Transcript 6.1 Equilibrium

Equilibrium
The First Condition of Equilibrium
In a situation involving equilibrium, there is no acceleration
(no change in velocity). Thus the net force (sum of all the
forces) is zero.
First Condition of Equilibrium : The Net Force is Zero
In equilibrium, the sum of all force vertical components is
zero and the sum of all force horizontal components is
zero.
The Balancing Force
The force that balances given forces is the force opposite
and equal to the sum of the given forces. This balancing
force produces a sum that is a point, zero force.
In the top diagram, the
red force is the sum of
the black and blue forces.
In the bottom diagram, the
red force is the force
balancing the black and
blue forces.
An Equilibrium Problem
Given the weight of w is 100.0 N, find the tension forces in the
ropes A and B.
An Equilibrium Problem : Method 1 (Easiest)
Given the weight of w is 100.0 N, find the tension forces in the
ropes A and B.
A
Solution using the Sine Law:
A
= 100.0 N
sin 45
sin 75
and B
= 100.0 N
sin 60
sin 75
A = 100.0Nsin45 = 73.20508…= 73.20 N
sin 75
B = 100.0Nsin60 = 89.6575… = 89.66 N
sin 75
100.0 N
B
Free Body Diagram
An Equilibrium Problem : Method 2
Given the weight of w is 100.0 N, find the tension forces in the
ropes A and B.
Bx – Ax = 0
Bcos30 – Acos30 = 0
.707107B - .866025A = 0
.707107B = .866025A
B = .866025A = 1.224745A
.707107
Ay[up] + By[up] – 100.0[up] = 0
Asin30 + Bsin30 = 100.0
.5A + .707107B = 100.0
{2 equations with 2 unknowns}
.5A + .707107(1.224745A) = 100.0 {substitution from above}
1.366025A = 100.0
A = 73.20 N
B = 1.224745(73.205) = 89.66 N
A Second Equilibrium Problem
An object is suspended as shown. Given T1 is 50.0 N, find
the object’s weight and T2.
T1
T1y
T2
T1x
W
Free Body Diagram
Since the weight is to be determined, T2 and T1x can be
ignored.
T1[up] + W (weight down) = 0
T1[up] = - W[down] = W[up]
50.0sin30 = W {cancelling the [up] directions}
W = 25.0 N [down]
T2 = 50.0cos30 = 43.3 N [left]
The Centre of Gravity
The point at which an object would balance on a pivot point is
referred to as the centre of gravity. For a uniform object,
the centre of gravity is at its mid point. For equilibrium
problems, all the mass and all the weight of the object can
be considered to be concentrated at the centre of gravity.
The Centre of Gravity for Irregular-Shaped Objects
The centre of gravity for irregularly-shaped objects is
determined by suspending an object from 2 different points
with plumb lines. Where the plumb lines intersect is the
centre of gravity. Many times the centre of gravity is
physically outside of the irregularly-shaped object. Objects
rotate about their centre of gravity.
Centre of Gravity and Balance
Objects balance at their centres of gravity. Objects
suspended on a pivot will move so that their centre of
gravity lines up beneath the pivot.
Torque
Torque measures the turning ability of an object about a pivot
point. Torque is measured by the product of the rotating
force and its perpendicular distance from the pivot.
Torque (
)=fxd
. The unit of torque is a Nm
is the Greek
letter tau
(small letter)
A General Formula for Torque
For forces not perpendicular to the torque arm (left below),
= rFsinθ = Frsinθ = FLcosa (right below).
θ
r
F
θ
Positive and Negative Torques
A torque causing a counter-clockwise rotation is given a
positive value while a torque causing a clockwise rotation
is given a negative value.
Force causing +
torque
Force causing torque
The Second Condition of Equilibrium : CT and CCT = 0
In a given plane for a pivoting object, there can be either a
clockwise torque (positive: +) or a counter-clockwise
torque (negative: -). In an equilibrium situation, the
clockwise torque equals the counter-clockwise torque.
CT = CCT.
30 N Force
2 m from
fulcrum
causing +
torque
60 Nm of +
torque
20 N Force
3 m from
fulcrum
causing torque
60 Nm of torque
The Centre of Gravity and Torque
In problems with a uniform beam, the centre of gravity is in
the centre of the object and causes no rotation (torque) if
the pivot is also around the centre. If the centre of gravity
is not at the pivot point, it causes a torque like any force
Centre of
Gravity at
Fulcrum
causes no
torque
Centre of Gravity
treated as a force
causing torque
Centre of Gravity to left of
fulcrum (non-uniform object)
causes + torque
Arbitrary Position of the Point of Rotation
For many torque problems, the point of rotation is placed at a
hinge or fulcrum. But in some problems the point of
rotation is arbitrarily assigned. In the example below,
either A or B could be assigned as fulcrums or pivots.
A
B
Levers
A lever is a bar that pivots about a point called the fulcrum.
In a lever clockwise torque equals counterclockwise torque
and the sum of the forces down equals the sum of the
forces up (the net forces add to zero).
Archimedes and the Lever
The Greek mathematician, Archimedes, discovered the law of
the lever. Many of the machines he designed used levers.
To make the point about the power of a lever for lifting he
said, “Give me a place to stand on, and I will move the
earth.”
Classes of Levers: Determined by Fulcrum Position
First class levers have the fulcrum between the effort and
load forces.
Second class levers have the fulcrum to one side of both the
the effort and load forces with the effort force farther from
the fulcrum than the load force.
Third class levers have the fulcrum to one side of both the
the effort and load forces with the load force farther from
the fulcrum than the effort force.
Advantages of Levers
First class levers can produce force or motion advantages.
Second class levers produce force advantages.
Third class levers produce motion advantages (force
disadvantage).
Example 1
A torque of 24.0 Nm is needed to tighten a nut. If a person
applies a force of 100 N at a right angle to the wrench, what
is the minimum length of wrench required?
From τ = rFsinθ, r = τ/Fsinθ so r = 24.0 Nm/100 N = 0.240 m =
24.0 cm
Example 2
A 25.0 N uniform beam is attached to a wall by means of a
hinge. Attached to the other end of this beam is a 100 N
weight. A rope also helps to support the beam as shown.
What is the tension in the rope?
T=?
L
30.0 o
W = 100 N
Example 2 Continued
The tension in the rope, T,
H
Is obtained by considering torque
Using the free body diagram (right),
The sum of all torques is zero so
LTsin30 - .5L25sin90 – L100sin90=0
Tsin30 - 12.5sin90 – 100sin90 = 0/L = 0
.5T – 12.5 - 100 = 0
.5T = 112.5
T = 225 N
T
V
Wb
100 N
Example 2 Continued
T
The forces of the wall on the beam
H
can be calculated from the relation
that the sum of the forces up and
100 N
down equals zero. The tension force,T, V
W
Can be resolved into its horizontal and vertical components.
Since T has a horizontal component
30
right of 195 N, the wall must exert
225sin30 =
112.5 N
a force of 195 N left.
225 N
225cos30 =
The downward forces add to 125N
194.86 N
so the upward forces must add to 125 N.
Given that T has an upward component
of 112.5 N, the wall must also exert an upward force of 12.5 N
to balance the downward force of 125N.
112.5[up] + x[wall up] = 100N[down] + 25N[down]
b
o
Another Equilibrium Problem
A uniform beam 5.0 m long has a weight of 200 N on it and is
suspended by three ropes, as shown. If an 800 N object is
placed as shown in the diagram, what is the tension in
each of the ropes?
T2
T1
1.0 m
30o
5.0 m
Equilibrium Problem Continued
The sum of torques is zero so
800(1) + 200(2.5) – 5Tsin30 = 0
1.0 m
T = 1300/5sin30 = 520 N
Resolving T into its vertical
and horizontal components,
Arbitrary
vert. = 260 N, horiz. = 450 N. Rotation
Point
The wall’s force T3 to the left
is 450 N.
2.5 m
T2
T1
T3
520 N
520cos30 =
450 N
30o
800 N
520sin30 =
260 N
200 N
Beam
Weight
Equilibrium Problem Continued
Since the sum of all forces
up and down is zero,
800 + 200 down = T2 + 260 up.
1000 – 260 = T2 up
740 N = T2
2.5 m
1.0 m
T2
T1
30o
Arbitrary
Rotation
Point
800 N
520 N
520cos30 =
450 N
200 N
Beam
Weight
520sin30 =
260 N
Centre of Gravity Problem
Two steel balls are joined by a “massless” bar. Find the
centre of gravity from the 1.2 kg mass.
Consider the centre of gravity
1.2 kg
2.2 kg
as a pivot and let x be the
distance from the 1.2 kg mass
to this pivot.
1.8 m
Then 1.8 – x is the distance
from the 2.2 kg mass to the
pivot. Since the torques are equal:
1.2(x) = 2.2(1.8 – x)
1.2x = 3.96 – 2.2x
3.4x = 3.96
X = 3.96/3.4 = 1.165 = 1.2 m from the 1.2 kg mass
Centre of Gravity Problem
Find the centre of gravity for these three ball joined by
“massless” bars. (Measure from the centre of the 2.8 kg
mass.)
4.1 m
Consider the centre of mass
4.1 - x
2.2 kg
between the 2.8 kg and other
masses. Then:
2.9 m
2.8x = 4.8(4.1 – x)
2.8 kg
x
y
2.8x = 19.68 – 4.8x
2.6 kg
2.9 - y
7.6 x = 19.68 and x = 2.6
Consider the centre of mass between the 2.6 kg and other
masses. Then:
5y = 2.6(2.9 – y)
5y = 7.54 -2.6y
7.6y = 7.54 and y = 1.0 Treating the centre of the 2.8 kg mass
as the origin, the centre of mass is located at (2.6, -1.0)
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