05B: STATICS

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05B: STATICS. San Juanico Bridge. Objectives: After completing this lecture, you should be able to:. State and describe with examples your understanding of the first and second conditions for equilibrium . - PowerPoint PPT Presentation

Transcript of 05B: STATICS

  • 05B: STATICS

  • Objectives: After completing this lecture, you should be able to:State and describe with examples your understanding of the first and second conditions for equilibrium.Write and apply the first and second conditions for equilibrium to the solution of physical problems similar to those in this lecture.

  • Translational EquilibriumThe linear speed is not changing with time. There is no resultant force and therefore zero acceleration. Translational equilibrium exists.

  • Rotational EquilibriumThe angular speed is not changing with time. There is no resultant torque and, therefore, zero change in rotational velocity. Rotational equilibrium exists.

  • EquilibriumAn object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.

  • Does Equilibrium Exist?A sky diver who reaches terminal speed?A fixed pulley rotating at constant speed? Yes Yes Is the system at left in equilibrium both translationally and rotationally?YES! Observation shows that no part of the system is changing its state of motion.

  • Statics or Total EquilibriumStatics is the physics that treats objects at rest or objects in constant motion.In this lecture, we will review the first condition for equilibrium (treated in Part 5A); then we will extend our treatment by working with the second condition for equilibrium. Both conditions must be satisfied for true equilibrium.

  • Translational Equilibrium OnlyIf all forces act at the same point, then there is no torque to consider and one need only apply the first condition for equilibrium:

  • Review: Free-body DiagramsRead problem; draw and label sketch.Construct force diagram for each object, vectors at origin of x,y axes.Dot in rectangles and label x and y compo-nents opposite and adjacent to angles.Label all components; choose positive direction.

  • Example 2. Find tension in ropes A and B.Recall: SF = 0SFx = Bx - Ax SFy = By + Ay - WBx - Ax = 0Bcos B Acos A = 0By + Ay W = 0Bsin B + Asin A W = 0

  • Example 2 (Cont.) Simplify by rotating axes:B = 410 NA = 287 NB cos B Acos A = 0B sin B + Asin A W = 0

  • Total EquilibriumIn general, there are six degrees of freedom (right, left, up, down, ccw, and cw):

  • General Procedure:Draw free-body diagram and label.Choose axis of rotation at point where least information is given.Extend line of action for forces, find moment arms, and sum torques about chosen axis:St = t1 + t2 + t3 + . . . = 0 Sum forces and set to zero: SFx= 0; SFy= 0 Solve for unknowns.

  • Center of GravityThe center of gravity of an object is the point at which its weight is concentrated. It is a point where the object does not turn or rotate.The single support force has line of action that passes through the c. g. in any orientation.

  • Examples of Center of GravityNote: C. of G. is not always inside material.

  • Finding the center of gravity01. Plumb-line Method02. Principle of MomentsSt(ccw) = St(cw)

  • Example 5: Find the center of gravity of the system shown below. Neglect the weight of the connecting rods.m1 = 4 kgm2 = 1 kgm3 = 6 kg3 cm2 cmSince the system is 2-D: Choose now an axis of rotation/reference axis:Trace in the Cartesian plane in a case that your axis of rotation is at the origin (0,0)(0, 2cm)(-3 cm, 2cm)CG

  • Example 6: Find the center of gravity of the apparatus shown below. Neglect the weight of the connecting rods.Choose axis at left, then sum torques:x = 2.00 mC.G.

  • Example 4: Find the tension in the rope and the force by the wall on the boom. The 10-m boom weighing 200 N. Rope is 2 m from right end.Since the boom is uniform, its weight is located at its center of gravity (geometric center)

  • Choose axis of rotation at wall (least information)St(ccw):rSt(cw):r1 = 8 mr2 = 5 mr3 = 10 mT = 2250 N

  • 300TyTxSF(up) = SF(down):SF(right) = SF(left):F = 1953 N, = 356.3o

  • Example 3: Find the forces exerted by supports A and B. Neglect the weight of the 10-m boom.40 N80 N2 m3 m7 mABDraw free-body diagramRotational Equilibrium:Choose axis at point of unknown force.At A for example.

  • Example 3: (Cont.)St = t1 + t2 + t3 + t4 = 0Rotational Equilibrium:orSt(ccw) = St(cw) With respect to Axis A:CCW Torques:Forces B and 40 N. CW Torques:80 N force. Force A is ignored:Neither ccw nor cw

  • Example 3 (Cont.)First: St(ccw)t1 = Br3t2 = w1r1Next: St(cw)t3 = w2r2

    St(ccw) = St(cw) B = 48 NSt = 0

  • Example 3 (Cont.)SF(up) = SF(down) Translational EquilibriumSFx= 0; SFy= 0Recall that B = 48 NA = 72 N

  • Example 3 (Cont.)Check answer by summing torques about right end to verify A = 72.0 NSt(ccw) = St(cw) (40 N)(12 m) + (80 N)(3 m) = A (10 m)480 Nm + 240 Nm = A (10 m)A = 72.0 N

  • Reminder on Signs:

  • SummaryAn object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.Conditions for Equilibrium:

  • Summary: Procedure

    *San Juanico Bridge (Marcos Highway)Connects Tacloban City on the Leyte side and Santa Rita town on the Samar sideRotational and translational equilibrium must be maintained**B = 409.5760.. NA = 286.7882.. N

    *x = 2.00 m

    *x = 2.00 m

    *T = 2250 N

    *Fx = 1948.5571, = 3.66966.. = 356.3F = 1953.0043 1953 N, 356.30

    **B = 48.0 N

    *A = 72.0 N

    *A = 72.0 N

    *