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Unformatted text preview: 5.54 A crane is used to lift a pile of logs with center of mass at point D.
Make a freebody diagram of the logs and discuss whether or not the
system is statically determinate and/ or properly constrained. ‘i5iiii13ienris‘iiet.sr Solution: FIGURE 85.54 There are 3 unknown scalar forces TA, TB, and T 0, provided the
geometry is given. Thus there will be 3 unknowns and potentially
six equations. However, the 3 forces pass through a common point
so the moment equations yield no information. Thus there are 3
equations in 3 unknowns and the system is statically determinant but
improgerly constrained as rotation could occur about the point of
intersection of the three forces. 418 5.56 Sketch the free body diagram of the circus tent stake. The stake is mounted
on a hard surface by a ball and socket arrangement at point C. Is this system properly constrained? Is it statically indeterminate?
F = 100 N y D (0,5,0) (4.1.0.03) (0.3.0,—0.1)
FIGURE P556
Solution: From the free body diagram there are 5 unknown magnitudes:
T1, T2, 0x, 0'3,r and Oz, which, with geometry assumed given, can be
determined from the 5 equations of equilibrium 2F :0, Sage, 25:0, ZM =0,ZMZ=0. Thus the system is statically determinant. The sixth equation, 2 M 2 0 is
automatically satisﬁed because all the forces intersect the yaxis, implying
that the system is im re erl constrained as nothing prevents rotation about the y—axis. You may want to point out that this is a two force member.
y . A 1.0.1.033) 10.30.41” FIGURE 55.56 420 5.62 A winch system consists of a 0.1—m diameter drum, shaft and motor.
Compute the reactions at A and C, and the motor torque Mz required
to keep the 100kg mass in equilibrium. The spindle supports at A and
C are thmstless bearings. Neglect any moments at A and C. 0.2m : Q Motor 100 kg FIGURE 135.62 Solution: The freabody is given and the equilibrium equations are
(unrestrained in z direction) FIGURE 55.62
ZFI=0:AI+CI=U (1)
ZFy=0:Ay+C =981 (2) 2 Mc = 0 = M21} + (.05i + 0.212) x (—981j)+ 0.412 x (Axi + A93) = 0 Further manipulation of the moment equation yields (continued) 428 M21} — (.05)(981)l; + (0.2) (931)i + (04143)} — 0.4Ayi = 0 or in scalar form I: ; M2 2 49.05 Nm
3' : 0.4241 = 0 1‘ : 196.2 — 0.4Ay = 0 These 5 equations yield Ax = Cm = 0, A3, = 490.5 N, Cy = 490.5 In. 429 5.64 A mounting platform is secured in place by a frictionless support at A,
a ball and socket at B and a rope at C. The A 100 N gravitational force
acts at its geometric center and two boxes sit 0n the platform modeled by the force F1 = 500 N and F2 = 50 N. Calculate the components of
the reaction forces at the supports. FIGURE P564 Solution: A free body diagram yields the following equilibrium
equations FIGURE 55.64 2 F3 : B:E = 0 (1)
(continued) 432 :13: Tc+By+Ay—650=0 (2) zn=n=o m 2 M0 = (0.53) xch+(l}xAgj)+ix (Byj)+12x (By3)+.5(i+12) x (—1003)
+(.75i + .512) x (—5003) + (0.33 + 0.61;) x (—505) = o multiplying out the moment terms (with BE = B; = 0) yields from
iz—Ay—By+330=0 01'
A3, + By = 330 (4). From «
k : .5Tc + By = 440 (5) Solving 15 yields
Tc=320N, 33:0, By:280N, 32:0, Ay=50N Note that while stable for the given load, there is no support against
twisting about the y axis so this is improperly constrained. 433 5.69 Compute the reactions at the ball and socket support at point D and
the tensions in the support ropes (T1 and T2) for the Sign support
system. The weight of the sign exerts a. force of 300 N in the down
direction (—y) at point E, 0.25 m from D and at point F, 1.75 m from
D Note that DC is not constrained from rotation about its axis. 3 (0.1.50) rn A 0,1.5.1.5)m o (0.0.5.015)
c (2.0.5.015) rn FIGURE P5.69 Solution: A free body diagram of the sign support is given in the ﬁgure C (anagram FIGURE $5.69 First note that all the forces intersect at common axis (that of the
(continued) 443 line D—C), thus there will be only 5 independent equations and only 5
variables can be determined (Dz, Dy, Dz1 T1 and T2). The vectors T1
and T2 must ﬁrst be written in terms of unit vectors along CB and CA respectively: —B A i A ICBI on = 2  T2 = T2: = T2(—0.8481 + 0.424J + 0.318k). ICAI
Equilibrium in the coordinate directions becomes
2 F2 = 0: DE — 0.848T1 — 0.848% = 0 (1)
SF, = 0 : Dy + 0.424T1 + 0.424T2 = 600 (2)
ER, = 0: D,z — 0.318T1 + 0.31813 2 0 (3) The moments about point 0 yield
(—0.25i) x (—3003)+(—1.75i) >< (H3003)+(—2i) x (Dxi+Dyj +0211) = 0 or by component i: 0:0 (4)
j: 2Dz=0 ' (5)
1E: Dy=300 ' (6) Solving equations 1, 2, 3 with Dz=OandDy=300N yields
Dz=600N T1=354N T2=354N 444 5.71 The total force acting on a. telephone pole due to the wires attached to
it is computed to be F = 100i — 50j + IOkN. Compute the reaction at
the ﬁxed connection at pointA. FIGURE P5.71 Solution: The free body diagram is given in the ﬁgure. The equations
of equilibrium become simply ' F=1oof50j‘+1ow° FIGURE 85.71 (continued) 447 2F =0:Am+100=00rAx=—100N
ZFz=osz—50=OorAy=50N
ZFz=02Az+10=00rAz=—10N
2MB =0 : Mzi+Myj+leE+(3j—12)m x (1003—503+1012)N=0
= (Mz+30—50)i+(My —1oo)j+ (Mz — 30ml}: 0 or
Mz=20Nm, My=100Nm, Mz=300Nm 448 5.73 A folding platform is used to hold parts as well as conserve ﬂoor space when
not in use. The platform is supported by a. hinge at G which is assumed to
support negligible moments, a leg, at B modeled as a frictionless support and
a removable pin at A modeled as a thrustless bearing again with negligible
moments. If the platform is loaded as illustrated compute the reaction forces
at A, B and C. Ignore the thickness of the platform. FIGURE P5.73 Solution: The free body diagram is given in ﬁgure $5.73.
The equations of equilibrium become: (continued) 451 oLm.
’h—r‘ zcx—’—————K
/ 04m
12m/4‘m 6: I ‘
A: x i z
y’ _ _ _ _ __ aoouu
1m
,Az {
I
2 By FIGURE 35.73 Force Summation: 2F1=0:Cm=0 (1)
2Fy=o:Ay+Cy+B =3000 (2)
EFZ=0:AZ+CZ=O (3) The moment equation is 2 Mo = (.5? + .412) x (—3000,?) + (.55) x (03+ Cy} + 0,12) + (i + 1.212) x By}
+1.21} x (ij + Azll) 2 0 Which yields the 3 component equations i:1200 m 1.235, — 1.2A,, 2 0 j : 41.50,, = 0 12 : 4500 + .501” +33, = u “Al—N
65min. Equations (1), (3) and (5) yield
Oz 2 A; = 03 = 0
by inspection. The remaining 3 equations (2, 4, 6) (solved in Mathcad) yield A” = 500 N,
By = 500 N
and 452 ...
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This note was uploaded on 07/25/2008 for the course ME 221 taught by Professor Buch during the Summer '08 term at Michigan State University.
 Summer '08
 Buch
 Statics

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