Lecture Notes Statics 3.4

Mechanics for Engineers: Statics, 13th SI EditionR. C. Hibbeler and Kai Beng Yap Pearson Education South Asia Pte Ltd 2013. All rights reserved.

THREE-DIMENSIONAL FORCE SYSTEMSIn-class Activities:Check HomeworkReading QuizApplications Equations of EquilibriumConcept QuestionsGroup Problem SolvingAttention Quiz


READING QUIZ1. Particle P is in equilibrium with five (5) forces acting on it in 3-D space. How many scalar equations of equilibrium can be written for point P? A) 2 B) 3 C) 4 D) 5E) 6 2. In 3-D, when a particle is in equilibrium, which of the following equations apply?A) ( Fx) i + ( Fy) j + ( Fz) k = 0 B) F = 0C) Fx = Fy = Fz = 0D) All of the above.E) None of the above.


APPLICATIONS


APPLICATIONS (continued)


THE EQUATIONS OF 3-D EQUILIBRIUMThis vector equation will be satisfied only when Fx = 0 Fy = 0 Fz = 0These equations are the three scalar equations of equilibrium. They are valid for any point in equilibrium and allow you to solve for up to three unknowns.


EXAMPLE I1) Draw a FBD of particle O.2) Write the unknown force as F5 = {Fx i + Fy j + Fz k} N3) Write F1, F2 , F3 , F4 , and F5 in Cartesian vector form.4) Apply the three equilibrium equations to solve for the three unknowns Fx, Fy, and Fz.


EXAMPLE I (continued)F4 = F4 (rB/ rB) = 200 N [(3i 4 j + 6 k)/(32 + 42 + 62)] = {76.8 i 102.4 j + 153.6 k} NF5 = { Fx i Fy j + Fz k} NSolution:


Equating the respective i, j, k components to zero, we have Fx = 76.8 600 + Fx = 0 ; solving gives Fx = 523.2 N Fy = 240 102.4 + Fy = 0 ; solving gives Fy = 137.6 N Fz = 180 900 + 153.6 + Fz = 0 ; solving gives Fz = 566.4 NThus, F5 = {523 i 138 j + 566 k} NUsing this force vector, you can determine the forces magnitude and coordinate direction angles as needed. EXAMPLE I (continued)


EXAMPLE II1) Draw a free-body diagram of Point A. Let the unknown force magnitudes be FB, FC, FD .2) Represent each force in its Cartesian vector form.3) Apply equilibrium equations to solve for the three unknowns.


FB = FB (sin 30 i + cos 30 j) N = {0.5 FB i + 0.866 FB j} NFC = FC i NFD = FD (rAD /rAD) = FD { (1 i 2 j + 2 k) / (12 + 22 + 22) } N = { 0.333 FD i 0.667 FD j + 0.667 FD k } NEXAMPLE II (continued)


Solving the three simultaneous equations yields FC = 646 N (since it is positive, it is as assumed, e.g., in tension) FD = 900 N FB = 693 NEXAMPLE II (continued)


CONCEPT QUIZ1. In 3-D, when you know the direction of a force but not its magnitude, how many unknowns corresponding to that force remain?A) One B) Two C) Three D) Four2. If a particle has 3-D forces acting on it and is in static equilibrium, the components of the resultant force ( Fx, Fy, and Fz ) ___ . A) have to sum to zero, e.g., -5 i + 3 j + 2 k B) have to equal zero, e.g., 0 i + 0 j + 0 k C) have to be positive, e.g., 5 i + 5 j + 5 k D) have to be negative, e.g., -5 i - 5 j - 5 k


1) Draw a free-body diagram of Point A. Let the unknown force magnitudes be FB, FC, F D.2) Represent each force in the Cartesian vector form.3) Apply equilibrium equations to solve for the three unknowns.GROUP PROBLEM SOLVING


W = load or weight of unit = 17500 k N FB = FB (rAB/rAB) = FB {(1.2 i 0.9 j 3 k) / (3.354)} N FC = FC (rAC/rAC) = FC { (0.9 j 3 k) / (3.132) } NFD = FD (rAD/rAD) = FD { ( 1.2 i + 0.3 j 3 k) / (3.245) } NGROUP PROBLEM SOLVING (continued)


GROUP PROBLEM SOLVING (continued)The particle A is in equilibrium, henceFB + FC + FD + W = 0Now equate the respective i, j, k components to zero (i.e., apply the three scalar equations of equilibrium). Fx = (1.2/ 3.354)FB (1.2/ 3.245)FD = 0 Fy = ( 0.9/ 3.354)FB + (0.9/ 3.132)FC + (0.3/ 3.245)FD = 0 Fz = ( 3/ 3.354)FB (3/ 3.132)FC (3/ 3.245)FD + 17500 = 0Solving the three simultaneous equations gives the tension forcesFB = 7337 NFC = 4568 NFD = 7098 N


ATTENTION QUIZ2. In 3-D, when you dont know the direction or the magnitude of a force, how many unknowns do you have corresponding to that force? A) One B) Two C) Three D) Four


Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Answers:1. B2. DStatics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Answers:1. A2. B

Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 3.4*Answer1. D2. C

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Lecture Notes Statics 3.4