Unformatted text preview: By definition, an object is in equilibrium when it is either
at rest or moving with constant velocity , i.e., with no
CHAPTER 12 acceleration. The following are examples of objects in
static equilibrium ... STATIC EQUILIBRIUM AND ELASTICITY
• Conditions for static equilibrium o Center of gravity
o Examples of equilibrium
• Couples (you study)
Garden of the Gods
Colorado Springs • Elasticity o Stress and strain
o Young’s modulus
o Hooke’s Law and the elastic limit
o Shear modulus (you study) Alexei Therefore, two conditions are necessary for a body to be
in static equilibrium ...
• The net external force acting on the body is zero,
i.e., ∑i Fi = 0, Question 1: A seesaw consists of a board of length 4.0m so there is no translational motion pivoted at its center. A 28kg child sits at one end of the
board. Where should a 40kg child sit to balance the • The net external torque about any point is zero,
i.e., ∑i τi = 0,
so there is no rotational motion
Let’s do a problem as an illustration ... seesaw? First, we must identify all the forces ... We assumed the weight force acts through the center of 40kg y mass ... strictly speaking it should be the center of
weight. What’s the connection between the two? 28kg center of gravity
(center of weight) x
4m d xCG For static and rotational equilibrium: r
∑i Fyi = 0 and ∑i τi = 0.
Therefore, the pivot must supply an upward force so that
the net force on the board is zero, i.e.,
F − (28 × g + 40 × g) = 0 ∴ F = 68 × g = 666.4 N.
Define ccw torques as positive and taking torques about
the pivot point we have: (28 × g) × 2 − (40 × g) × d = 0
= 1.40 m .
So, for equilibrium Cathy must stand 1.40m from the
pivot point. O r
W = ∑ i mi g i O xi r
w i = mi g i Take some origin, O, and “break” the object into small
elements of mass mi . The total torque about O is: τo = Wx CG = ∑i mi gi x i ,
where x CG is the center of gravity (weight) defined as
x CG = i i i = i i i .
(This is similar to the definition of the center of mass.)
The center of mass is at the same point as the center of
weight, i.e., x CM ≡ x CG , only if g is constant, then
τo = ( ∑i mi x i )g = Mgx CM = Wx CM . CLASS DISCUSSION PROBLEM [12.1]: O r
W = ∑ i mi g i If the origin is taken at the center of gravity, then the
object produces zero torque about that point. (a) (b) The center of gravity is then ... the point about which the Why is it you cannot touch your toes without falling over gravitational force on the object produces zero torque if you have your heels against a wall as in (a) ... yet, no matter what the orientation of the object. It could ordinarily, you have no problem, as in (b)? also be called ... the center of weight!
Use the proper “physics” terminology, please! If g is constant over the object then the center of gravity
is located at the center of mass. 1m r
Tx y 2m x
1m Fy 2m 4kg Fx
Force of the wall
on the rod
20kg The rod must supply a force ... how do we know that?
However, we have no idea in what direction so choose it
arbitrarily to begin with. Note, by Newton’s 3rd Law Question 2: A sign hangs in front of a store. The sign the force the rod exerts on the wall is equal and opposite has a mass of 20kg and it hangs at the end of a horizontal to the force the wall exerts on the rod. rod of length 2m and mass 4kg, which is hinged at the
wall. The rod is supported by a wire attached to a point
on the wall 1m above the rod. (a) What is the tension in
the wire? (b) What is the magnitude and direction of the
force the rod exerts on the wall? (a) For static and rotational equilibrium: r
∑i Fxi = 0, ∑i Fyi = 0 and ∑i τi = 0. Take torques about the hinge ... a great idea since we
don’t know anything about F . Then
( Ty × 2) − (4 × g × 1) − (20 × g × 2) = 0
∴ Ty = 215.6 N. r
F 1m r
Tx 2m y Tx y 2m x
Fy 4kg Fx 4kg
physics! Fx 20kg Ty Ty θ x
F 1m r
= tan θ = .
2 20kg (b) For static and rotational equilibrium: r
∑i Fxi = 0, ∑i Fyi = 0 and ∑i τi = 0. ∴ Tx = 2 Ty , i.e., Tx = −431.2 N.
∴ T = Tx2 + Ty2 = ( −431.2)2 + (215.6)2 So ∑i Fxi = Fx + Tx = Fx − 431.2 N = 0
∴ Fx = 431.2 N, ∑i Fyi = Fy + Ty − (20 × g) − (4 × g) = 0 and ∴ Fy = 19.6 N.
F = ( 431.2ˆ + 19.6ˆ ) N
j = 482.1N r
Note ... we found T even though we
know nothing about F! I
physics! Fy φ
Fx 19.6 φ = tan −1 = 2.6o 431.2 The force of the rod on the wall is therefore
F = −(431.2ˆ + 19.6ˆ ) N.
F 1m r
Tx y 2m Question 3: A square plate is made by welding together
x Fy four smaller square plates, each of side l. Each of the
smaller squares is made from a different material so they 4kg
physics! have different weights, as shown in the figure. (a) Find 20kg Fx suspended from the point O, what is the angle between r
Note: we could get F another way ... (a) by taking
torques about the right hand end of the rod, since the
system is in static equilibrium, ∑i τi = 0 about any point, the position of the center of gravity. (b) If the plate is
the vertical and the left-hand side of the plate?
O l 60N 30N l 40N 50N l l i.e., (4 × g × 1) − (Fy × 2) = 0. ∴ Fy = 19.6 N.
(b) By taking torques about the top end of the wire,
i.e., Fx × 1 − 4 × g × 1 − 20 × g × 2 = 0 .
∴ Fx = 431.6 N. (a) By definition, the center of O
l 60N 30N gravity (weight), with respect
to the lower left corner is given l 40N 50N l l (0, 0) by x CG = ∑i wi x i
W (b) When hung from O, a vertical line through O passes
through the CG. (60 × l 2 + 40 × l 2 + 30 × 3l 2 + 50 × 3l 2)
lφ (60 + 40 + 30 + 50) = 0.944l ,
and y CG = i i i
W = (40 × l 2 + 50 × l 2 + 60 × 3l 2 + 30 × 3l 2)
(60 + 40 + 30 + 50) = 1.00l.
Therefore, the coordinates of the CG are (0.944l, l) . l
0.944l CG ∴ tan φ = 0.944l
l i.e., φ = 43.3o. Let the ladder, length l, rest at an angle θ against the
wall. Identify all of the force acting on the ladder.
F1 Since the wall is
frictionless, the force of l r
Fn Question 4: A uniform ladder rests against a frictionless,
vertical wall. If the coefficient of static friction between
the bottom of the ladder and the floor is 0.3, what is the
smallest angle at which the ladder will remain stationary? l sin θ r
w the wall acting on the
ladder is F1, which is ⊥ to
the wall, i.e., a normal θ
fs force. The force exerts two forces on the ladder; a
normal force Fn and a
static frictional force f s. The weight of the ladder is w,
l cos θ l cos θ which acts through the CG, i.e., the middle of the ladder.
For static and rotational equilibrium: v
∑i Fxi = 0, ∑i Fyi = 0 and ∑i τi = 0.
∴ ∑i Fxi = F1 − f = 0 ... ... ... (1) and ∑i Fyi = Fn − w = 0 ... ... ... (2) A couple Taking torques about the foot of the ladder:
∑i τi = w cos θ − F1 (l sin θ) = 0 ... (3)
w l sin θ = 59.0o .
2 l cos θ 2 O• l r
F2 Two forces form a couple if they are equal but opposite F1 = f s = µ sFn = µ s w
∴ θ = tan −1 12µ
s ( θ
fs l cos θ w
2F1 But from (1) and (2): l r
Fn ∴ tan θ = r
F1 ) and their line of action is separated (by a distance l):
i.e., F1 = F2 = F.
Then the torque about any point O is:
τ = ( x1 × F1 ) + ( x 2 × F2 )
i.e., τ = x1F − x 2F = ( x1 − x 2 )F = lF
So the torque produced by a couple is the same about
all points in space, i.e., no matter where O is located ...
it depends only on their perpendicular separation. Stress and strain (Young’s modulus)
F However, the strain is reversible over only a limited
range of stress ... F tensile
stress ∆L + L Objects deform when subjected to a force. In the case of
the length of a metal bar or wire ...
The stress ⇒ F A , and the strain ⇒ ∆L L .
Young’s modulus ⇒ Y =
() Units: Force/area ⇒ N ⋅ m−2 Elastic limit Fracture Plastic behavior Hooke’s Law
deformation strain ... up to the elastic limit . Beyond that the material
shows plastic behavior until it reaches the point of
fracture (tensile strength). • Thomas Young (1773-1829)
The example above is tensile stress. There is also The initial linear behavior is known as Hooke’s Law ... compressive stress. Young’s modulus is often the same after Robert Hooke (1635-1703). in both cases, but there are exceptions (e.g., bone). (a) By definition Y =
d = 1.00mm Question 5: A steel wire of length 1.50m an diameter (F A ) .
( ∆L L ) 1.50 m aluminum 1.50 m steel 1.00mm is joined to an aluminum wire of identical
dimensions to make a composite wire of length 3.00m.
(a) What is the length of the composite wire if it is used weights of the wires are negligible.
Aluminum: YAl = 70 × 109 N ⋅ m−2 . Tensile strength = 90 × 106 N ⋅ m−2 .
Steel: Ysteel = 200 × 109 N ⋅ m−2 .
Tensile strength = 520 × 106 N ⋅ m−2 . = 6.25 × 107 N ⋅ m−2. ∴ ∆L = 6.25 × 107 × to support a mass of 5kg? (b) What is the maximum load
the composite wire could withstand? Assume the The stress in each wire is:
5 × 9.81
A π × (0.0005)2 5kg L
= 1.34 mm.
70 × 109
∆L2 = 6.25 × 107 ×
= 0.47 mm.
200 × 109 Aluminum: ∆L1 = 6.25 × 107 ×
Steel: ∴ Increase in length ⇒ (1.34 + 0.47)mm
i.e.,“new” length ⇒ 3.00181m.
(b) The tensile strength of aluminum is less than that of
steel. Consequently, as the load increases, it will be the
aluminum wire that fails.
The maximum stress is F A = Mg A = 90 × 106 N ⋅ m−2. Shear modulus
∆x F θ 2
6 π × (0.0005)
∴ M = 90 × 10 × = 90 × 10 ×
6 = 7.21kg . The shear strain is
The stress is F
= tan θ
L (note which area!) Shear modulus ⇒ Ms = shear stress
shear strain (F A ) = (F A ) .
(∆x L) tan θ
Units: Force/area ⇒ N ⋅ m−2. =A ...
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