − = −
∴ line of action of the wrench intersects the yz plane at
3.68 ft, 1.031 fty z= − =
