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Unformatted text preview: Example 5— 16 The Windlass shown in Fig. 5—25a is supported by two smooth journal
bearings A and B. which are properly aligned on the shaft. Determine the
magnitude of the vertical force P that must be applied to the handle to
maintain equilibrium of the lOO—kg crate. Also calculate the reactions at the
bearings. ' Fig. 5—25 SOLUTION (SCALAR ANALYSIS) FreeBody Diagram. Since the bearings at A and B are aligned correctly,
only force reactions occur at these supports, Fig. 5—2517. Why are there no
moment reactions?’ Equations of Equilibrium. Summing moments about the x axis yields a
direct solution for P. Why? For a scalar moment summation, it is necessary to compute the moment of each force as the product of the force magnitude
and the perpendicular distance from the x axis to the line of action of the
force. Using the righthand rule and assuming positive moments act in the
+i direction, we have V ' 2114,, = 0; 981 N(O.1 m) — P(0.3 cos 20° m) = 0
‘ ' P = 348.0 N Ans. Using this result and summing moments about the y and z axes yields 2M), = 0; —981 N(O.5 m) 4 Az(0.8 m) + (348.0 N)(0.45 m) = 0
A2 = 417.4 N Ans.
2M2 = 0; — y(0.8 m) = O Ay = 0 Ans. The reactions at B are obtained by a force summation, using the results
computed above. EFy=0; , OlBy=0 By=0 Ans.
2Fz = 0; 417.4‘ — 981+ B, — 3480 = 0 B, = 911.6 N Ans.
As shown on the freebody diagram, the supports do not provide resist ance against translation in the x direction. Hence, the Windlass is only
partially constrained. Determine the tension in cables BC and BD and the reactions at
the ball—andsocket joint A for the mast shown in Fig. 5—26a. soLUTION (VECTOR ANALYSIS)
FreeBody Diagram. There are five unknown force mag—
nitudes shown on the freebody diagram, Fig. 5—26b. Equations of Equilibrium. Expressing each force in
Cartesian vector form, we have
F = {—1000j} N
FA = Axi + ij + Azk
TC = 0.707Tci — 0.707TCk
1'30 x
TD = TD (7) = — 0.333TDi + 0667ij — 0.667TDk
BD Applying the force equation of equilibrium gives
2F=0; F+FA+TC+TD=0
(Ax + 0.707TC — 0.333TD)i + (—1000 + Ay + 0.667TD)j
+ (A, — 0.707TC — 0.667TD)k = 0
BF), = 0; Ax + 0.707TC — 0.333TD = 0 (1)
EFy = 0; Ay + 0.667TD — 1000 = 0 (2)
' 2F, = 0; AZ  0.707TC — 0.667TD = O (3) Summing moments about point A, we have 2MA=0; er(F+TC+TD)=0
6k X {—IOOOj + 0.707Tci — 0.707Tck
— 0.333TDi + 0667ij — 0.667TDk) = 0 Evaluating the cross product and combining terms yields (—4TD + 6000)i + (4.24TC — 2T0)j = 0
M, = 0; —4TD + 6000 = 0 (4)
EMy = 0; 4.24Tc — 2TB = O (5) The moment equation about the z axis, 2M2 = O, is automatically satis
ﬂed. Why? Solving Eqs. (1) to (5) we have TC = 707 N TD = 1500 N Ans.
Ax = 0.0 N A, = 0.0 N A, = 1500 N  Ans. Since the mast is a two—force member, note that the values of
Ax = Ay = 0.0 could have been determined by inspection. memuwmm W at Ch 5 ~ Equilibrium of a Rigid Body Example 5—18 Rod AB shown in Fig. 5—2 7a is used to support the ZOO—N force. Deter
mine the reactions at the ball and—socket joint A and at the smooth collarB. Equatzons of Equzlzbnum Representing each force on the freebody dia
gram in Cartesian vector form we have
FA—Axl‘i‘ij‘i‘Azk ()
a
FB = Bxi + By j
; F = {—ZOOk} N
Applying the force equation of equilibrium,
i 2F=0; FA+FB+F=0
(Ax + Bx)i + (A3, + By) j + (AZ — 200)k = 0
g EFX = Ax + Bx = 0
ZFy = ; Ay + By = 0
2Fz=' Az—200=0
i Summing moments about point A yields
2MA=0; (erF)+(erFB)=0
Since rC = érB, then
i (1i+1j— 0.5k) x (—200k) + (2i + 2j—1k)x(Bxi+ Byj) = 0
Expanding and reananging terms gives
(By — 200)i + (‘Bx + 200)j + (28y f 28x)k = 0
2M)r = 0; By — 200 = 0 (4)
EM), = 0; —Bx + 200 = 0 '4 “
2M2 = 0; 2By — 28x = 0 Solving Eqs. (1) to (6), we get Example 5—19 The bent rod in Fig. 5—28a is supported at A by a journal bearing, at D
by a ball—and—socket joint, and at B by means of cable BC. Using only one
equilibrium equation, obtain a direct solution for the tension in cable BC.
The bearing at A is capable of exerting force components only in the z and
i y directions, since it is properly aligned on the shaft. SOLUTION ( VECTOR ANALYSIS) ‘ FreeBody Diagram. As shown in Fig. 5—2812, there are six unknowns: the three force components at the balland—socket joint, two at the bearing,
and the tension force in the cable. Equations of Equilibrium. The cable tension TB may be obtained di
rectly by summing moments about an axis passing through points D andA.
Why? The direction of the axis is defined by the unit vector u, where _ I‘D/1 1 . j 1 .
rDA V5 VEJ
= —0.707i — 0.707j J Hence, the sum of the moments about this axis is zero provided
EMDA =uE(er) =0 Here 1' represents a position vector drawn from any point on the axis DA to any point on the line of action of force F (see Eq. 4—9). With reference to
Fig. 5—28b, we can therefore write u(erTB+rExW)=r0 (—0.707i — 0.707j)  [(—1 j) x (%%T3i — 343T). j + %T3k) Fi . 5—25.21?" + (—0.5j) x(—981k)] = 0 g ~i (—0.707i ~ 0.707 j)  [(—0.857TB + 490.5)i + 0.286TBk] = 0
—0.707(—0.857TB + 490.5) + 0 + 0 = 0 _ 490.5
0.857 (b) TB = 572 N _ Ans. The advantage of using Cartesian vectors for this solution should be noted.
It would be especially tedious to determine the perpendicular distance
from the DA axis to the line of action of TB using scalar methods.
Note: In Example 5—17, a direct solution for AZ is possible by summing
moments about an axis passing through the supports at C and D, Fig.
526a. If this is done only the moment of F and AZ must be considered. Try
and apply the above technique and determine the result AZ = 1500 N. ...
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 Spring '07
 Rodin
 Statics

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