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Announcements. Civil and Mechanical engineers: This week is for you! Statics: Oooh , so exciting! Please pay attention: We want you to build bridges that don’t fall down!. Lecture 17: Rotational Statics part 1. Today’s Concepts: a) Torque Due to Gravity b) Static Equilibrium. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. Announcements • Civil and Mechanical engineers: This week is for you! • Statics: Oooh, so exciting! • Please pay attention: We want you to build bridges that don’t fall down!

2. Lecture 17: Rotational Statics part 1 Today’s Concepts: a) Torque Due to Gravity b) Static Equilibrium

3. New Topic, Old Physics: As the name implies, “statics” is the study of systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges... Statics: The key equations are familiar to us: If an object doesn’t move: The net force on the object is zero The net torque on the object is zero (about any axis)

4. Statics: Example: What are all of the forces acting on a car parked on a hill? x y f N mg θ

5. Car on Hill: Use Newton’s 2nd Law: FNET = MACM = 0 Resolve this into xand ycomponents: • x:f-mg sinθ = 0 • f= mg sinθ x y f N • y:N-mg cosθ= 0 • N = mg cosθ mg θ

6. Torque Due to Gravity Magnitude: τ =RCM Mg sin (θ) Lever arm

7. Example: Consider a plank of mass Msuspended by two strings as shown. We want to find the tension in each string: T1 T2 M xcm L/4 L/2 y Mg x

8. Balance forces: T1+T2=Mg T1 T2 M x cm L/4 L/2 y Mg x

9. Balance Torques Choose the rotation axis to be out of the page through the CM: The torque due to the string on the right about this axis is: τ2 = T2L/4 T1 T2 + M xcm The torque due to the string on the left about this axis is: L/4 L/2 τ1= -T1L/2 y Mg Gravity acts at CM, so it exerts no torque about the CM x

10. Finish the problem The sum of all torques must be 0: T1 T2 We already found that T1+ T2= Mg M x cm L/4 L/2 y Mg x

11. What if you choose a different axis? Choose the rotation axis to be out of the page at the left end of the beam: The torque due to the string on the right about this axis is: T1 T2 + M x cm The torque due to the string on the left is zero L/4 L/2 y Mg Torque due to gravity: x

12. You end up with the same answer! The sum of all torques must be 0: T1 T2 We already found that T1+T2 =Mg M x cm L/4 L/2 y Mg x Same as before

13. Approach to Statics: Summary In general, we can use the two equations to solve any statics problem. When choosing axes about which to calculate torque, we can sometimes be clever and make the problem easier....

14. CheckPoint: Ball on Rod M 30o 90o M 2L L In Case 1 one end of a horizontal massless rod of length Lis attached to a vertical wall by a hinge, and the other end holds a ball of mass M. In Case 2the massless rod holds the same ball but is twice as long and makes an angle of 30owith the wall as shown. In which case is the total torque about the hinge biggest? A) Case 1B) Case 2C) Both are the same About 50% of the class got this right. gravity Case 1 Case 2

15. 2M 30o 90o M 2L L ACT: 2 Rods, part I In Case 1 one end of a horizontal rod of mass Mand length L is attached to a vertical wall by a hinge, and the other end is free to rotate. In Case 2a rod of mass 2M and length 2L makes an angle of 30owith the wall as shown. In which case is the torque about the hinge largest? A) Case 1B) Case 2C) Both are the same gravity Case 1 Case 2

16. 2M 30o 90o M 2L L ACT: 2 Rods, part II In Case 1 one end of a horizontal rod of mass Mand length L is attached to a vertical wall by a hinge, and the other end is free to rotate. In Case 2a rod of mass 2M and length 2L makes an angle of 30owith the wall as shown. Which rod has larger angular acceleration at the moment shown? A) Case 1B) Case 2C) Both are the same gravity Case 1 Case 2

17. ACT: Ball and Beam An object is made by hanging a ball of mass Mfrom one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below. How far to the right of the pivot is the center of mass of the plank only? A) L/4 B) L/2 C) 3L/4 L/4 gravity L M M

18. CheckPoint: Ball and Beam An object is made by hanging a ball of mass Mfrom one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below. Is the object balanced? A) Yes B) No, it will tip left C) No, it will tip right L/4 gravity L M M

19. CheckPoint: Sign M M L/2 L In Case 1, one end of a horizontal plank of mass Mand length Lis attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In Case 2 the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire biggest? A) Case 1 B) Case 2 C) Both are the same About 40% of the class got this right. gravity Case 1 Case 2

20. ACT: Sign L Which equation correctly expresses the fact that the total torque about the hinge is zero? A) B) C) T L sin θ θ Mg The L cancels out:

21. CheckPoint: Revote M M L/2 L In Case 1, one end of a horizontal plank of mass Mand length Lis attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In Case 2 the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire biggest? A) Case 1 B) Case 2 C) Same

22. 2L/3 L L Lever Arm Lever Arm: The perpendicular (shortest) distance between the line of action of a force and the rotation axis. T T T

23. ACT In which of the static cases shown below is the tension in the supporting wire bigger? In both cases Mis the same, and the blue strut is massless. T1 T2 300 300 L L/2 M M Case 2 Case 1 A) Case 1 B) Case 2 C) Same

24. It’s the same. Why? Case 1 Case 2 d1 d2 θ θ L/2 Balancing Torques T1 T2 L M M