The pipe assembly supports the vertical loads shown. Determinethe components of reaction at the ball-and-socket joint A and thetension in the supporting cables BC and BD.Equations of Equilibrium:Converting all forces in vector form;Given:TBp = TBDUBD = TBD2mF = 6 kNTBD =|T80 i+TBD j+F2 = 4 kN2mTBCTBc = TBCUBC = TgcIacl,1.5m1.5mTBC =Tsc i+Tscj+Tsck1m2mThe reaction forces at point A can be eliminated by writing themoment equation of equilibrium about point A.3m6 kN4 kNim 1.5m.y3mΣΜΑ-0TAB X (TBp + Tạc) + (r, x F,) + (r; x F;)= 0Free body diagram (FBD):-r, is the distance from point A to F,. Since r, is directed along y axis,therefore, r, = 0i + 4j + Okr, is the distance from point A to F2. Since r, is directed along y axis,therefore, r, = 0i + 5.5j + Ok r, is the distance from point A to F,. Since r, is directed along y axis,therefore, r, = 0i +5.5j + Okl'AB =i +IkTABX"ABY"ABzriyrzyΣΜΑ-|(TBD + TBc)x (TEp + Tgc)y (Typ + Tgc)=|*|F1x Fay F12Fayo-Apply the equation of equilibrium, EM,=0, EM,-0 and EM,=0and solve the equation simultaniouslyTBDkNIkNRepresenting force at point A as F,=A, i +A, j+A,kThen, applying the force equation of equilibrium, EF,=0, EF,=0 andEF,=0 and solve the equation simultaniouslyA,kNAy =kNkN

Fro the given Free body diagram(FBD) F1 =6 kN and F2=4 kN Now by considering A as origin the…

The pipe assembly supports the vertical loads shown. Determine Equations of Equilibrium: the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Converting all forces in vector form; Given: Tạp = TapUBD = TBD F.= 6 kN 2m Tạp = Ta0 i+ F2= 4 kN 2m BC Tạc = TạcUBc = 1.5m Tạc =| 1m The reaction forces at point A can be eliminated by writing the moment equation of equilibrium about point A. 3m 6 kN 2m 4 kN 1m 1.5m. 3m IMA = 0

The pipe assembly supports the vertical loads shown. Determine Equations of Equilibrium: the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Converting all forces in vector form; Given: Tạp = TapUBD = TBD F.= 6 kN 2m Tạp = Ta0 i+ F2= 4 kN 2m BC Tạc = TạcUBc = 1.5m Tạc =| 1m The reaction forces at point A can be eliminated by writing the moment equation of equilibrium about point A. 3m 6 kN 2m 4 kN 1m 1.5m. 3m IMA = 0

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

The pipe assembly supports the vertical loads shown. Determine
the components of reaction at the ball-and-socket joint A and the
tension in the supporting cables BC and BD.
Equations of Equilibrium:
Converting all forces in vector form;
Given:
TBp = TBDUBD = TBD
2m
F = 6 kN
TBD =
|T80 i+
TBD j+
F2 = 4 kN
2m
TBC
TBc = TBCUBC = Tgc
Iacl,
1.5m
1.5m
TBC =
Tsc i+
Tscj+
Tsck
1m
2m
The reaction forces at point A can be eliminated by writing the
moment equation of equilibrium about point A.
3m
6 kN
4 kN
im 1.5m.
y
3m
ΣΜΑ-0
TAB X (TBp + Tạc) + (r, x F,) + (r; x F;)= 0
Free body diagram (FBD):-
r, is the distance from point A to F,. Since r, is directed along y axis,
therefore, r, = 0i + 4j + Ok
r, is the distance from point A to F2. Since r, is directed along y axis,
therefore, r, = 0i + 5.5j + Ok

r, is the distance from point A to F,. Since r, is directed along y axis,
therefore, r, = 0i +5.5j + Ok
l'AB =
i +
Ik
TABX
"ABY
"ABz
riy
rzy
ΣΜΑ-
|(TBD + TBc)x (TEp + Tgc)y (Typ + Tgc)=|*|F1x Fay F12
Fay
o-
Apply the equation of equilibrium, EM,=0, EM,-0 and EM,=0
and solve the equation simultaniously
TBD
kN
IkN
Representing force at point A as F,=A, i +A, j+A,k
Then, applying the force equation of equilibrium, EF,=0, EF,=0 and
EF,=0 and solve the equation simultaniously
A,
kN
Ay =
kN
kN

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