Chapter 9
The Giancoli onLine Tutor

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Bodies in Equilibrium

I. Strength of Materials

Strength of Materials deals mainly with stress and strain. Stress is the force applied to a solid "bar" to stretch (tensile stress) or compress (compressive stress) it divided by the cross sectional area of the bar.

Stress= F/A

Strain is the change in length of the bar under stress divided by its original length.

Strain= Dl/l

The ratio of stress to strain is the Elastic modulus, E, usually a given.

E= (F/A)/(Dl/l)

Elastic modulus for wood (Oak) = 13x10⁹ N/m²

Elastic modulus for Copper = 110x10⁹ N/m²

Elastic modulus for Wrought Iron = 210x10⁹ N/m²

Elastic modulus for Cast Iron = 110x10⁹ N/m²

Elastic modulus for Steel = 200x10⁹ N/m²

Elastic modulus for Aluminum = 70x10⁹ N/m²

Combining the above relations you can obtain

F=(EA/l)*Dl

Providing that Dl is not too large the term in () above is a constant and the relationship is identical to Hooke's Law F=-kx. (the "-" sign is because the force in Hooke's Law is the force applied by the "spring" not the force applied to it as it is here.)

Virtually all problems can be worked by manipulating the above equation. Usually the only complication is finding the cross sectional, A, from the given data. There are three general cases:

Rectangular as in the support for a modern desk; A=a*b where "a" and "b" are the side lengths.
Square as in a stool leg; A=a², where "a" is the side length.
Circular as in a stool leg; A=p*R², where "R" is the radius.

Interactive Examples

A 50 kg mass is hung from a steel wire 2 mm in diameter and 2 meters long. What is its tensile stress to the nearest N/m²? Use 10 for g. The breaking point for steel is 5x10⁸ N/m²
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To the nearest mm how far is the wire stretched?
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What is the Hooke's law constant for this wire?
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II. Statics

Conditions

Statics specifies the conditions under which an object is in equilibrium - not moving and not rotating.

The sum of the vertical forces or components=0
The sum of the horizontal forces or components=0
The sum of the torques about any axis=0

(These are identical to the conditions in the previous chapter except that the acceleration and the angular acceleration are both zero.)

Strategy

Interactive Activity
Complete the following steps to learn how to set up typical statics problems.

The basic steps are as follows:

Draw in all forces acting on the object .
1. The force between an object and a surface will, in general, have two components, a normal force (perpendicular to the surface) and friction (parallel to the surface and against the slipping tendency).
2. The tension in a rope (or wire) is always along the rope.
3. Weight is always down.
4. A hinge force will have a vertical and horizontal component the direction (up or down, left or right) of each will not always be obvious. However, if you choose wrong you will still get the right answer with a minus sign in front of it.
Replace all forces which are neither vertical or horizontal with their vertical and horizontal component forces.
Add all the forces in the "y" direction and set them equal to zero.
Add all the forces in the "x" direction and set them equal to zero.
Choose an axis and calculate the sum of the torques equal to zero.
1. There are no absolute rules for picking the axis but in general pick the point where the most forces act as your axis since this will eliminate these forces from your torque equation.
2. Remember clockwise torques are positive and counter-clockwise torques are negative.
3. Usually it is easiest to calculate the torque by multiplying the perpendicular ( to the body) component of the force by the distance from the axis to its point of application. This is not always true (see the ladder example below).
Solve the equations for the unknowns algebraically.

Interactive Examples

If the mass of the bar in the above figures is 50 kg and its length is 3 m, to the nearest N what is the Tension in the wire? Use 10 for g.
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What is F_d?
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What is F_u?
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II-c. Ladder Problems

Play the movie frame by frame by using the two buttons on the right. The red circle represents a person climbing a ladder. To simplify the problem we assume that the wall is frictionless and the ladder weightless. If we choose the axis of rotation at the base of the ladder than the only two forces exerting torques are the wall (negative) and the weight (positive). It is the weight which is holding the ladder against the wall but only so long as the line of action of the weight is to the right of the base of the ladder. Otherwise the torque exerted would be negative and the ladder would tip over backwards. The first frame shows that this is perilously close to happening. However, there is a danger in moving the bottom too far out from the base of the wall, since, as this is done the required force of friction between the ladder and the ground rapidly grows and the ladder will slip if it gets too large.

Note if you do not have the quicktime plugin you can view a GIF animation version of the movie here .

There are only two vertical forces ( the weight and the normal force from the ground) so they must be equal and opposite.

There are only two horizontal forces ( The normal force from the wall and the friction) so the must be equal and opposite.

The torque exerted by the normal force from the wall will equal this force multiplied by the perpendicular distance between the line of action of the force and the axis of rotation. This is obviously the vertical distance from the base of the wall to where the ladder touches the wall, that is, the lever arm.

If the person is 1/3 of the way up the ladder then the lever arm for calculating the torque from the weight will equal 1/3 the distance between the bottom of the ladder and the base of the wall or if x is the distance climbed along the ladder, the lever arm would be xcos(q).

The Key for solving a problem like this will be to finding the height and the base. They will usually be given but, if not,

if you know the length of the ladder and either the base or height use the Pythagorean Theorem to find the other.
If you know the length of the ladder and the angle its makes with the ground use the sine function to find the height and the cosine function to find the base.

Interactive Activities

A 100 kg man climbs a 5 meter long massless ladder ( purchased at Home Depot ). The bottom of the ladder is positioned 3 m from the bottom of the wall the ladder is set against. The maximum force of static friction between the ladder and the ground is 500 N. To the nearest tenth of a meter how far up the ladder does the man gets before the ladder slips?
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