Experimental Physics - Mechanics - Static Equilibrium 1 Experimental Physics EP1 MECHANICS - Static Equilibrium - Rustem Valiullin https://bloch.physgeo.uni - leipzig.de/amr/
Experimental Physics - Mechanics - Static Equilibrium 1
Experimental Physics EP1 MECHANICS
- Static Equilibrium -
Rustem Valiullin
https://bloch.physgeo.uni-leipzig.de/amr/
Experimental Physics - Mechanics - Static Equilibrium 2
O
r! iF!
Conditions for equilibrium
FdrFFr ==´= qt sin!!!
0=netF!F
!
O
r!
F!'r!
The net external force must be zero
0=nett!The net external torque about any axis must be zero
'r!
'O'OOr
!
å ==i
inet FF 0!!
å ´=i
iiOOnet Fr!!!
,t
å ´=i
iiOnet Fr!!! '',t ..'..' 11 +´++´= ii FrFr
!!!!
..)(..)( '1'1 +´-++´-= iOOiOO FrrFrr!!!!!!
åå ´-÷ø
öçè
æ ´=i
iOOii
i FrFr!!!!
' Onet ,t!=
If an object is in translational equilibrium and the net torque iszero about some point, then the net torque is zero about any point.
Experimental Physics - Mechanics - Static Equilibrium 3
Center of gravity
gm!
CM
CM
òò
åå ==
dm
dmrmmr
ri
iiCM
!!!
O
gmi
Mg
gmrMgrgmr iiCGiCG åå ==!!!
If g is constant over an object, then the center of massand the center of gravity of the object do coincide.
Center of mass:
Experimental Physics - Mechanics - Static Equilibrium 4
The ladder problem
Mg
L
0=wµ
fµ
afN
wN
sf
0=netF!
0=nett!îíì
=-=-00
MgNfN
f
sw
0cos2
sin =- aa LMgLNw
sw fMgN == acot2
ffss Nff µ=£ max,
ffff MgNMg µaµµa 2cotcot2
£Þ=£
-1,5 -1,0 -0,5 0,0 0,5 1,0 1,5
-3,0
-2,5
-2,0
-1,5
-1,0
-0,5
0,0
0,5
1,0
1,5
2,0
2,5
3,0
tan(a)
angle
fµa
21tan ³
°» 25cra
°-°» 4560cra
O
Experimental Physics - Mechanics - Static Equilibrium 5
The ladder climbing problem
Mg
L
0=wµ
fµ
a
fN
wN
sf
mgl
0=netF!
0=nett!îíì
=+-=-
0)(0gmMN
fN
f
sw
0cos2
sin =÷øö
çèæ +- aa mglLMgLNw
ffss Nff µ=£ max,sw fLmlMgN =÷
øö
çèæ +=2
cota
( )gmMNf fffs +== µµmax,
( )22
2 LlgmMLmlMg ff £Þ+£÷
øö
çèæ + µµ
( )gmMLmlMg f +£÷
øö
çèæ + µa2
cot
( )mLM
mLmM
l f
2tan -
+£ aµ
aµ tan: LlMm f£>>
general case
Experimental Physics - Mechanics - Static Equilibrium 6
Inclined domino set
L
23
45
l llL -
022=-
- crcr lAglLAg rr
CM
2Llcr =
0
1
aL /
21L
aL
x +=2
22L
aL
x +=2L
aLixi +=
åå
å=
=
=úûù
êëé +==
n
in
ii
n
iii
nCML
aLi
nm
xmX
1
1
1, 2
1
( )aLnLnn
aL
nLn
nX nCM 2
)1(22
112
1, ++=++»
LX nCM £,an £+156 =Þ= na
The sixth domino will crush the system.
5
Experimental Physics - Mechanics - Static Equilibrium 7
Ø The two conditions have to be fulfilled for a body to be
in static equilibrium:
- the net external force must be zero;
- the net external torque must be zero.
Ø If an object is in static equilibrium under action of
three non-parallel forces, the lines of action of these
forces must intersect at one point.
Ø The force of gravity can be replaced by
single force acting at the center of gravity.
To remember!