l05 - equilibrium
TRANSCRIPT
-
8/18/2019 L05 - Equilibrium
1/25
January 20/21, 2014
ECOR 1101 Mechanics I
Sections C and F
Jack Vandener!
Lecture 04 – Equilibrium and FBDs
(Chapter 3 – Sections 3.!3.4"
-
8/18/2019 L05 - Equilibrium
2/25
O"ecti#es
Learn the concept o# #ree!bod$ dia%rams
Learn to sol&e problems in&ol&in% particles in equilibrium
Learn to use equilibrium equations usin% cartesian &ector
coordinates
2
'D and 3D Equilibrium o# a particle
ECOR1101 –Mechanics I
-
8/18/2019 L05 - Equilibrium
3/25
Equilibrium o# a article
) particle is in equilibrium i#*
+t remains at rest under action o# a s$stem o# #orces, or, +t continues in its state o# motion -ith constant &elocit$
under action o# a s$stem o# #orces.
For a particle to be in equilibrium the resultant o# all
#orces actin% on it must be ero.
Satis#ies /e-tons st la- o# motion
1he abo&e equation represents the necessar$ and
su##icient condition #or equilibrium o# a particle in
space.
3
F R= F x ∑ + F y + F z = 0∑∑
-
8/18/2019 L05 - Equilibrium
4/25
)ccordin% to /e-tons 'nd la- o# motion, i# ΣF = ma = 0,
the particle is in equilibrium since a = 0 and ΣF = 0 i.e. the particle is under constant &elocit$ or is at rest
1he equilibrium equation can be used to sol&e problems
dealin% -ith equilibrium o# a particle in&ol&in% no more
than three un2no-ns
4
Equilibrium o# a article
ECOR1101 –Mechanics I
-
8/18/2019 L05 - Equilibrium
5/25
1o appl$ the equations o# equilibrium to a particle all
#orces (2no-n and un2no-n" must be accounted #or. 1he best -a$ to do this is to isolate the particle #rom its
surroundin%s to #orm a #ree!bod$ dia%ram (FBD".
1hen appl$ all the #orces (2no-n and un2no-n" actin%
on the particle
5
Free!bod$ Dia%rams (FBD"
ECOR1101 –Mechanics I
-
8/18/2019 L05 - Equilibrium
6/25
+solate particle #rom its surroundin%
S2etch outline shape o# particle -ith all #orces (acti&e
and reacti&e" indicated
Label all #orces (2no-n and un2no-n" -ith both their
ma%nitudes and directions
+# $ou 2no- that an un2no-n #orce is in tension, do $ou
dra- it a-a$ or to-ards the particle5
6
rocedure #or dra-in% FBD
ECOR1101 –Mechanics I
-
8/18/2019 L05 - Equilibrium
7/25
Free Bod$ Dia%rams (FBDs"
7
-
8/18/2019 L05 - Equilibrium
8/25
FBDs
8
6i%id Bodies
Sprin%s 7oo2es la-
F 8 2 s
F 8 sprin% constant 9 displacement
Cables (assumptions"
:ust be in tension /e%li%ible -ei%ht
Do not stretch
Frictionless pulle$
-
8/18/2019 L05 - Equilibrium
9/25
FBDs
9
Draw a FBD of the cable AB and of the joint C.
-
8/18/2019 L05 - Equilibrium
10/25
Sample roblem
) '00!2% c$linder is hun% b$ means o# t-ocables AB and AC, -hich are attached to the
top o# a &ertical -all. ) horiontal #orce $
perpendicular to the -all holds the c$linder in
the position sho-n. Determine the ma%nitude
o# P and tension in each cable
10ECOR1101 –Mechanics I
!
10 !
1" !1." !
" !
A
B
C
A
B
C
P
P
w
TAC
TABk
i j
+ntroduce unit &ectors i, ", k alon%
ortho%onal a9es and resol&e #orces
P # P i $ 0 j $ 0k
% = 0i $ 0 j % mg k 8
8 0i $ 0 j ! '00×;.
-
8/18/2019 L05 - Equilibrium
11/25
11ECOR1101 –Mechanics I
r AB # &%1."!'i % &!' j $ &10!' k
r AC # &%1."!'i $ &10!' j$ &10!'k
( ) ( )( ) ( )
mk, jir
u
mk, jir
u
m
m
ACAC
ABAB
7046.07046.00846.0
778.0622.0093.0
193.1410102.1
862.121082.1222
222
++−==
+−−==
=++−=
=+−+−=
AC
AB
AC
AB
r
r
r
r
0
0
=+++∴
=++=∑ ∑ ∑ ∑
ACAB
zyx
uuWP
FFFF
AC ABT T
(ince the c)linder is *nder e+*ilibri*!,
X AB AC AB AC
y AB AC AB AC
z Ab AC AB AC
AB
AC
F 0 P 0 0.093T 0.0846T 0 P 0.093T 0.0846T
F 0 0 0 0.622T 0.7046T 0 T 1.133T
F 0 0 1962 0.778T 0.7046T 0 T 0.9056T 2521.851
T 1401.6 N 1.40 kN
T 1237.1 N 1.24 kN
P 235 N
= ⇒ + − − = ⇒ = +
= ⇒ + − + = ⇒ =
= ⇒ − + + = ⇒ = − +
= =
= =
=
∑
∑∑
-
8/18/2019 L05 - Equilibrium
12/25
Cables, Sprin%s and ulle$s
Cables (or cords", in %eneral,
are assumed to ha&e the
#ollo-in% properties
>ei%htless
Supports onl$ tension in the
direction o# the cable (cannotbe pushed"
Cannot stretch (i.e. increase
in len%th under load"
) cable passin% o&er a#rictionless pulle$ has a constant
ma%nitude
12ECOR1101–Mechanics I
-
8/18/2019 L05 - Equilibrium
13/25
Cables, ulle$s and Sprin%s
Sprin%s, -hen de#ormed, e9ert
a #orce proportional to the
amount o# de#ormation.
Sprin%s are o#ten de#ined b$ the
sprin% constant or sti##ness k
1he ma%nitude o# #orce e9ertedon a linearl$ elastic sprin% -ith
sti##ness k is %i&en b$* F = ks
s = l − l o ,
l o = unstretched len%th,l = stretched len%th
13ECOR1101–Mechanics I
-
8/18/2019 L05 - Equilibrium
14/25
Coplanar Force S$stems ('D"
14ECOR1101–Mechanics I
F1
F2
x
y
F 1x
F 2x
F 2y
F 1y
1he t-o equations o# equilibrium can
be sol&ed #or at most t-o un2no-ns.
)pplication o# the equation must ta2e
into account direction o# components
o# the #orce
+# a particle sub?ected
to a s$stem o# coplanar#orces (9!$ plane", then
the #orces can be
resol&ed and
equilibrium equations
applied.
-
8/18/2019 L05 - Equilibrium
15/25
Establish x-y a9es
Dra- a #ree!bod$ dia%ram
Dra- and label all #orces (2no-n and un2no-n" -ith
ma%nitudes, sense and direction
Choose an arbitrar$ direction #or un2no-n #orces
6esol&e #orces in x-y a9es
)ppl$ equations o# equilibrium
)ssume a @&e direction #or the purpose o# -ritin% $our
equation o# equilibrium
Sol&e #or un2no-n #orces Compare $our ans-ers to $our ori%inal assumption (not to
the @&e direction -hen -ritin% $our equations"
6edra- $our FBD -ith all Forces as positi&e numbers
15
rocedure For )nal$sis o# Coplanar ('D" Force Equilibrium
ECOR1101–Mechanics I
-
8/18/2019 L05 - Equilibrium
16/25
1hree!Dimensional (3D" Force S$stems
Conditions #or equilibrium
6esol&e #orces into respecti&e
Cartesian components, i, j, k
16ECOR1101–Mechanics I
x
z
F1
F1zF1x
F1y
F2
F2y
F2z
F2x
0 F =∑
00
0
0
0
=+ + =
=
=
=
∑∑ ∑ ∑
∑
∑
∑
F
i j k x y z
x
y
z
F F F
F
F
F 1he three equation o# equilibrium are al%ebraic sums
o# #orce components and can be used to #ind at most
three un2no-ns (coordinate direction an%les or
ma%nitudes o# #orces actin% on a particle"
-
8/18/2019 L05 - Equilibrium
17/25
Establish x-y-z a9es
Dra- a #ree!bod$ dia%ram
Dra- and label all #orces (2no-n and un2no-n" -ith
ma%nitudes, sense and direction
Choose an arbitrar$ direction #or un2no-n #orces
6esol&e #orces in x-y-z a9es
)ppl$ equations o# equilibrium
)ssume a @&e direction #or the purpose o# -ritin% $our
equation o# equilibrium
Sol&e #or un2no-n #orces Compare $our ans-ers to $our ori%inal assumption (not to
the @&e direction -hen -ritin% $our equations"
17
rocedure For )nal$sis o# 3D Force S$stems
ECOR1101–Mechanics I
-
8/18/2019 L05 - Equilibrium
18/25
Sample roblem
1he shear le% derric2 is
used to haul the '00!2% net
o# #ish onto the doc2.
Determine the compressi&e
#orce alon% each o# the
le%s )B and CB and the
tension in the -inch cable
DB. )ssume the #orce in
each le% acts alon% its a9is.
18ECOR1101–Mechanics I
-
8/18/2019 L05 - Equilibrium
19/25
19ECOR1101–Mechanics I
W!"# "$# %&&d!na"#' (& )&!n"' A, B, C, and
*, )&'!"!&n +#%"&', n!" +#%"&', F&%#
-#%"&'
A2m, 0, 0/
B0, 4m, 4m/
C2m, 0, 0/
*0, 5.6m, 0/
W
FBD
x
y
z
FBA
FBC
B
A
C
D
"
-
-./
"
-
8/18/2019 L05 - Equilibrium
20/25
20ECOR1101–Mechanics I
rBA= 2mi 4m j 4mk
rBC= 2mi 4m j 4mk
rBD= 0mi 9.6m j 4mk
W = 0i 0 j 200×9.81/k
= 0i 0 j 1962Nk
,BA = BA
r BA
=2!− 4− 4k
2/2
+ −4( )2
+ −4( )2
= 0.333!−0.667−0.667k
( ) ( ) , m
r BC k ji
k jiru BCBC 667.0667.0333.0
44/2
442
222−−−=
−+−+−
−−−==
( ) ( ) , m
r BDk ji
k jiru BDBD 385.0923.00
46.9/0
46.90
222−−=
−+−+
−−==
-
8/18/2019 L05 - Equilibrium
21/25
21ECOR1101–Mechanics I
kN
kN
kN N
WuuuF BDBCBA
65.3
52.2
52.285.2521
01962385.0667.0667.00
445.1
0923.0667.0667.00
0333.0333.00
00
=
−=
−=−=
=−−−−⇒=
−=∴
=−−−⇒=
=∴
=−⇒=
=+++⇒=
∑
∑
∑∑
DB
BC
BA
BD BC BA z
BA BD
BD BC BA y
BC BA
BC BA x
BD BC BA
F
F
F
F F F F
F F
F F F F
F F
F F F
F F F
W
FBD
x
z
FBA
FBC
B
A
C
D
-
8/18/2019 L05 - Equilibrium
22/25
roblem F3!<
22
Deter!ine the tension deeloed in cables AB2 AC2 and AD.
-
8/18/2019 L05 - Equilibrium
23/25
roblem 3!A;
23
Deter!ine the !a3i!*! wei4ht of the crate that can be s*orted
fro! cables AB2 AC2 and AD so that the tension deeloed in an)one of the cables does not e3ceed "0 lb.
-
8/18/2019 L05 - Equilibrium
24/25
roblem 3!=
24
If cable AD is ti4htened b) a t*rnb*c5le and deelos a tension of
12600 lb2 deter!ine the tension deeloed in cables AB and ACand the force deeloed alon4 the antenna tower AE at oint A.
-
8/18/2019 L05 - Equilibrium
25/25
roblem 3!
25
7he joint of a sace fra!e is s*bjected to fo*r !e!ber forces.
Me!ber OA lies in the 3%) lane and !e!ber OB lies in the )%8lane. Deter!ine the forces actin4 in each of the !e!bers re+*ired
for e+*ilibri*! at the joint.