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Lecture #5

Lecture #5. Translational Equilibrium (2D Coplaner Force Systems, Reference Chapter 3, section 3). Newton’s First Law of Motion. A body at rest will stay at rest and a body in motion will stay in motion unless acted upon by an unbalanced force . Therefore, sum of all forces must be zero:.

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Lecture #5

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  1. Lecture #5 Translational Equilibrium (2D Coplaner Force Systems, Reference Chapter 3, section 3)

  2. Newton’s First Law of Motion • A body at rest will stay at rest and a body in motion will stay in motion unless acted upon by an unbalanced force. • Therefore, sum of all forces must be zero:  F = 0

  3. Newton’s Second Law of Motion • The acceleration a body undergoes when experiencing an unbalanced force is proportional to the magnitude of the unbalanced force, in the direction of the unbalanced force, and inversely proportional to the mass of the object – Basis of Dynamics  F = ma

  4. Motion Definitions • Rectilinear Motion - Motion in a straight line. • Translational Motion - Motion where the body does not rotate • Again, this is studied in dynamics!

  5. 2D Particle Static Equilibrium (i.e. Section 2.3, Coplanar Force Systems) • When a body is in this state there are no unbalanced forces acting on it. • The Resultant Force has a magnitude of zero. In general, for a particle in equilibrium,  F = 0or  Fxi +  Fy j = 0 = 0 i +0 j (a vector equation) Or, written in a scalar form, • Fx= 0and  Fy= 0 These are two scalar equations of equilibrium (E-of-E). They can be used to solve for up to two unknowns.

  6. 3D Particle Static Equilibrium • Because the resultant force is balanced, the following situation occurs • ΣFx = 0 • ΣFy = 0 • ΣFz = 0

  7. Summary: 2D Static Equilibrium: • Using Components: • Graphically:

  8. Static Equilibrium Problems • In 2D, have 2 equations, so can solve for 2 unknowns • Find magnitudes of two forces with known directions • Find magnitude and direction of one force, knowing magnitude and direction of other force(s) • In 3D have 3 equations, so can solve for 3 unknowns Do simple 2D

  9. 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. 2D COPLANAR FORCE SYSTEMS (Section 3.3) To determine the tensions in the cables for a given weight of the cylinder, you need to learn how to draw a free body diagram and apply equations of equilibrium.

  10. 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. Why ? - It is key to being able to write the equations of equilibrium—which are used to solve for the unknowns (usually forces or angles).

  11. 1. Imagine the particle to be isolated or cut free from its surroundings. 2. Show all the forces that act on the particle. 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables . y FBD at A FB 30˚ A x FD FC = 392.4 N (What is this?) Note : Cylinder mass = 40 Kg How ? Active forces: They want to move the particle. Reactive forces: They tend to resist the motion. A

  12. y FBD at A FB 30˚ A A x FD FC = 392.4 N EQUATIONS OF 2-D EQUILIBRIUM Since particle A is in equilibrium, the net force at A is zero. So FB + FC +FD = 0 or  F = 0 A FBD at A In general, for a particle in equilibrium,  F = 0or  Fxi +  Fy j = 0 = 0 i +0 j (a vector equation) Or, written in a scalar form, • Fx= 0and  Fy= 0 These are two scalar equations of equilibrium (E-of-E). They can be used to solve for up to two unknowns.

  13. y FBD at A FB 30˚ A A x FD FC = 392.4 N EXAMPLE Note : Cylinder mass = 40 Kg Write the scalar E-of-E: +   Fx = FB cos 30º – FD = 0 +   Fy = FB sin 30º – 392.4 N = 0 Solving the second equation gives: FB = 785 N → From the first equation, we get: FD = 680 N ←

  14. SPRINGS, CABLES, AND PULLEYS T1 T1 T2 Spring Force = spring constant * deformation, or F = k * s With a frictionless pulley, T1 = T2= T See HO

  15. Given: Cylinder E weighs 30 lb and the geometry is as shown. Find: Forces in the cables and weight of cylinder F. Plan: EXAMPLE 1. Draw a FBD for Point C. 2. Apply E-of-E at Point C to solve for the unknowns (FCB & FCD). 3. Knowing FCB , repeat this process at point B.

  16. EXAMPLE(continued) y FCD A FBD at C should look like the one at the left. Note the assumed directions for the two cable tensions. x 30 15 FBC 30 lb The scalar E-of-E are: +   Fx = FBCcos 15º – FCD cos 30º = 0 +   Fy = FCDsin 30º – FBC sin 15º – 30 = 0 Solving these two simultaneous equations for the two unknowns FBC and FCD yields: FBC = 100.4 lb FCD = 112.0 lb

  17. EXAMPLE(continued) y FBA FBC =100.4 lb Now move on to ring B. A FBD for B should look like the one to the left. 45 15 x WF The scalar E-of-E are:    Fx = FBA cos 45 – 100.4 cos 15 = 0    Fy = FBA sin 45 + 100.4 sin 15 – WF = 0 Solving the first equation and then the second yields FBA = 137 lb and WF = 123 lb

  18. Given: The box weighs 550 lb and geometry is as shown. Find: The forces in the ropes AB and AC. GROUP PROBLEM SOLVING Plan: 1. Draw a FBD for point A. 2. Apply the E-of-E to solve for the forces in ropes AB and AC.

  19. y FBD at point A FC FB 5 3 30˚ 4 A x FD = 550 lb GROUP PROBLEM SOLVING (continued) Applying the scalar E-of-E at A, we get; +   F x = FB cos 30° – FC (4/5) = 0 +   F y = FB sin 30° + FC (3/5) - 550 lb = 0 Solving the above equations, we get; FB = 478 lb and FC = 518 lb

  20. 1. Select the correct FBD of particle A. A 30 40 100 lb F1 F2 A A) B) 30 40° 100 lb A F2 F F1 D) C) 30° 40° 30° A A 100 lb 100 lb ATTENTION QUIZ

  21. 2. Using this FBD of Point C, the sum of forces in the x-direction ( FX)is ___ . Use a sign convention of +  . A) F2 sin 50° – 20 = 0 B) F2 cos 50° – 20 = 0 C) F2 sin 50° – F1 = 0 D) F2 cos 50° + 20 = 0 F2 20 lb 50° C F1 ATTENTION QUIZ

  22. READING QUIZ 1) When a particle is in equilibrium, the sum of forces acting on it equals ___ . (Choose the most appropriate answer) A) A constant B) A positive number C) Zero D) A negative number E) An integer 2) For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as _____ . A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin  T1 T2

  23. Determine the forces in cables AC and AB needed to hold the 20-kg ball D in equilibrium. Take F = 300 N and d = 1 m Practice Problem 1 Reference 3-18/19.

  24. Determine the tensions developed in wires CD, CB and BA and the angle q required for equilibrium of the 30 lb cylinder E and 60 lb cylinder F. Practice Problem 2 Reference 3-26.

  25. Determine the stretch in springs AC and AB for equilibrium of the 2-kg block. The springs are shown in the equilibrium position. Practice Problem 3 Reference 3-14.

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