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Moments and Summation of Forces and assignment.pdf

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Engineering Management Science
EMS 302
Ron Babin

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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