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EMS 302
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Ron Babin
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Lecture

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Engineering Management Science

EMS 302

Ron Babin

Fall

Description

Loads and Reactions
Summation and Moments of Forces
1
Loads and Reactions
• When an object is suspended it exerts force
on all the elements that support it (load)
• Elements include:
– All components of the rigging system
– Structural members of the building
• Force: strength or energy trying to cause
change
2
Static Equilibrium
• Newton's third law of physics: For every action there
is an equal and opposite reaction
• Meaning: Rigging components and supporting
members resist the applied force of the load with a
reactive force exactly equal and opposite to the
applied force 100lbs
100 lbs
• Two distinct branches of mechanics, Statics and
Dynamics apply to rigging
3
1 Statics and Dynamics
• Statics: The study of • Dynamics deals with
forces and the effect of motion and the effect
forces acting on rigid of forces acting on rigid
bodies at rest bodies in motion
• No movement in object
= Static equilibrium
• There is a balance of all
of the forces acting on
the system
4
The Force Law of Equilibrium
• The algebraic sum of all the forces acting on
an object in static equilibrium is zero
• Therefore all the forces acting on the body are
equal and opposite
100lbs
100 lbs + = 0
100lbs
5
Free Body Diagram (FBD)
• Analyzing the forces acting on a body can be complicated
• all applied and reactive forces acting on itf its support members, showing
• FBD contain:
– L• Feet, inches, meters, etcbers and the distances between forces
– Magnitude of all of the known forces
Force acting on object – lbs, kg, etc
– Point of application of all the forces
• Where force is being applied (end, middle, 2’ from left etc)
– Direction of all the forces (at Point of Application)
• Up, down, etc
– Sense of all forces
• Positive or negative
6
2 FDB: Example
• A 16' long seesaw with two – 200 lbs people
at each end, supported directly in the middle
• Forces are indicated by arrows
• Reactive forces are indicated by the letter R
200 lbs 200 lbs
16’
8’ 8’
R
7
Summation of Forces
• Σ F=0 (sum of all the forces acting on an object equals zero)
• Sometimes necessary to determine all of the horizontal and vertical forces
acting on an object
– Σ H = 0 for the horizontal forces
– Σ V = 0 for the vertical forces
• Labeling the forces in the FBD,
– downward force of -200 lbs
– downward force of -200 lbs
– an unknown positive reaction
• All vertical applied forces - no horizontal forces in this example
• P indicates all applied point load forces
• R indicates all reactive forces
8
200 lbs 200 lbs
16’
8’ 8’
R
-200 lbs -200 lbs
16’
8’ 8’
+R
9
3 Solving for R
P l -200 lbs 16’ Pr =-200 lbs
8’ 8’
+R
ΣV=0
ΣV=P lP rR
Pl+P rR=0
R+(-200lbs)+(-200lbs)=0
R+(-400lbs)=0
R=400lbs
10
Moment of Force
• Moment of force is the force's tendency to produce rotation on the object
on which it acts
• Rotation is about some axis or point of rotation
• Point of rotation is called the centre of moments
• The distance between the applied force and axis is called the lever arm or
moment arm
• The length of the lever arm is measured from a line drawn through the axis
perpendicular to the line of action of the force
• Units are expressed in distance and force (foot-pounds etc.)
Applied force
Lever or moment art
Point of rotation or
Centre of moments 11
Moment of Force
• In previous example the beam is 16' long, the fulcrum, or axis, is in the
centre, 8' away from the applied force of 200 lbs
• To calculate the moment of force applied, multiply the length of the lever
arm (distance) by the applied force (point load)
• Result ??
200 lbs
8’
R
M= Distance X Point load
= D x P
= 8’ x 200lbs
= 1600 ft/lbs
12
4 Moment of Force
• Law of equilibrium applies to moments as well as to vertical and
horizontal forces
– ΣM=0
• Moment on the right end of the beam is 1600 ft-lbs in a clockwise
direction
• Moment on the left is also 1600 ft-lbs but in a counterclockwise
direction
• Moments must be equal and opposite in direction to maintain
static equilibrium
• Clockwise is positive and counterclockwise is negative
L.S. R.S.
-200 lbs +200 lbs
16’ M= D X P M= D X P
= 8’ x -200lbs = 8’ x (+200lbs)
8’ 8’ = -1600 ft/lbs = +1600 ft/lbs
R
ΣM=0
13
Moment of Force
• If different forces applied at each end of the beam
the fulcrum must be moved to accommodate the
different weights and maintain balance
• Example: Applied force on th

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