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
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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)
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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:
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Editor's Notes
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
x = 2.00 m
x = 2.00 m
T = 2250 N
Fx = 1948.5571…, φ = 3.66966.. Θ = 356.3 F = 1953.0043 … 1953 N, 356.3 0
