Physics 211 Lecture 17

1. Physics 211Lecture 17 Today’s Concepts: a) Torque Due to Gravity b) Static Equilibrium NOOO MATS COME BACK! STELZER FORCES US TO GET TO KNOW EACH OTHER! I JUST WANT TO BE ANTI-SOCIAL AND DO PHYSICS!

2. Your Comments I want to see examples of different axes of rotation being used to calculate the tension or net torque Can a force exert both a force and a torque at the same time? SPRING BREAK!!!! I didn't sign up for TAM 210 This is the first time I've been completely confused all semester. I just don't understand how you know if you take the cosine or sine of an angle when you're calculating the torque. Whats the greatest force on Earth? Love :D Before spring break this might be my last opportunity to get my words on the wall of FAME in the lecture Can you please explain the thing about tension, the length of the plank and the angle that it makes with the wall? Pretty please?! Can you explain how to determine the axis? play fleetwoodmack and the Joker by Steve Miller band :) por favor? Please give hints to the homework at the end of the lecture; that really helps. Thanks! Why is the direction of the torque in the plank and string problem in and out of the screen? Shouldn't it just be clockwise or counterclockwise? I would just like to thank you for not also scheduling this exam for this week because I would have had to fail my calculus and chemistry exams due to spending all my time focusing on physics. Much appreciated.

3. 2nd Exam is the week after break: Covers lectures 7-13 (not the stuff we are doing now) Sign up for the conflict if you need to before the end of this week. As before, email Dr. Jain if you have a double conflict (rdjain1@illinois.edu). Tuesdays class right after break will be a review(we will work out Fall 2010 exam) . Don’t forget that there are lots of extra office hours during exam week.

4. As the name implies, “statics” is the study of systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges... New Topic, Old Physics: 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 (for any axis)

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

6. Use Newton’s 2nd Law: FNET = MACM = 0 Resolve this into x and y components: Car on Hill: • x:f-mg sinq = 0 • f= mg sinq x y f N • y:N-mg cosq= 0 • N = mg cosq mg q

7. Torque Due to Gravity Magnitude: t =RCMMg sin (q ) Lever arm Evaluate as if gravity acts at the Center of Mass

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

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

10. Choose the rotation axis to be out of the page through the CM: Balance Torques The torque due to the string on the right about this axis is: t2 = T2L/4 T1 T2 M x cm The torque due to the string on the left about this axis is: L/4 L/2 t1 = -T1L/2 y Mg Gravity acts at CM, so it exerts no torque about the CM x

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

12. Choose the rotation axis to be out of the page at the left end of the beam: What if you choose a different axis? 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

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

14. 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.... Approach to Statics: Summary

15. CheckPoint 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 2 the 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 gravity Case 1 Case 2

16. CheckPoint 2L L In which case is the total torque about the hinge biggest? A) Case 1 B) Case 2 C) Both are the same M 30o 90o M Case 1 Case 2 A) Torque is at a maximum at 90 degrees. B) As case 2 has the longer length, under same force, it has higher total torque. C) since sin(30) is 1/2, the expanded length is neutralized by the angle

17. CheckPoint An object is made by hanging a ball of mass M from 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 fall left C) No, it will fall right L/4 gravity L M M

18. Clicker Question How far to the right of the pivot is the center of mass of just the plank. A) L/4 B) L/2 C) 3L/4 x M

19. CheckPoint L/4 L/4 x Is the object balanced? A) Yes B) No, it will fall left C) No, it will fall right M M The pivot is L/4 away from the center of mass and the ball is L/4 away from the pivot. Both the ball and beam weigh the same, so they will be balanced.

20. CheckPoint M M L/2 L In Case 1, one end of a horizontal plank of mass M and length L is 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 gravity Case 1 Case 2

21. Clicker Question L Which equation correctly expresses the fact that the total torque about the hinge is zero? A) B) C) T L sin(q ) q Mg The L cancels out:

22. Clicker Question M M L/2 L In which case is the tension in the wire biggest? A) Case 1 B) Case 2 C) Same

23. T Hy Mg mg Hx Use hinge as axis Why does the force due to the hinge have both horizontal and vertical components?

24. T Hy Mg mg Hx y x

25. m Solve for m