Chap. 11: Static Equilibrium - Physics and Astronomy at...
Transcript of Chap. 11: Static Equilibrium - Physics and Astronomy at...
Chap. 11: Static
Equilibrium
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It doesn’t move!
0
0
net
net
F
Static Equilibrium
Equations
Calculate the torque along with the rotational axis!
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Static Equilibrium
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Real Life Phys 218
Static Equilibrium
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Real Life Phys 218
Static Equilibrium
Static Equilibrium Analysis
0 n e t
i
Identify each force on a rigid body or a system
Calculate the torque (magnitude and direction) by each force
Not rotating
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Static Equilibrium
We assume:
Center of mass (CM or cm)
= Center of gravity
(if uniform gravity)
? F r
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Torque due to Gravity
Static Equilibrium
What is the unknown mass? BALANCEING
in mass x distance
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Static Equilibrium
torque = zero
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What is the unknown mass?
Static Equilibrium
M
m
FT = ?
A metal advertising sign (mass M) is suspended from
the end of a horizontal rod of length L and mass m.
ISEE – Where is physics?
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Static Equilibrium
FRx – FT cosq = 0
FRy + FT sinq – mg – Mg = 0
FT sinq L – mg (L/2) – Mg L= 0
0P
M, m, L, h , q : given FRy?, FRx?, FT?
ISEE – Static Equilibrium Eqs.
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Static Equilibrium
Now You Can Do This!
M = 30.61 kg m = 5.00 kg
You design this and use a string that sustain up to 50 N.
Will your customer be happy?
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Static Equilibrium
Try This!
M
m
l
FT = ?
You design this and use a string that sustain up to 500 N.
Will your customer be happy?
M = 30.61 kg m = 5.00 kg
l = 4.00 m
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Static Equilibrium
You design this and use a string that sustain up to 50 N.
Will your customer be happy?
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Static Equilibrium
M, m, l, q , ms: given
x?
Sm
oo
th su
rface
Rough surface
y
Note: sinq = 4/5, cosq = 3/5
l
Particle
(mass M)
x
x
q
… Model and F.B.D.
Mg
ISEE – F.B.D. for Ladder
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Static Equilibrium
M, m, l, q , ms: given
x? (FW?)
Sm
oo
th su
rface
Rough surface x
y
FGx – FW = 0
FGy – mg – Mg = 0
FGx = ms Fgy
FW sinq l
– mg (l/2) cosq – Mg x = 0 q
Note: sinq = 4/5, cosq = 3/5
l
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ISEE – Equilibrium Eqs.
Static Equilibrium
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)about (rotation 0
0
0
P
F
F
ii,P
ii,y
ii,x
ISEE – Equilibrium Eqs.
P P
P
Static Equilibrium
Try this!
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Static Equilibrium
Try this!
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Try this!
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You want to raise a bicycle wheel (mass m and radius R) up over a curb of height h. What is the smallest magnitude of F that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel and (b) at the top of the wheel?
r
r
F
(cw) )((cw) Fh-RFr F r F
mg
mg
hR R
22)( hRR
(ccw) )((ccw) 22
mghRRmgr gm r g
Chap. 11: Static
Equilibrium
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Strain, stress, and elastic moduli
Strain, Stress, and Elastic Moduli • Stretching, squeezing, and twisting a real body causes it to
deform, as shown in Figure 11.12 below. We shall study the
relationship between forces and the deformations they cause.
• Stress is the force per unit area and strain is the fractional
deformation due to the stress. Elastic modulus is stress divided
by strain.
• The proportionality of stress and strain is called Hooke’s law.
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Tensile and Compressive Stress and Strain
• Tensile stress = F /A and Tensile strain = l/l0.
• Compressive stress and Compressive strain are defined in a
similar way. (See Figures 11.13 and 11.14 below.)
• Young’s modulus is tensile stress divided by tensile strain, and is
given by Y = (F/A)/(l / l0).
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Tensile Stress and Strain
• In many cases, a body can experience both tensile and
compressive stress at the same time, as shown in Figure 11.15
below.
• Follow Example 11.5.
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Stress = F/A
A
Bulk Stress and Strain • Pressure in a fluid is force per
unit area: p = F/A.
• Bulk stress is pressure change
p and bulk strain is fractional
volume change V/V0. (See
Figure 11.16.)
• Bulk modulus is bulk stress
divided by bulk strain and is
given by B = –p/(V/V0).
• Follow Example 11.6.
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Sheer Stress and Strain
• Sheer Stress is F||/A and
Sheer Strain is x/h, as shown
in Figure 11.17.
• Sheer modulus is sheer stress
divided by sheer strain, and
is given by S = (F||/A)/(x/h).
• Follow Example 11.7.
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Some Values of Elastic Moduli
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Problem 1 A wire of length l0 and cross-sectional area A supports a hanging
weight W. (a) Show that if the wire obeys Eq. (11.7), it behaves like a
spring of force constant AY/l0, where Y is Young’s modulus for the
material of which the wire is made. (b) What would the force constant
be for a 75.0-cm length of 16-gauge (diameter = 1.291 mm) copper
wire? See Table 11.1. (c) What would W have to be to stretch the wire
in part (b) by 1.25 mm?
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Problem 2 In lab tests on a 9.25-cm cube 0f a certain
material, a force of 1375 N directed at 8.50o to
the cube causes the cube to deform through an
angle of 1.24o. What is the shear modulus of
the material?
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qq tanh
x
q
S = (F||/A)/(x/h).
Elasticity and Plasticity
• Hooke’s law applies up to point a in Figure 11.18 below.
• Table 11.3 shows some approximate breaking stresses.
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Quizzes
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A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same
strain.
Q11.5
Two rods are made of the
same kind of steel and have
the same diameter.
F F length 2L
F F length L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
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Y = (F/A)(l0/l)
A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same
strain.
Q11.6
Two rods are made of the
same kind of steel. The
longer rod has a greater
diameter. F F length 2L
F F length L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
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Y = (F/A)(l0/l)
Which of the following situations satisfies both the first
condition for equilibrium (net force = 0) and the second
condition for equilibrium (net torque = 0)?
A. an automobile crankshaft turning at an increasing
angular speed in the engine of a parked car
B. a seagull gliding at a constant angle below the
horizontal and at a constant speed
C. a thrown baseball that does not rotate as it sails
through the air
D. more than one of the above
E. none of the above
Q11.1
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A. T = w sin q
B. T = w cos q
C. T = w/(sin q)
D. T = w/(cos q)
E. none of the above
Q11.3
A metal advertising sign (weight w)
is suspended from the end of a
massless rod of length L. The rod is
supported at one end by a hinge at
point P and at the other end by a
cable at an angle q from the
horizontal.
What is the tension in the cable?
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A. the weight of the sign
B. the tension in the cable
C. the vertical force component exerted on the rod by hinge
P
D. two or more of these are tied for least
A11.4
A metal advertising sign (weight w)
is suspended from the end of a
massless rod of length L. The rod is
supported at one end by a hinge at
point P and at the other end by a
cable at an angle q from the
horizontal.
Which of these forces is least?
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