Equilibrium, Elasticity, and Hooke’s Law. Conditions for equilibrium:. Exam Example 26: Ladder against wall (example 11.3, problem 11.10). d/2. h . x . Static equilibrium:. y. State with is equilibrium but is not static. Data: m, M, d, h, y, μ s Find: (a) F 2 , (b) F 1 , f s ,. d. θ. - PowerPoint PPT Presentation
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Equilibrium, Elasticity, and Hooke’s La 0 0 ) 2 ( 0 0 ) 1 ( i i i a F nditions for equilibrium: atic equilibrium: 0 v with uilibrium but is not static. 0 const v rategy of problem solution: e of the axis of rotation: y - the simpler the better. e-body diagram: all external forces and ints of action. lculate lever arm and or each force. ve for unknowns. 0 F 0 Exam Example 26: Ladder against wall (example 11.3, problem 11.10) (c) y man when ladder starts to slip Data: m, M, d, h, y, μ s Find: (a) F 2 , (b) F 1 , f , 1 F 2 F s f g m g M h θ x d/2 y Solution: equilibrium equations yield (a) F 2 = Mg + mg ; (b) F 1 = f s Choice of B-axis (no torque from F 2 and f s ) F 1 h = mgx + Mgd/2 → F 1 = g(mx+Md/2)/h = f s (c) Ladder starts to slip when μ s F 2 = f s , x → μ s g (M+m) = g (my man d/h+Md/2)/h → d h d md Mh d h m Mh md h m M y s s s man 2 2 ) ( 2 2 2
Transcript
Equilibrium, Elasticity, and Hooke’s Law
00)2(
00)1(
i
ii aF
Conditions for equilibrium:
Static equilibrium: 0v
State with is equilibrium but is not static.
0constv
Strategy of problem solution:(0)(i) Choice of the axis of rotation:arbitrary - the simpler the better.(ii) Free-body diagram: identify all external forces and their points of action.(iii) Calculate lever arm and torque for each force.(iv) Solve for unknowns.
0F
0
Exam Example 26: Ladder against wall (example 11.3, problem 11.10)
(c) yman when ladder starts to slip
Data: m, M, d, h, y, μsFind: (a) F2, (b) F1, fs,
1F
2F
sf
gmgM h
θ
x
d/2
y
Solution: equilibrium equations yield(a) F2= Mg + mg ; (b) F1 = fs Choice of B-axis (no torque from F2 and fs) F1h = mgx + Mgd/2 → F1= g(mx+Md/2)/h = fs
(c) Ladder starts to slip when μsF2 = fs, x = yd/h→ μsg (M+m) = g (mymand/h+Md/2)/h →
d
hd
mdMh
dh
mMh
mdhmMy s
ssman 22
)( 222
Center of GravityDefinition. The center of gravity of a rigid body isthe point at which its weight can be considered to actwhen calculating the torque due to gravity:
gMrcg
Derivation:
i i
cmii
iii gMrgMMrm
gmr
)(
Mrm
rr iicmcg
(if has the same value at all points on a body)g
cmcg
cgiz
xgmmgmxgmxx
gmmx
gmxgmx
)(
)(
21
2211
21
2211
Center of gravitycan be located bysuspending objectfrom different points.
Note: Center of gravity may lie “outside” the object. gM
The larger the area of support and the lower the center of gravity,the more difficult it is to overturn a body.
Stress, Strain, and Elastic ModuliStress = (Elastic modulus) · Strain
