05B: STATICS

1. 05B: STATICS

2. San Juanico Bridge

3. 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.

4. Constant speed Car at rest Translational Equilibrium The linear speed is not changing with time. There is no resultant force and therefore zero acceleration. Translational equilibrium exists.

5. Constant rotation Wheel at rest Rotational Equilibrium The angular speed is not changing with time. There is no resultant torque and, therefore, zero change in rotational velocity. Rotational equilibrium exists.

6. First Condition: Second Condition: Equilibrium • An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.

7. T 300 Does Equilibrium Exist? 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. Yes A sky diver who reaches terminal speed? Yes A fixed pulley rotating at constant speed?

8. Statics or Total Equilibrium Statics 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.

9. Construct free-body diagram. • Sum forces and set to zero: SFx= 0; SFy= 0 • Solve for unknowns. Translational Equilibrium Only If all forces act at the same point, then there is no torque to consider and one need only apply the first condition for equilibrium:

10. Review: Free-body Diagrams • Read 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.

11. 550 B 350 By A Ay A B 550 350 Ax Bx w 500 N Example 2.Find tension in ropes A and B. Recall: SF = 0 SFx = Bx - Ax Bcos θB – Acos θA = 0 Bx - Ax = 0 By + Ay – W = 0 SFy = By + Ay - W Bsin θB + Asin θA – W = 0

12. B By A Ay 550 350 Ax Bx w Example 2 (Cont.)Simplify by rotating axes: B cos θB – Acos θA = 0 B sin θB + Asin θA – W = 0 A = 287 N B = 410 N

13. SFx= 0 Right = left SFx= 0 Up = down S t = 0 S t (ccw)= S t (ccw) ccw (+) cw (-) Total Equilibrium In general, there are six degrees of freedom (right, left, up, down, ccw, and cw):

14. 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.

15. Center of Gravity 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. The single support force has line of action that passes through the c. g. in any orientation.

16. Examples of Center of Gravity Note: C. of G. is not always inside material.

17. Finding the center of gravity 01. Plumb-line Method 02. Principle of Moments St(ccw) = St(cw)

18. Example 5:Find the center of gravity of the system shown below. Neglect the weight of the connecting rods. m3 = 6 kg m2 = 1 kg Since the system is 2-D: Choose now an axis of rotation/reference axis: 3 cm CG 2 cm m1 = 4 kg Trace in the Cartesian plane in a case that your axis of rotation is at the origin (-3 cm, 2cm) (0, 2cm) (0,0)

19. x 4 m 6 m 5 N 10 N 30 N Example 6:Find the center of gravity of the apparatus shown below. Neglect the weight of the connecting rods. C.G. Choose axis at left, then sum torques: x = 2.00 m

20. T 300 800 N T T Fx 300 300 3 m 2 m 5 m 200 N Fy 800 N 200 N 800 N 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, it’s weight is located at its center of gravity (geometric center)

21. Example 4 (Cont.) T T Fx 300 300 5 m 3 m 2 m Fy w1 w2 200 N 800 N r Choose axis of rotation at wall (least information) r1= 8 m St(ccw): r2= 5 m St(cw): r3= 10 m T = 2250 N

22. Example 4 (Cont.) T 300 Fx 300 3 m 2 m 5 m Fy T 200 N 800 N 200 N 800 N Ty Tx 300 SF(up) = SF(down): SF(right) = SF(left): F = 1953 N, θ = 356.3o

23. B A 2 m 7 m 3 m 80 N 40 N Example 3:Find the forces exerted by supports A and B. Neglect the weight of the 10-m boom. 2 m 7 m 3 m Draw free-body diagram A B 40 N 80 N Rotational Equilibrium: Choose axis at point of unknown force. At A for example.

24. 7 m 2 m 3 m A B 40 N 80 N B A 2 m 7 m 3 m 80 N 40 N Example 3: (Cont.) Rotational Equilibrium: St = t1 + t2 + t3 + t4 = 0 or St(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

25. Example 3 (Cont.) 7 m 2 m 3 m First: St(ccw) A B 40 N 80 N t1 = Br3 A B r3 t2 = w1r1 r1 Next: St(cw) r2 w1 t3 = w2r2 w2 St = 0 St(ccw) = St(cw) B = 48 N

26. Example 3 (Cont.) 7 m 2 m 3 m Translational Equilibrium A B 40 N 80 N B SFx= 0; SFy= 0 A w1 SF(up) = SF(down) w2 Recall that B = 48N A = 72N

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

28. 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. Summary: Procedure