Lecture 05 - Lecture 05 * COPLANAR FORCE SYSTEMS *...

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Statics: Lecture Notes for Sections 3.3-3.4 1 Lecture 05 * COPLANAR FORCE SYSTEMS * THREE-DIMENSIONAL FORCE SYSTEMS Section 3 3 34 Section 3.3, 3.4 Ehab Zalok Objectives : Students will be able to solve 3-D particle equilibrium problems by a) Drawing a 3-D free body diagram, and, b) Applying the three scalar equations (based on one vector equation) of equilibrium. 2
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Statics: Lecture Notes for Sections 3.3-3.4 2 COPLANAR FORCE SYSTEMS (Section 3.3) This is an example of a 2-D or coplanar force system. If the whole assembly is in equilibrium, then particle A is also in equilibrium. To determine the tensions in the 3 cables for a given weight of the engine, we need to learn how to draw a free body diagram and apply equations of equilibrium. THE WHAT, WHY AND HOW OF A FREE BODY DIAGRAM (FBD) Free Body Diagrams are one of the most important things for you to know how to draw and use. What ? - It is a drawing that shows all external forces acting on the particle. 4 Why ? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).
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Statics: Lecture Notes for Sections 3.3-3.4 3 How ? 1. Imagine the particle to be isolated or cut free from its surroundings. 2. Show all the forces that act on the particle. Active forces: They want to move the particle. Reactive forces: They tend to resist the motion. 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables . 5 FBD at A Note : Engine mass = 250 Kg A EQUATIONS OF 2-D EQUILIBRIUM Since particle A is in equilibrium, the net force at A is zero. So T AB + T AC + T AD = 0 or T = 0 FBD at A A In general, for a particle in equilibrium , F = 0 or F x i + F y j = 0 = 0 i +0 j (A vector equation) 6 Or, written in a scalar form , F x = 0 and F y = 0 These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns.
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Statics: Lecture Notes for Sections 3.3-3.4 4 EXAMPLE Write the scalar E of E: +  F x = T B cos 30º – T D = 0 Note : Engine mass = 250 Kg FBD at A 7 +  F y = T B sin 30º – 2 . 452 kN = 0 Solving the second equation gives: T B = 4 . 90 kN From the first equation, we get: T D = 4 . 25 kN SPRINGS, CABLES, AND PULLEYS Cables and Ropes Can’t push on a rope. Neglect self-weight Cables, ropes: Always in tension 8 Force on a rope: Magnitude : As required for equilibrium Direction: Same as rope P of A: Point of contact Sense: Always AWAY from the body ( Pull )
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Statics: Lecture Notes for Sections 3.3-3.4 5 SPRINGS, CABLES, AND PULLEYS 9 Spring Force = spring constant *
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Lecture 05 - Lecture 05 * COPLANAR FORCE SYSTEMS *...

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