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- 1. 05B: Equilibrium<br />
- 2. San Juanico Bridge<br />
- 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. 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. 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. 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. General Procedure:<br /><ul><li>Draw free-body diagram and label.
- 8. Choose axis of rotation at point where least information is given.
- 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. 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. Examples of Center of Gravity<br />Note: C. of G. is not always inside material.<br />
- 12. Finding the center of gravity<br />01. Plumb-line Method<br />02. Principle of Moments<br />St(ccw) = St(cw)<br />
- 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. 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. 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. 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. SF(up) = SF(down):<br />SF(right) = SF(left):<br />F = 1950 N, 356.3o<br />
- 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. 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. <ul><li>Draw free-body diagram and label.
- 21. Choose axis of rotation at point where least information is given.
- 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. Solve for unknowns.</li></ul>Summary: Procedure<br />

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