Physics statics
- 3. Translational Equilibrium The linear speed is not changing with time. There is no resultant force and therefore zero acceleration. Translational equilibrium exists. Car at rest Constant speed
- 4. 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. Wheel at rest Constant rotation
- 5. Equilibrium An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque. First Condition: Second Condition:
- 6. Total Equilibrium In general, there are eight degrees of freedom (right, left, up, down, forward, backward, ccw, and cw): F x = 0 Right = left F y = 0 Up = down ccw (+) cw (-) (ccw) = (cw)
- 7. 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: Sum forces and set to zero: F x = 0; F y = 0 Solve for unknowns.
- 8. 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.
- 9. Examples of Center of Gravity Note: C. of G. is not always inside material.
- 10. Finding the center of gravity 01. Plumb-line Method 02. Principle of Moments (ccw) = (cw)
- 11. Example 5: Find the center of gravity of the system shown below. Neglect the weight of the connecting rods. m 1 = 4.00 kg m 2 = 1.00 kg m 3 = 6.00 kg 3.00 cm 2.00 cm Since 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, 2.00 cm) (-3.00 cm, 2.00 cm) CG
- 12. 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 m C.G. 30 N 10 N 5 N 4 m 6 m x
- 13. Example 4: Find the tension in the rope and the force by the wall on the boom. The 10.00-m boom weighing 200.0 N . Rope is 2.00 m from right end. Since the boom is uniform, it’s weight is located at its center of gravity (geometric center) 30 0 T 800.0 N 30 0 T 800.0 N 200.0 N 30 0 800.0 N 200.0 N T F x F y 2.00 m 3.00 m 5.00 m
- 14. Choose axis of rotation at wall (least information) (ccw): r (cw): r 1 = 8.00 m r 2 = 5.00 m r 3 = 10.00 m T = 2250 N 30 0 T 800.0 N 200.0 N 30 0 w 2 w 1 T F x F y 2.00 m 3.00 m 5.00 m Example 4 (Cont.)
- 15. F(up) = F(down): F(right) = F(left): F = 1950 N, 356.3 o
- 16. Example 3: Find the forces exerted by supports A and B . The weight of the 12-m boom is 200 N. 40 N 80 N 2 m 3 m 7 m A B Draw free-body diagram Rotational Equilibrium: Choose axis at point of unknown force. At A for example. 40 N 80 N 2 m 3 m 7 m A B
- 17. Summary An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque. Conditions for Equilibrium:
- 18. Summary: 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: Sum forces and set to zero: F x = 0; F y = 0 Solve for unknowns.
Editor's Notes
- #3: San Juanico Bridge (Marcos Highway) Connects Tacloban City on the Leyte side and Santa Rita town on the Samar side Rotational and translational equilibrium must be maintained
- #12: x = 2.00 m
- #13: x = 2.00 m
- #15: T = 2250 N
- #16: Fx = 1948.5571…, φ = 3.66966.. Θ = 356.3 F = 1953.0043 … 1953 N, 356.3 0