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Statics. Equilibrium Moves at constant velocity V =C Moves at constant velocity V =C At rest V = 0 At rest V = 0 Constant Velocity.

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Presentation on theme: "Statics. Equilibrium Moves at constant velocity V =C Moves at constant velocity V =C At rest V = 0 At rest V = 0 Constant Velocity."— Presentation transcript:

1 Statics

2 Equilibrium Moves at constant velocity V =C Moves at constant velocity V =C At rest V = 0 At rest V = 0 Constant Velocity

3 Equilibrium condition  Satisfying Newtown’s 1 st Law of Motion: ∑ F = 0  This condition implies that ma = 0. here m could not equal zero and so a = 0 (i.e. not accelerating or decelerating)

4 Free Body Diagram (FBD) Definition:  Is a diagram for a particle or a body isolated from its surrounding.  It shows all the forces acting on the particle

5 Procedures to draw FBD  Drawing outlined shapes  Show all the forces  Identify each force

6 Example[1] F1 F2 Mg M

7 Example[2] A B C D T1 M1g B T2 C M2g T2 T3 M1 M2

8 Connection types: linear springs LoLo LoLo L F δ Applying Load  deflection ( δ) = L – L o  F = K δ where K is the spring stiffness  increasing K makes the spring stiffer.  Stiffer springs needs more force to deflect it

9 Forces in Linear Springs  As the load affects the spring, an internal resistance load creates.  The relation between the external load and the internal force of the spring is proportional and the proportional factor is the stiffness (K)  Linear springs could be: tension spring compression springs

10 Equivalent of Linear Springs (parallel) K1 K2 K3 Kn Force Keq = K1 + K2 + …. + Kn

11 Equivalent of Linear Springs (series) Force K1K2 Kn

12 Connection types: cables and pulleys  it has always tension force in the direction of the cable.  In most of engineering problems, cables are assumed massless and unable to stretch. T T

13 Example[3] T

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16 Equilibrium conditions: ∑F = 0 ∑F x i + ∑F y j = 0 ∑F x = 0 ∑F y = 0

17 Procedures for analysis FBD Coordinate system (x and y axes) Force Labeling (magnitude and direction) Assuming the sense of the unknown forces Equilibrium equations Positive and negative assigning Define the direction of the solution ∑ F x = 0 and ∑ F y = 0

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24 Equilibrium conditions: ∑F = 0 ∑F x i + ∑F y j + ∑F z k= 0 ∑F x = 0 ∑F y = 0 ∑F z = 0

25 Procedures for analysis FBD Coordinate system (x, y and z axes) Force Labeling (magnitude and direction) Assuming the sense of the unknown forces Equilibrium equations Positive and negative assigning Define the direction of the solution ∑ F x = 0, ∑ F y = 0 and ∑ F z = 0

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