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Statics Introduction to statics
Static Equilibrium Equations
Examples:
Suspended beam
Hanging lamp
Ladder
Australian Institute of Sport
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Statics:
Statics is the study of systems that dont move.
Ladders, Stability of solid objects
Balanced objects
Buildings, Suspension bridges
Clifton Suspension BridgeChinese Golden Dragon Acrobats
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Statics:
Example: Evaluate the forces acting on a car parked on a hill.
xy
N
mg
f
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Car on Hill:
Use Newtons 2nd Law: FNET
= MACM
= 0
xy
N
mg
f
F 0
x: f - mgsin= 0
f = mg sin
y: N - mg cos = 0
N = mg cos
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Suspended beam:
Now consider a beam of mass Msuspended by two strings as
shown. We want to find the tension in each string:
L/2
L/4
Mx cm
T1 T2
Mg
F
0 First use
T1 + T2= Mg
This is no longer enough to
solve the problem!1 equation, 2 unknowns.
We need more information!!
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Suspended beam
We dohave more information:
We know the beam is not rotating!
NET=I= 0
The sum of all torques is zero.
This is true about any axiswe choose.
0
L/2
L/4
Mx cm
T1 T2
Mg
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Using Torque...
Choose the rotation axis to be along the zdirection (out of the
page) through the CM:
2 2 4 TL
The torque due to the stringon the right about this axis is:
1 12
TL
The torque due to the string onthe left about this axis is:
Gravity exerts no torque about the CM
L/2
L/4
Mx cm
T1 T2
Mg
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Using Torque...
Now
TL
TL
2 14 2
0
T T2 12
We already found that
T1 + T2= MgT Mg1
1
3
T Mg22
3
L/2L/4
Mx cm
T1 T2
Mg
0
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Approach to Statics:
0F 0
In general, we can use the two equations
to solve any statics problem.
When choosing axes about which to calculate torque, we can
sometimes be clever and make the problem easier....
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Juan and Bettina are carrying 80 kg block on a
massless 4 m board. Find the force in newtonsexerted by each to carry the block.
1. Juan 400 N, Bettina 400 N
2. Juan 350 N, Bettina 450 N
3. Juan 300 N, Bettina 500 N
4. Juan 250 N, Bettina 550 N
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Juan and Bettina are carrying 80 kg block on amassless 4 m board. Find the force in newtons
exerted by each to carry the block.
Force: Torque:
mg
FJ FB
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Center of Mass & Statics
The center of mass is at the point where the system
balances!
Sum of all gravitational torques about an axis through the
center of mass is 0!
02211 dmdm
m1m2
+
d1 d2
CM
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02211 gdmgdm
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Mobile
A mobile hangs as shown below.
The rods are massless.
The mass of the ball at the bottom right is 1kg.
Find the unknown masses.
1 kg
1 m 3 m
1 m 2 m
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Solution
First figure out M1.
1 kg
1 m 3 m
1 m 2 m
M1
0)1()1)(3( 1 Mmkgm
kg3M1
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Solution
Now M2.
0)2()4)(1(
4
2
Mmkgm
kgT
M kg2 2
M2
1 kg
1 m 3 m
1 m 2 m
3 kg
T
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Solution
So finally
1 kg
1 m 3 m
1 m 2 m
3 kg
2 kg
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Statics
A 1 kgball is hung at the end of a rod 1 m long. The system balances
at a point on the rod 0.25 m from the end holding the mass.What is the mass of the rod?
1. 0.5 kg 2. 1 kg 3. 2 kg
1 kg
1 m
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Solution
The total torque about the pivot must be zero.
1 kg
The center of mass of the rod is at its center, 0.25 m to theright of the pivot.
X
CM of rod
Since this must balance the ball, which is the same distanceto the left of the pivot, the masses must be the same!
same distancemROD = 1 kg
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Example: Hanging Lamp
A lamp of mass Mhangs from the end of plank of mass m and
length L. One end of the plank is held to a wall by a hinge, and
the other end is supported by a massless string that makes an
angle with the plank. (The hinge supplies a force to hold the
end of the plank in place.)
What is the tension in the string?
What are the forces supplied by the
hinge on the plank?
hinge
M
m
L
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M
m
L/2
Fx
Fy
T
L/2
Mg
mg
y
x
x: T cos + Fx = 0
y: T sin + Fy - Mg - mg= 0
0 Then in the zdirection. choose the rotation axis tobe through the hinge so hinge forces
Fx and Fy will not be included
0LTsin-mg2
LLMg
F
0First
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Hanging Lamp...
So we have three equations and three unknowns:
Fx = -T cos
Fy = Mg + mg - T sin
M
m
L/2
Fx
Fy
T
L/2
Mg
mg
mg
2
LLMgLTsin
which we can solve to find:
sin
g2mM
T+
=
tan
g2mMFx
F mgy 1
2
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What are the two forces F1 and F2 exerted bya hand on a 5m long rod of mass m? Compare
them to a total force exerted by the hand.
5 m
0.1 m
m
F1
F2 mg
2.4m
Torque:
Force:
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Example: Ladder against smooth wall
A ladder (length L, mass m) leans against a smooth wall (no
friction between wall and ladder).
A static frictional force Fbetween the ladder and the floorkeeps it from slipping (S = 0.70)
The angle between the ladder and the floor is .
What angle is needed to prevent the ladder
from slipping?
L m
F
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Example: Ladder against smooth wall...
Consider all of the forces.
Gravity
Friction
Normal forces Nfand Nw by the floor and wallrespectively on the ladder.
UseF = 0
x: -Nw + F = 0
y: Nf - mg = 0
F = Nw
Nf = mg
m
F
mg
Nw
Nf
y
x
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Example: Ladder against smooth wall
Now using base of ladder as axis.
0
L/2
m
F
mg
Nw
Nf
)sin()90sin(2
0 WLNmgL
W
W
N
mg
Nmg
2
1)tan(
)sin()cos(2
10
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Example: Ladder against smooth wall
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L/2
m
F
mg
Nw
Nf
WNmg
21)tan(
mgNF
FN
SfS
W
36
71.7.02
1
2
1
2
1)tan(
SSmg
mg
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Tipping a Box
A box is placed on a ramp in the configurations shown below. Friction
prevents it from sliding. The center of mass of the box is indicated bya blue dot in each case.
In which cases does the box tip over?
1. all 2. 2 & 3 3. 3 only 4. none
1 2 3
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Box
We have seen that the torque due to gravity acts as though all themass of an object is concentrated at the center of mass.
Consider the bottom right corner of the box to be a pivot point.
If the box can rotate in such a way that the center of mass is
lowered, it will!
1 2 3
updown
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Box
1 23
We have seen that the torque due to gravity acts as though all themass of an object is concentrated at the center of mass.
Consider the bottom right corner of the box to be a pivot point.
If the box can rotate in such a way that the center of mass is
lowered, it will!