Transcript Static Equilibrium. Supports, Loads, Driven Oscillations

Equilibrium
Forces and Torques
9/11/07
Topics to Cover
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Components of forces and trigonometry
Force examples
Center of Mass
Torques/Moments
Torque Examples
Equilibrium
Terminology
http://archone.tamu.edu/rburt/cosc321/cosc321whole.htm
Force Components
Fx  F cos
Fy  F sin 
F  Fx2  Fy2
tan 
Fy
Fx
soh sine (angle) = opposite/hypotenuse
cah cosine (angle) = adjacent/hypotenuse
toa tangent (angle) = opposite/adjacent
Trigonometry
• Fx is negative
– 90° to 270°
• Fy is negative
– 180° to 360°
• tan is positive
– quads I & III
• tan is negative
– quads II & IV
Components of Force
• F = 10 lbs
• Fx = F * cos (30°)
• Fx = 10 * 0.87 = 8.7 lbs
y
10 lbs
30°
• Fy = F * sin (30°)
• Fy = 10 * 0.5 = 5 lbs
8.7 lbs
5 lbs
x
Think – Pair - Share
• F = 20 lbs
• Find Fx
• Fx = 20 * cos (60°) = 10 lbs
y
20 lbs
17 lbs
60°
• Find Fy
• Fy = 20 * sin (60°) = 17 lbs
10 lbs
x
Components of Force
Similar Triangles
• Find Fx
• Hypotenuse
Find Fy
– Use pythagorean theorem
– √(32 +42) = 5
3
5

Fx 20
4
5

Fy 20
3 * 20  5 * Fx
4 * 20  5 * Fy
3 * 20 5 * Fx

5
5
Fx  12lbs
4 * 20 5 * Fy

5
5
Fy  16lbs
y
20 lbs
4
3
x
Think - Pair - Share
• Find Fx
• Hypotenuse
–
√(32
+72)
Find Fy
y
= 7.6
3 7.6

Fx 20
7 7.6

Fy 20
3 * 20  7.6 * Fx
7 * 20  7.6 * Fy
3 * 20 7.6 * Fx

7.6
7.6
Fx  8lbs
7 * 20 7.6 * Fy

7.6
7.6
Fy  18.4lbs
20 lbs
7
3
x
How can we tell if forces are
balanced?
• The x and y components cancel
Do these forces balance?
y
y
10 lbs
20 lbs
10 lbs
7
30°
x
x
15 lbs
•
No
3
No
Do these forces balance?
•
•
•
•
y
10 lbs
10 lbs
30°
X components cancel
y components:
10 * sin (30°) = 5
(* 2 for each force)
30°
x
• YES!
10 lbs
What is the balancing force?
• 10 lb force
– Fx = 10 * cos (30°) = -8.7 lbs
– Fy = 10 * sin (30°) = 5 lbs
y
20 lbs
7
10 lbs
• 20 lb force
– Fx = 3*20/7.6 = 8 lbs
– Fy = 7*20/7.6= 18.4 lbs
• Combine components
– X: -8.7 + 8 = -0.7 lbs
– Y: 5 + 18.4 = 23.4 lbs
30°
3
88°
23 lbs
• Negative of these components represent balancing force
– X = 0.7 lbs, y = -23.4 lbs
– Resultant = √(.72 + 23.42) = 23.4 lbs
– Angle: tan-1(-23.4/0.7) = 88° in fourth quadrant
x
Force Example
• Find the MAGNITUDE, DIRECTION & SENSE of the 3
Coplanar Forces
– Find the hypotenuse of the 2 triangles using pythagorean theory.
Force Example – con’t
10 # Force
Fy = Sin 30º * 10
Fy = 0.5 * 10
Fy = + 5 #
20 # Force
Fy = (4/5) * 20
Fy = 0.8 * 20
Fy = 16 #
Fx = Cos 60º * 10
Fx = 0.866 * 10
Fx = - 8.66 #
Fx = (3/5) * 20
Fx = 0.6 * 20
Fx = 12 #
30 # Force
Fy = (7/7.616) * 30
Fy = 0.919 * 30
Fy = - 27.57 #
Fx = (3/7.616) * 30
Fx = 0.394 * 30
Fx = 11.82 #
Force Example – con’t
Sum up all the components
Ry = 5 + 16 - 27.57
Rx = -8.66 + 12 + 11.82
Ry = -6.57 #
Rx = 15.16 #
The Resultant Force (R) is found using Pythagorean theory.
R  6.572 15.162  272.99  16.52#
The Direction of the Force (q ) is found using trigonometry
Force – Final Result
•
The Magnitude of the Resultant is 16.52 #
•
The Direction is 23.41o to the horizontal (X) axis
•
The Sense of the force is way from the origin
Lamp Post Forces Example
• Cable Tension = mg / sin 
• Pull of cable = push of strut
• Cable is in Tension, the strut in Compression!
cable

mg

strut
Lamp post
mg
sin  
cable tension
Extended Objects – Center of Mass
• Draw the forces (the free-body diagram) on a
stack of books.
– How much does a book weigh?
– What is diagram for the top book? For middle? For
bottom?
• Note that the bottom of the structure must support the most
weight!
• A book applies a force over some area
(distributed load). Why am I justified in drawing
only one arrow?
N
• Center of Mass!
mg
top
middle
bottom
Center of mass of triangular wedge
• Toward the thicker end - more mass on that side.
– exact value is a third of the way from thicker end
• For an ideal, rigid object that does not change shape or
break, we can say that gravity just acts at one point, the
center of mass (or center of gravity, or centroid).
Center of mass con’t
• Where can you hold a meter stick without
causing a rotation? What if there is a
weight on one end?
• How about a see-saw? How to balance?
Mechanical Equilibrium
• Mechanical equilibrium
– Total translational and rotational forces equal
zero
– No acceleration
– Static equilibrium means not moving
– Dynamic equilibrium means constant velocity,
i.e. not accelerating
Translational Equilibrium
• In order to be in translational equilibrium, the
total forces of the object must be zero.
• Set the horizontal and vertical components of
the forces to zero.
F
x
0
F
y
0
Rotational Equilibrium
Torque or Moment
• In order to be in rotational equilibrium, the
total torques acting on an object must be
zero.
  0
or
M  0
  F  d or
M  F d
Moments – rotation
• Defined by magnitude and direction
• Units N*m or ft*lb
• Direction – represented by arrows
– counterclockwise positive
– clockwise negative
• Value found from F and perpendicular
distance
– M = F*d
– d is also called “lever” or “moment” arm
Another Torque Example
• Same F
• Which torque is larger? Why?
Tools that use torque
• Torques allows us to move objects heavier than
we can actually lift
• What if we needed 1000 pounds of force to raise
a block?
– None of us can just bend down and lift with that force
– What tool would we use?
• Problem:
– Say we dig a little hole, and slide our 2x4 under the
block, and then place a small wedge 6” away from the
end (board is 6’, remember). How much can we exert
now?
Pole Vaulter
•Find the forces FL and FR.
FL
pivot
c.g.
2 ft.
FG= 5 lbs
FR
6 ft.
The pole vaulter is holding pole so that it does not
rotate and does not move. It is in equilibrium.
Pole vaulter – con’t
Rotation = 0
 net   L   G
 net  rL FL  rG FG
 net  0
Translation = 0
Fnet  FL  FR  FG
Fnet  0
Pole vaulter – con’t
 net   L   G
Fnet  FL  FR  FG
0  2 ft  FL  6 ft  5lb
0  15lbs  FR  5lbs
FL  15lbs
FR  10lbs
FL=15 lbs
pivot
c.g.
2 ft.
FG= 5 lbs
FR=10 lbs
6 ft.
Torques or Moments - Rotation
• Torques are also arrows, just like forces
– clockwise or counterclockwise
cable
strut
Lamp post
Activity 2 - Equilibrium
a) Class 1
b) Class 2
Terminology
• A metal column deforms when it supports
weight, but it is so small you can’t see it.
• This deformation is called COMPRESSION if
it pushes inwards and TENSION if it pulls out,
and SHEAR if it makes different parts of the
object go different ways.
Terminology - con’t
“Loads”
• “Static loads” are approximately constant
• Some of these are “dead loads” and don’t vary at all
– Weight of the structure
• “Live loads” vary, but only change very slowly
– snow
– number of people
– Furnishings
Terminology - con’t
Dynamical loads
• Dynamic loads change rapidly
– Wind and Earthquakes
– Can be very dangerous to buildings
• Demo: Impact load. Weigh an object on a scale, show
that dropping from equilibrium gives double the force of
the settled value.
• Oscillations
– Everything has a natural frequency it will prefer to
oscillate at. If driven near this frequency, the
response is large.