1. Equilibrium Forces and Torques 9/11/07

2. Topics to Cover Components of forces and trigonometry Force examples Center of Mass Torques/Moments Torque Examples Equilibrium Terminology http://archone.tamu.edu/rburt/cosc321/cosc321whole.htm

3. Force Components soh sine (angle) = opposite/hypotenuse cah cosine (angle) = adjacent/hypotenuse toa tangent (angle) = opposite/adjacent

4. Trigonometry F x is negative –90° to 270° F y is negative –180° to 360° tan is positive –quads I & III tan is negative –quads II & IV

5. Components of Force F = 10 lbs F x = F * cos (30°) F x = 10 * 0.87 = 8.7 lbs F y = F * sin (30°) F y = 10 * 0.5 = 5 lbs 10 lbs 30° 8.7 lbs 5 lbs x y

6. Think – Pair - Share F = 20 lbs Find F x F x = 20 * cos (60°) = 10 lbs Find F y F y = 20 * sin (60°) = 17 lbs 20 lbs 60° 10 lbs 17 lbs x y

7. Components of Force Similar Triangles Find F x Find F y Hypotenuse –Use pythagorean theorem –√(3 2 +4 2 ) = 5 20 lbs 3 4 x y

8. Think - Pair - Share Find F x Find F y Hypotenuse –√(3 2 +7 2 ) = 7.6 20 lbs 3 7 x y

9. How can we tell if forces are balanced? The x and y components cancel

10. Do these forces balance? No No 20 lbs 3 7 x y 30° 10 lbs 15 lbs x y 10 lbs

11. Do these forces balance? X components cancel y components: 10 * sin (30°) = 5 (* 2 for each force) YES! 10 lbs x y 30°

12. What is the balancing force? 10 lb force –F x = 10 * cos (30°) = -8.7 lbs –F y = 10 * sin (30°) = 5 lbs 20 lb force –F x = 3*20/7.6 = 8 lbs –F y = 7*20/7.6= 18.4 lbs Combine components –X: -8.7 + 8 = -0.7 lbs –Y: 5 + 18.4 = 23.4 lbs Negative of these components represent balancing force –X = 0.7 lbs, y = -23.4 lbs –Resultant = √(.7 2 + 23.4 2 ) = 23.4 lbs –Angle: tan -1 (-23.4/0.7) = 88° in fourth quadrant 20 lbs 3 7 x y 30° 10 lbs 88° 23 lbs

13. Force Example Find the MAGNITUDE, DIRECTION & SENSE of the 3 Coplanar Forces –Find the hypotenuse of the 2 triangles using pythagorean theory.

14. Force Example – con’t 10 # Force Fy = Sin 30º * 10 Fy = 0.5 * 10 Fy = + 5 # Fx = Cos 60º * 10 Fx = 0.866 * 10 Fx = - 8.66 # 20 # Force Fy = (4/5) * 20 Fy = 0.8 * 20 Fy = 16 # Fx = (3/5) * 20 Fx = 0.6 * 20 Fx = 12 # 30 # Force Fy = (7/7.616) * 30 Fy = 0.919 * 30 Fy = - 27.57 # Fx = (3/7.616) * 30 Fx = 0.394 * 30 Fx = 11.82 #

15. Force Example – con’t Sum up all the components Ry = 5 + 16 - 27.57Rx = -8.66 + 12 + 11.82 Ry = -6.57 #Rx = 15.16 #

16. The Resultant Force (R) is found using Pythagorean theory. The Direction of the Force (q ) is found using trigonometry

17. Force – Final Result The Magnitude of the Resultant is 16.52 # The Direction is 23.41 o to the horizontal (X) axis The Sense of the force is way from the origin

18. Lamp Post Forces Example Cable Tension = mg / sin  Pull of cable = push of strut Cable is in Tension, the strut in Compression! cable strut Lamp post mg  

19. Extended Objects – Center of Mass Draw the forces (the free-body diagram) on a stack of books. –How much does a book weigh? –What is diagram for the top book? For middle? For bottom? Note that the bottom of the structure must support the most weight! A book applies a force over some area (distributed load). Why am I justified in drawing only one arrow? Center of Mass! top middle bottom N mg

20. Center of mass of triangular wedge Toward the thicker end - more mass on that side. – exact value is a third of the way from thicker end For an ideal, rigid object that does not change shape or break, we can say that gravity just acts at one point, the center of mass (or center of gravity, or centroid).

21. Center of mass con’t Where can you hold a meter stick without causing a rotation? What if there is a weight on one end? How about a see-saw? How to balance?

22. Mechanical Equilibrium Mechanical equilibrium –Total translational and rotational forces equal zero –No acceleration –Static equilibrium means not moving –Dynamic equilibrium means constant velocity, i.e. not accelerating

23. Translational Equilibrium In order to be in translational equilibrium, the total forces of the object must be zero. Set the horizontal and vertical components of the forces to zero.

24. Rotational Equilibrium Torque or Moment In order to be in rotational equilibrium, the total torques acting on an object must be zero.

25. Moments – rotation Defined by magnitude and direction Units N*m or ft*lb Direction – represented by arrows –counterclockwise positive –clockwise negative Value found from F and perpendicular distance –M = F*d –d is also called “lever” or “moment” arm

26. Another Torque Example Same F Which torque is larger? Why?

27. Tools that use torque Torques allows us to move objects heavier than we can actually lift What if we needed 1000 pounds of force to raise a block? –None of us can just bend down and lift with that force –What tool would we use? Problem: –Say we dig a little hole, and slide our 2x4 under the block, and then place a small wedge 6” away from the end (board is 6’, remember). How much can we exert now?

28. Pole Vaulter The pole vaulter is holding pole so that it does not rotate and does not move. It is in equilibrium. c.g. 2 ft. F G = 5 lbs FRFR FLFL pivot 6 ft. Find the forces F L and F R.

29. Pole vaulter – con’t Rotation = 0 Translation = 0

30. Pole vaulter – con’t c.g. 2 ft. F G = 5 lbs F L =15 lbs pivot 6 ft. F R =10 lbs

31. Torques or Moments - Rotation Torques are also arrows, just like forces –clockwise or counterclockwise cable strut Lamp post

32. Activity 2 - Equilibrium a) Class 1 b) Class 2

33. Terminology A metal column deforms when it supports weight, but it is so small you can’t see it. This deformation is called COMPRESSION if it pushes inwards and TENSION if it pulls out, and SHEAR if it makes different parts of the object go different ways.

34. Terminology - con’t “Loads” “Static loads” are approximately constant Some of these are “dead loads” and don’t vary at all –Weight of the structure “Live loads” vary, but only change very slowly –snow –number of people –Furnishings

35. Terminology - con’t Dynamical loads Dynamic loads change rapidly –Wind and Earthquakes –Can be very dangerous to buildings Demo: Impact load. Weigh an object on a scale, show that dropping from equilibrium gives double the force of the settled value. Oscillations –Everything has a natural frequency it will prefer to oscillate at. If driven near this frequency, the response is large.