1. Static Equilibrium; Torque, and Elasticity
Chapter 5
Static Equilibrium; Torque, and Elasticity

2. Units of Chapter 5 The Conditions for Equilibrium
Center of Mass; Center of Gravity
Solving Statics Problems
Simple Machines
Applications to Muscles and Joints
Elasticity; Stress and Strain
Fracture
Spanning a Space: Arches and Domes

3. 5-1 The Conditions for Equilibrium
An object with forces acting on it, but that is not moving, is said to be in equilibrium.
In this next unit, we will look at the condition when the net force, those acting on and within an object or structure are balanced. If they are balanced what does that tell you about the acceleration is (zero). Statics is the calculation of the forces acting on and within a structure that is in equilibrium. We are also going to look at the affects of rotation and deformation of objects due to forces. We will look at two new things called torque and elasticity.
Statics looks at the forces on a structure and whether or not the structure can take the forces without breaking. We will use this when we build your bridges.
Let’s start when what “Equilibrium” is. All objects on the earth have at least one force acting on it all the time. What is that force? If this book is not moving, then the forces on it must be equal and the force that counteracts the gravity force is the normal force that is from the table. Since both of these forces are equal and opposite, the book is in equilibruim. Equilibrium in latin means “equal forces” or “balance”. This is not newton’s third law though. That acts on different objects not the same one.

4. 5-1 The Conditions for Equilibrium
The first condition for equilibrium is that the forces along each coordinate axis add to zero.
There are two conditions that must be met for equilibruim to occur. The first is that the net force in each direction must be zero. If the chandeler is not moving, the vertical component of Fa must be equal to 1960 N. The force Fb is equal to the horizontal component of Fa.

5. 5-2 Torque
To make an object start rotating, a force is needed; the position and direction of the force matter as well.
The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.
The second condition we need for equilibrium has to do with rotation. Let’s look at a door to see how this works. To make the door swing on the hinge, you need to apply a force to the door. The greater the force the quicker the door opens. If you push at a point closer the the hinge, the door will open slower. The effect of the force is less. In fact, how fast the door opnes is proportional to the magnitude of the force and to the distance from the hinge. This distance is called the lever arm or moment arm. In the diagram, it is rA and rB. If rA is 3 times bigger than rB thaen the door will open at an acceleration 3 times greater.

6. 5-2 Torque A longer lever arm is very helpful in rotating objects.
It is easier for a plumber to turn this pipe with a longer handle or for a person to loosen the lug nuts using a longer tire iron.

7. 5-2 Torque
Torque is how well a force does at changing or accelerating a rotation.
We use the greek letter  (tau) for torque.
Acceleration is proportional to torque
The units of torque are Newtons time meter.

8. 5-2 Torque
Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.
Remember I said that the lever arm is the perpendicular distance from the axis of rotation to the line of force. So let’s look at how we figure that out. Which of these forces is going to make it easiest to open the door? Which one will not make the door open at all? The lever arm is found by drawing a line in the direction of the force and another line perpendicular to that where the hinge is. The lever arm for FA is the distance from the hinge to the door since the force is perpendicular.

9. 5-2 Torque
The torque is defined as:
(5.4)

10. 5-2 The Conditions for Equilibrium
The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.
So now let’s talk about the second condition of equilibruim. Look at this ruler. The first condition of equilibrium is met the forces add up to give zero but what will happen to the ruler? It will rotate. But in order to be static it must not move so we need to net torque on the object to also be zero. So the secon condition is that the sum of the torque must be zero. This will mean there is no angular rotation about any axis.

11. Cases we will consider
We will look at cases where all of the forces act in one plane (the xy plane).
In those cases, the torque acts about an axis perpendicular to the xy plane.
Use the axis that makes it easiest to work the problem.

12. Example problem
The biceps muclse exerts a verticla force on the lower arm bent as shown. For each case, calculate the torque about the axis of rotation through the elbow joint, assuming the muscle is attached 5.0 cm from the elbow.

13. Solution a) The lever arm is given. F = 700 N r = 0.05m
 = r F = (0.050m)(700N) = 35 m N
b) The lever arm is shorter since the arm is bent
F = (700 N)(sin60º)
r = 0.050m
 = (0.050m)(700N) (0.866) = 30 m N
You can also solve by using the component of the moment arm

14. 5-3 Center of Mass There is one point that moves in the same path a
particle would take if subjected to the same force as the diver. This point is called the center of mass (CM).
Notice that these two divers are doing two different kinds of dives but that there is a point in their bodies that moves the same no matter what is happening around it. This point is called the center of mass.

15. 5-3 Center of Mass
This wrench is acted on by a zero net force which is making it rotate and translate along a horizontal surface.
If you looked at all of the atoms in the wrench and added up the mass and their position, you could find the center of the mass.

16. 5-3 Center of Mass
Most objects are not point particles but instead take up space.
The center of mass is a single point that is the weighted average of the mass distributed in a body.

17. 5-3 Center of Mass
For two particles, the center of mass lies closer to the one with the most mass:
where M is the total mass.
How do you figure out where the center of mass is? You can calculate it by taking the mass of A times the distance to mass a and add that to the mass of B times the distance to the mass and divide by the total mass.

18. 5-3 Center of Mass
The center of gravity is the point where the gravitational force can be considered to act. It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object.

19. 5-3 Center of Mass
The center of gravity can be found experimentally by suspending an object from different points. The CM need not be within the actual object – a doughnut’s CM is in the center of the hole.

20. 5-5 Solving Statics Problems
Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act.
Choose a coordinate system and resolve forces into components.
Write equilibrium equations for the forces.
Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously.
Solve.

21. 5-5 Solving Statics Problems
The previous technique may not fully solve all statics problems, but it is a good starting point.

22. 5-5 Solving Statics Problems
A uniform 1500-kg beam, 20.0 m long supports a 15,000-kg printing press 5.0 m from the right support column. Calculate the force on each of the vertical support columns.

23. Solution Need to look at the forces in the beam. Label them
The weight of the beam acts at its center of gravity
Pick an axis for the torque that is convenient
Use the torque equation to solve for FB

24. 5-5 Solving Statics Problems
 = - (10.0m) (1500kg) g - (15.0m) (15,000kg) (g)+ (20.0m) FB=0
Solve for FB = 118,000 N
Use sum of forces = 0 to solve for FA
F = FA+FB-(1500kg)g-(15,000kg)g=0
FA= -118,000+14, ,000 = 43,700 N

25. 5-2 Solving Statics Problems
If a force in your solution comes out negative (as FA will here), it just means that it’s in the opposite direction from the one you chose. This is trivial to fix, so don’t worry about getting all the signs of the forces right before you start solving.

26. 5-2 Solving Statics Problems
If there is a cable or cord in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend.

27. 5-7 Applications to Muscles and Joints
These same principles can be used to understand forces within the body.

28. Problem
How much force must the biceps muscle exert when a 5.0 kg mass is held in the hand (a) with the arm horizontal and (b) when the arm is at 45 degree angle as in the lower picture? Assume that the mass of the forearm and hand together is 2.0 kg and their CG is as shown.

29. Solution
a) calculate torque about the point where FJ acts in the upper picture. The sum of the torque = 0 so
(0.050m)FM – (0.15m)(2.0kg)g – (0.35m)(5.0kg)g = 0
Solve for FM:
FM = (0.15m)(2.0kg)g + (0.35m)(5.0kg)g = m
FM = (41 kg)g = 400 N

30. B) The lever arm, as calculated about the joint is reduced by sin 45 degrees for all three forces. Our torque equation will look the same as the one above except that each term will have it lever arm reduced by the same factor, which will cancel out. So the same result is obtained,
FM = 400 N
The point of insertion of a muscle varies from person to person. A light increase in the distance of the joint to the opint of insertion of the biceps muscle from 5.0 to 5.5 cm can be a considerable advantage for lifting and throwing. Champion athletes are often found to have muscle insertions farther from the joint than the average person, and if this applies to one muscle it usually applies to all.

31. 5-7 Applications to Muscles and Joints
The angle at which this man’s back is bent places an enormous force on the disks at the base of his spine, as the lever arm for FM is so small.
As another example of the large fores acting within the human body, we consider the muscles used to support the trunk when a person bends forward. The lower vertebra on the spinal column (fifth lumbar vertebra) acts as a fulcurm for this bending position. The muscles in the back that support the trunk act at an effective angle of about 12 degrees to the axis of the spine. The upper figure shows the forces on the upper body. We assume that the trunk makes an angle of 30 degrees with the horizontal. The force exerted by the back muslces is represented by Fm, the force exerted on the base of the spine at the lowest vertebra is Fv and wH, wZ and Wr represent the weights of the head, free hanging arms and trunk. The values of the weights are shown in the table. The distances refer to a person that is 180 cm tall but are usually in a ratio of 1:2:3 for the average person. When you calculate this out, Fm= 2.4w, Fv=2.6w
The force on the lowest vertebra is over 2 ½ times the total body weight. This force is exerted by the scral bone at the base of the spine thruoght a fluid filled and somewhat flexibel intervertebral disc. The discs at the base of the spine are clearly being compressed under a very large forces. If the body was less bent over then the stress is less. If the man was holidng a mass of 20 kg and he weighed 90 kg (200 lb) the force is almost 4 times his weight.

32. 5-4 Stability and Balance
If the forces on an object are such that they tend to return it to its equilibrium position, it is said to be in stable equilibrium.
If I pull this ball away from center, it will go back to the middle once I let go. That is because it is equilibrium. However, if I drop this pencil if is not in equilibrium and will not go back to vertical. I can balance it by keeping its mass directly over the tip, it will be in equilibrium until anything light applies a torque on it and make the pencile fall. In most structures, we have to have stable Equilibrium.

33. 5-4 Stability and Balance
If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.

34. 5-4 Stability and Balance
An object in stable equilibrium may become unstable if it is tipped so that its center of gravity is outside the pivot point. Of course, it will be stable again once it lands!
The critical point is when the CG shifts from one side of the pivot point to the other. In general, an object whose center of gravity is above its base of support will be stable if a vertical line projected downward from the CG falls within the base of the support. A brick standing on its flat side is more stable than if it is on the end. The larger the base and the lower the CG the more stable the object.

35. 5-4 Stability and Balance
People carrying heavy loads automatically adjust their posture so their center of mass is over their feet. This can lead to injury if the contortion is too great.
Humans are much less stable than four-legged mammals, which not only have a larger base of support because of their four legs but also have a lower center of gravity. We constantly try to keep our CG over our feet. Shift our hips backward to bend over. Try bending over with back and hieels against the wall.

36. 5-5 Elasticity; Stress and Strain
Hooke’s law: the change in length is proportional to the applied force.
(5-3)
If a force is exerted on an object, such as this rod, the length of the rod changes. If the amount that it lengthens is small compared to the length of the object, then the change in the length delta L is proportional to the force. If the force is too great, the bar will stretch too much and break.

37. 5-5 Elasticity; Stress and Strain
This proportionality holds until the force reaches the proportional limit. Beyond that, the object will still return to its original shape up to the elastic limit. Beyond the elastic limit, the material is permanently deformed, and it breaks
at the breaking point.

38. 5-5 Elasticity; Stress and Strain
The change in length of a stretched object depends not only on the applied force, but also on its length and cross-sectional area, and the material from which it is made.
The material factor is called Young’s modulus, and it has been measured for many materials.
The Young’s modulus is then the stress divided by the strain.

39. Young’s Modulus Young’s Modulus is Stress/Strain or
E = F/A or ∆L = 1 F L0
∆L/L E A
Discuss some of the number is talble 5.1

40. 5-5 Elasticity; Stress and Strain
In tensile stress, forces tend to stretch the object.
If you can look at the lower half of the rod, because it is in equilibrium, their must be a force acting upward equal to the downward force. So external forces lead to internal forces.

41. 5-5 Elasticity; Stress and Strain
Compressional stress is exactly the opposite of tensional stress. These columns are under compression.

42. 5-5 Elasticity; Stress and Strain
Shear stress tends to deform an object:
An object under shear stress has equal and opposite forces applied across its opposite faces.

43. Shear Modulus
An object under shear has a constant called Shear Modulus S or G that is about ½ to 1/3 of the Young’s Modulus E
∆L = 1 F L0
S A

44. 5-6 Fracture
If the stress is too great, the object will fracture. The ultimate strengths of materials under tensile stress, compressional stress, and shear stress have been measured.
When designing a structure, it is a good idea to keep anticipated stresses less than 1/3 to 1/10 of the ultimate strength.
The ultimate strengths for tension, compression and shear for a bunch of materials are measured. These values give the maxiumum force per unit of area or stress that an object can withstand under each of the types of stress. They can vary however so we use a safety fact of 3 to 10 or more.

45. 5-6 Fracture
A horizontal beam will be under both tensile and compressive stress due to its own weight.

46. 5-6 Fracture
What went wrong here? These are the remains of an elevated walkway in a Kansas City hotel that collapsed on a crowded evening, killing more than 100 people.

47. 5-6 Fracture
Here is the original design of the walkway. The central supports were to be 14 meters long.
During installation, it was decided that the long supports were too difficult to install; the walkways were installed this way instead.

48. 5-6 Fracture
The change does not appear major until you look at the forces on the bolts:
The net force on the pin in the original design is mg, upwards.
When modified, the net force on both pins together is still mg, but the top pin has a force of 2mg on it – enough to cause it to fail.

49. 5-7 Spanning a Space: Arches and Domes
The Romans developed the semicircular arch about 2000 years ago. This allowed wider spans to be built than could be done with stone or brick slabs.

50. 5-7 Spanning a Space: Arches and Domes
The stones or bricks in a round arch are mainly under compression, which tends to strengthen the structure.

51. 5-7 Spanning a Space: Arches and Domes
Unfortunately, the horizontal forces required for a semicircular arch can become quite large – this is why many Gothic cathedrals have “flying buttresses” to keep them from collapsing.

52. 5-7 Spanning a Space: Arches and Domes
Pointed arches can be built that require considerably less horizontal force.

53. 5-7 Spanning a Space: Arches and Domes
A dome is similar to an arch, but spans a two-dimensional space.

54. Summary of Chapter 5
An object at rest is in equilibrium; the study of such objects is called statics.
In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis.
An object in static equilibrium can be either in stable, unstable, or neutral equilibrium.

55. Summary of Chapter 5
Materials can be under compression, tension, or shear stress.
If the force is too great, the material will exceed its elastic limit; if the force continues to increase the material will fracture.