Chapter 9- Static Equilibrium
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The following changes will take place until the end of the semester
Office-hours:
- Monday, 12:00-13:15h
- Wednesday, 13:00-14:15h
Old assignments and midterm exams
(solutions have been posted on
the web)
can be picked up in my office
(LB-212)
Chapter 9
• The Conditions for Equilibrium
• Solving Statics Problems
Recalling Last Lectures
Angular Momentum and Its Conservation
We have defined several angular quantities in analogy to linear motion.
Type
Linear
Rotational
Relation
Displacement
x
θ
x = rθ
Velocity
v
ω
v = rω
Acceleration
a
α
a = rα
Dynamics
F (force)
τ (torque)
τ = Fd
Inertia
m (mass)
I (moment of inertia)
Kinetic Energy
Newton’s 2nd Law
Equations of motion
Angular Momentum and Its Conservation
1) Angular momentum
(8-26)
L is called angular momentum.
Angular Momentum and Its Conservation
2) Work in angular momentum
(8-27)
Eq. 8-27 represents the work done by the torque τ when rotating an object
through an angle ∆θ.
Power P (rate with it the work is done):
(8-28)
Angular Momentum and Its Conservation
3) Impulse in angular momentum
Impulse in rotational motion as
(8-29)
Angular Momentum and Its Conservation
4) Generalized for of Newton’s 2nd Law
General form of the rotational equivalent of Newton’s 2nd law:
(8-30)
Changes in the momentum of inertia ( L = Iω ) are taken into account in the
formalism of Newton’s 2nd law.
Angular Momentum and Its Conservation
5) Conservation of Angular Momentum
The total angular momentum is conserved if there is NO external net torque
acting on the system.
(8-31)
Angular Momentum and Its Conservation
Since NO external torque is applied on her, her
total angular momentum should be conserved:
Therefore, if her momentum of inertia decreases,
her angular velocity has to increase to maintain
the total angular momentum constant.
Today
Problem 8-62 (textbook): A 4.2-m-diameter merry-go-round is rotating freely with
an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 Kg.m2. Four
people standing on the ground, each of mass 65 kg, suddenly step onto the edge of
the merry-go-round. What is the angular velocity of the merry-go-round now?
What if the people were on it initially and then jumped off in a radial direction (relative
to the merry-go-round)?
.
Problem 8-62 (textbook):
The angular momentum of the merry-go-round and people combination will be
conserved because there are no external torques on the combination. This situation
is a totally inelastic collision, in which the final angular velocity is the same for both
the merry-go-round and the people.
Subscript 1 represents before the collision, and subscript 2 represents after the
collision. The people have no initial angular momentum.
L1 = L 2
ω 2 = ω1
→
I1
I2
I 1ω 1 = I 2 ω 2
= ω1
I m-g-r
I m-g-r + I people
→


I m-g-r
= ω1 
2 
I
+
M
R
4
person
 m-g-r



1760 kg m 2
= ( 0.80 rad s ) 
= 0.48 rad s
2 
2
 1760 kg m + 4 ( 65 kg )( 2.1 m ) 
If the people jump off the merry-go-round radially, then they exert no torque on the
merry-go-round, and thus cannot change the angular momentum of the merry-goround. The merry-go-round would continue to rotate at .0.80 rad/s.
The Conditions for Equilibrium
So far, we have worried about motion (velocity) and change of motion (net
acceleration).
However, there are many situations where we want to make sure objects remain
in their original configuration of motion: at rest or constant velocity.
In this lecture, we will deal with the situation where the original state of motion of
an object is such that it is at rest.
As we have seen, for an object to remain in its state of rest NO net force
should act on it in other words, it does not have any net acceleration.
If external or internal net forces acts on an object, it can deform or fracture it.
We want to find a state of equilibrium such that the object is not subjected to
fractures or deformations
The Conditions for Equilibrium
Static
Static is the area of physics concerned about the calculation of the forces acting
on structures that are in equilibrium.
Determining these forces allow engineers to build structures that can
significantly minimize effects due to deformation or fracture.
No matter what, there is always at least one force acting on a object: gravity.
If, for example, a book is at rest on a table,
the normal force has to balance the
gravitational force in order for the book to
be found in equilibrium.
The Conditions for Equilibrium
It is clear that if an object is to stay in its state of equilibrium:
NO net forces NOR net torque should act on the object
In other words NO changes in its translational or
rotational motion must occur.
The chandelier should not move in the vertical,
horizontal or circular directions for equilibrium
to be met.
The Conditions for Equilibrium
First condition of equilibrium
No change in translational motion
The first condition for equilibrium is that the forces
along each coordinate axis add to zero.
(9-1)
In the figure, the equilibrium condition implies that:
120o
The Conditions for Equilibrium
Second condition of equilibrium
No change in rotational motion
The second condition of equilibrium is that there be
no torque around any axis; the choice of axis is
arbitrary and we will assume that its direction is
perpendicular to the xy plane of rotation.
(9-2)
l
In the figure, the sign supported by a beam of
length l mounted by a hinge on a wall is in
equilibrium. Chosen the hinge as the center of
rotation, and counterclockwise as positive:
l/2
The Conditions for Equilibrium
Recalling an important note:
We consider that the mass of an object with
uniform mass distribution (or uniform object for
short) is such that it can be assumed to be
located at the centre of the object.
This point is know as either centre of mass or
centre of gravity of the object.
In the figure, assuming an uniform beam, its
centre of mass will be located at the middle of the
beam (and this is why the force of gravity is
shown to be acting on this point).
l
l/2
For a uniform sphere, its centre of mass is located in its center.
Solving Statics Problems
1. Choose one object at a time, and make a free-body diagram showing all the
forces on it and where they act.
2. Choose a coordinate system and resolve forces into components.
3. Write equilibrium equations for the forces.
4. Choose any axis perpendicular to the plane of the forces and write the torque
equilibrium equations. A clever choice here can enormously simplify the
problem.
5. Make sure you correctly identify the components of the forces that can
contribute to rotational motion. Forces that have zero lever arm do not
contribute to rotational motion.
6. Solve.
Solving Statics Problems
If you do not know the direction of a force, just assume a generic one.
For example, assume that FB points up and FA is unknown in the figure below..
You initially assume that FA points up and then try to find the conditions of
equilibrium for this problem. You also assume that the axis of rotation passes
through the point of application of FA and that counterclockwise is positive. Also
assume that the point CG (center of gravity) is 25 m from this axis
y
x
FA in your solution comes negative. It just means that it’s in the opposite direction
from the one you had originally chosen. This is trivial to fix, so don’t worry about
getting all the directions of the forces right before you start solving.
Solving Statics Problems
If there is a cable or cord in the problem, it can support forces only along its
length. Forces perpendicular to that would cause it to bend.
Problem 9-1 (textbook): Three forces are applied to a tree sapling, as shown in
Fig. 9–41, to stabilize it. If
and
, find
in magnitude and
direction.
Problem 9-1 :
If the tree is not accelerating, then the net force in all
directions is 0.
∑F
x
= FA + FB cos110 + FC x = 0 →
FC x = − FA − FB cos110 = −310 N − ( 425 N ) cos110 = −164.6 N
∑F
y
= FB sin + FC y = 0 →
FC y = − FB sin 110 = − ( 425 N ) sin 110 = −399.4 N
FC =
FC2 x + FC2 y =
θ = tan
−1
And so
FC y
FC x
= tan −1
( −164.6 N ) + ( −399.4 N ) = 432.0 N ≈ 4.3 × 10 2 N
2
−399.4 N
−164.6 N
2
= 67.6 o , φ = 180 o − 67.6 o = 112.4 o ≈ 112 o
is 430 N, at an angle of 112o clockwise from
.
Problem 9-18 (textbook): Calculate
(a) the tension FT in the wire that supports the 27-kg beam shown in Fig. 9–52, and
r
(b) the force FW exerted by the wall on the beam (give magnitude and direction).
Problem 9-18 :
(a)
The beam is in equilibrium, and so the net force and net torque
on the beam must be zero. From the free-body diagram for the
beam, calculate the net torquer (counterclockwise positive) about
the wall support point to find FT , and calculate the netr force in
both the x and y directions to find the components of FW .
∑ τ = F L sin 40
T
mg
FT =
2 sin 40 o
=
o
− mg L 2 = 0 →
( 27 kg ) ( 9.80 m
s2
2 sin 40 o
) = 205.8 N ≈
L
2.1 × 10 2 N
x
= FW x − FT cos 40o = 0 → FW x = FT cos 40o = ( 205.8 N ) cos 40o = 157.7 N
y
= FW y + FT sin 40o − mg = 0 →
∑F
∑F
(
)
FW y = mg − FT sin 40 o = ( 27 kg ) 9.80 m s 2 − ( 205.8 N ) sin 40 o = 132.3 N
Problem 9-18 :
(b)
x
= FW x − FT cos 40o = 0 → FW x = FT cos 40o = ( 205.8 N ) cos 40o = 157.7 N
y
= FW y + FT sin 40o − mg = 0 →
∑F
∑F
(
)
FW y = mg − FT sin 40 o = ( 27 kg ) 9.80 m s 2 − ( 205.8 N ) sin 40 o = 132.3 N
FW =
FW2 x + FW2 x =
θ = tan
−1
FW y
FW x
= tan
−1
(157.7 N )2 + (132.3 N )2
132.3
157.7
= 40 o
= 205.8 N ≈ 2.1 × 10 2 N
Problem 9-27 (textbook): Consider a ladder with a painter climbing up it (Fig. 9–
61). If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the
ladder begins to slip at its base when her feet are 70% of the way up the length of the
ladder, what is the coefficient of static friction between the ladder and the floor?
Assume the wall is frictionless.
Problem 9-27:
The ladder is in equilibrium, so the net torque and net force must be zero.
By stating that the ladder is on the verge of slipping, the static frictional force at the
ground, FC x , is at its maximum value and so:
FC x = µ s FC y
Since the person is standing 70% of the way up
the ladder, the height of the ladder is:
Ly = d y 0.7 = 2.8 m 0.7 = 4.0 m
The width of the ladder is:
Lx = d x 0.7 = 2.1 m 0.7 = 3.0 m
Problem 9-27
Torques are taken about the point of contact of the ladder
with the ground, and counterclockwise torques are taken
as positive. The three conditions of equilibrium are as follows:
∑F
x
= FC x − F W = 0
∑F
y
= FG y − M g − mg = 0 →
→
FC x = F W
(
)
FG y = ( M + m ) g = ( 67.0 kg ) 9.80 m s 2 = 656.6 N
∑τ
= FW L y − mg ( 12 L x ) − M gd x = 0
Solving the torque equation gives:
FW =
1
2
mLx + Md x
Ly
g=
1
2
(12.0 kg )( 3.0 m ) + ( 55.0 kg )( 2.1 m )
4.0 m
The coefficient of friction then is found to be
µs =
FG x
FG y
=
( 9.80 m s ) = 327.1 N
32 7 .1 N
65 6 .6 N
2
= 0 .5 0