Chapter 14 Statics

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  • SanJuanico Bridge (Marcos Highway)Connects Tacloban City on the Leyte side and Santa Rita town on the Samar sideRotational and translational equilibrium must be maintained
  • x = 2.00 m
  • x = 2.00 m
  • T = 2250 N
  • Fx = 1948.5571…,φ = 3.66966.. Θ = 356.3F = 1953.0043 … 1953 N, 356.30
  • Chapter 14 Statics

    1. 1. 05B: Equilibrium<br />
    2. 2. San Juanico Bridge<br />
    3. 3. Translational Equilibrium<br />Constant speed<br />Car at rest<br />The linear speed is not changing with time. There is no resultant force and therefore zero acceleration. Translational equilibrium exists.<br />
    4. 4. Constant rotation<br />Wheel at rest<br />Rotational Equilibrium<br />The angular speed is not changing with time. There is no resultant torque and, therefore, zero change in rotational velocity. Rotational equilibrium exists.<br />
    5. 5. First Condition:<br />Second Condition:<br />Equilibrium<br />An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.<br />
    6. 6. SFx= 0<br />Right = left<br />SFy= 0<br />Up = down<br />S t = 0<br />S t (ccw)= S t (cw)<br />ccw (+)<br />cw (-)<br />Total Equilibrium<br />In general, there are eight degrees of freedom (right, left, up, down, forward, backward, ccw, and cw):<br />
    7. 7. General Procedure:<br /><ul><li>Draw free-body diagram and label.
    8. 8. Choose axis of rotation at point where least information is given.
    9. 9. Extend line of action for forces, find moment arms, and sum torques about chosen axis:</li></ul>St = t1 + t2 + t3 + . . . = 0<br /><ul><li> Sum forces and set to zero: SFx= 0; SFy= 0
    10. 10. Solve for unknowns.</li></li></ul><li>Center of Gravity<br />The 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.<br />The single support force has line of action that passes through the c. g. in any orientation.<br />
    11. 11. Examples of Center of Gravity<br />Note: C. of G. is not always inside material.<br />
    12. 12. Finding the center of gravity<br />01. Plumb-line Method<br />02. Principle of Moments<br />St(ccw) = St(cw)<br />
    13. 13. Example 5:Find the center of gravity of the system shown below. Neglect the weight of the connecting rods.<br />m3 = 6.00 kg<br />m2 = 1.00 kg<br />Since the system is 2-D: Choose now an axis of rotation/reference axis:<br />3.00 cm<br />CG<br /> 2.00 cm<br />m1 = 4.00 kg<br />Trace in the Cartesian plane in a case that your axis of rotation is at the origin <br />(-3.00 cm, 2.00 cm)<br />(0, 2.00 cm)<br />(0,0)<br />
    14. 14. x<br />4 m<br />6 m<br />5 N<br />10 N<br />30 N<br />Example 6:Find the center of gravity of the apparatus shown below. Neglect the weight of the connecting rods.<br />C.G.<br />Choose axis at left, then sum torques:<br />x = 2 m<br />
    15. 15. T<br />300<br />800.0 N<br />T<br />T<br />Fx<br />300<br />300<br />3.00 m<br />5.00 m<br />2.00 m<br />200.0 N<br />Fy<br />800.0 N<br />200.0 N<br />800.0 N<br />Example 4: Find the tension in the rope and the force by the wall on the boom. The 10.00-mboom weighing 200.0 N. Rope is 2.00 m from right end.<br />Since the boom is uniform, it’s weight is located at its center of gravity (geometric center)<br />
    16. 16. Example 4 (Cont.)<br />T<br />T<br />Fx<br />300<br />300<br />5.00 m<br />2.00 m<br />3.00 m<br />Fy<br />w1<br />w2<br />200.0 N<br />800.0 N<br />r<br />Choose axis of rotation at wall (least information)<br />r1= 8.00 m<br />St(ccw):<br />r2= 5.00 m<br />St(cw):<br />r3= 10.00 m<br />T = 2250 N<br />
    17. 17. SF(up) = SF(down):<br />SF(right) = SF(left):<br />F = 1950 N, 356.3o<br />
    18. 18. Example 3:Find the forces exerted by supports A and B. The weight of the 12-m boom is 200 N.<br />B<br />A<br />2 m<br />3 m<br />7 m<br />80 N<br />40 N<br />2 m<br />3 m<br />7 m<br />Draw free-body diagram<br />B<br />A<br />40 N<br />80 N<br />Rotational Equilibrium:<br />Choose axis at point of unknown force.<br />At A for example.<br />
    19. 19. Summary<br />Conditions for Equilibrium:<br />An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.<br />
    20. 20. <ul><li>Draw free-body diagram and label.
    21. 21. Choose axis of rotation at point where least information is given.
    22. 22. Extend line of action for forces, find moment arms, and sum torques about chosen axis:</li></ul>St = t1 + t2 + t3 + . . . = 0<br /><ul><li> Sum forces and set to zero: SFx= 0; SFy= 0
    23. 23. Solve for unknowns.</li></ul>Summary: Procedure<br />

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