TR-4_Static_Equilibrium_2011C_student (1)

# TR-4_Static_Equilibrium_2011C_student (1) - 1 Static...

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Unformatted text preview: 1 Static Equilibrium. Forces & Moments. Equilibrium of a Particle. Equilibrium of a Rigid Body. TR-4 2 Learning Objectives ● When you complete this module you will be able to – Calculate the x-y components of a force given its magnitude and angle. – Find unknown forces by using the equations of force equilibrium. – Write the Procedure for calculating the Moment (or Torque) of a force around a point. ● This information will be useful in future courses: ENGR 221, CVEN 305, etc. 3 Standard Protocol for Finding Force Components 1. Mark all angles CCW from the positive x-axis (angle in standard position). 2. Then the force components are ALWAYS given by θ θ sin cos F F F F y x = = F θ F θ F θ θ F 4 While working with vectors, we have to keep in mind: x y F x > 0 F y > 0 F x > 0 F y < 0 F x < 0 F y > 0 F x < 0 F y < 0 F F F F II I III IV 5 Finding Force Components Exercise 1 . Individual ● Consider the four force vectors shown here and find their X and Y components. ● How did you find the signs of the components? 4 5 ° 5 lb 5 lb 6 0 ° 3 0 ° 6 0 ° 5 lb 5 lb A B C D 6 A. B. C. D. Exercise 1. Solution lb F negative is F quadrant lb F B lb F lb F A y x nd x y x 2 3 5 60 sin 5 , 2 2 5 60 cos 5 . 2 5 45 sin 5 2 5 45 cos 5 . = ° =- = °- = = ° = = ° = 45° 5 lb 5 lb 60° 3 0 ° 60° 5 lb 5 lb lb F negative is F quadrant lb F D negative are F and F quadrant lb F lb F C y y th x y x rd y x 2 3 5 60 sin 5 , 4 2 5 60 cos 5 . , 3 2 3 5 30 cos 5 2 5 30 sin 5 .- = °- = = ° = = °- =- = °- = 7 While working on Exercise 1, we had to keep in mind: ● If force F vector points toward positive direction of x-axis, then horizontal component of the force is positive: F x > 0 ● If force F vector points toward negative direction of x-axis, then horizontal component of the force is negative: F x < 0 ● If force F vector points toward positive direction of y-axis, then vertical component of the force is positive: F y > 0 ● If force F vector points toward negative direction of y-axis, then vertical component of the force is negative: F y < 0 x y x y x y x y 8 F1 F2 F3 Exercise 2. Individual Consider an eye-bolt acted upon by the forces shown. ● Using the Standard Protocol for Finding Force Components ○ Find Fx and Fy of the 20 kN force (F1). ○ Repeat for the 30 kN force (F2). ○ Repeat for the 42 kN force (F3). ● Find the horizontal component of the net force. ● Find the vertical component of the net force. 9 Exercise 2. Solution F1 F2 F3 Force: 20 @40 to positive x-axis : 1 20cos40 15.32 ; 1 20sin 40 12.86 ; x y kN F kN F kN = = = = Force: 30 @110 to x-axis : 2 30cos110 10.26 ; 2 30sin110 28.19 ; + = = - = = x y kN F kN F kN Repeat for the 42 kN force: 42 kN @ 160° to + x-axis . 10 F1 x = 15.32 kN, F1 y = 12.86 kN F2 x = -10.26 kN, F2 y = 28.19 kN F3 x = -39.47 kN, F3 y = 14.36 kN ● Horizontal component of the net force: Σ F x = F1 x + F2 x + F3 x = -34.41 kN ● Vertical component of the net force: Σ F y = F1 y + F2 y...
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## This note was uploaded on 11/08/2011 for the course ENGR 111 A taught by Professor Reddy during the Spring '10 term at Texas A&M.

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TR-4_Static_Equilibrium_2011C_student (1) - 1 Static...

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