Chapter 5 Static Equilibrium; Torque, and Elasticity

Chapter 5

Static Equilibrium; Torque, and Elasticity

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

5-1 The Conditions for Equilibrium

An object with forces acting on it, but that is not moving, is said to be in equilibrium.

5-1 The Conditions for Equilibrium

The first condition for equilibrium is that the forces along each coordinate axis add to zero.

5-2 TorqueTo 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.

5-2 Torque

A longer lever arm is very helpful in rotating objects.

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.

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.

5-2 Torque

The torque is defined as:

(5.4)

5-2 The Conditions for EquilibriumThe second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.

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.

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.

Solution

a) The lever arm is given.• F = 700 N• r = 0.05m = r F = (0.050m)(700N) = 35 m Nb) 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

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).

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.

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.

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.

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.

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.

5-5 Solving Statics Problems

1. Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act.

2. Choose a coordinate system and resolve forces into components.

3. Write equilibrium equations for the forces.

4. 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.

5. Solve.


The previous technique may not fully solve all statics problems, but it is a good starting point.

5-5 Solving Statics ProblemsA 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.

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

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+14,700+147,000 = 43,700 N

5-2 Solving Statics ProblemsIf 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.


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.

5-7 Applications to Muscles and Joints

These same principles can be used to understand forces within the body.

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.

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

b) Solve for FM:

FM = (0.15m)(2.0kg)g + (0.35m)(5.0kg)g = 0.050m

FM = (41 kg)g = 400 N

• 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

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.

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, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.


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!


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.

5-5 Elasticity; Stress and Strain

Hooke’s law: the change in length is proportional to the applied force.

(5-3)

5-5 Elasticity; Stress and StrainThis 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.


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.

Young’s Modulus

• Young’s Modulus is Stress/Strain orE = F/A or ∆L = 1 F L0

∆L/L0 E A


In tensile stress, forces tend to stretch the object.


Compressional stress is exactly the opposite of tensional stress. These columns are under compression.


Shear stress tends to deform an object:

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

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.

5-6 Fracture

A horizontal beam will be under both tensile and compressive stress due to its own weight.

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.

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.

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.

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.

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


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.


Pointed arches can be built that require considerably less horizontal force.


A dome is similar to an arch, but spans a two-dimensional space.

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.

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.

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Chapter 5 Static Equilibrium; Torque, and Elasticity