UNESCO-NIGERIA TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
MECHANICAL ENGINEERING TECHNOLOGY
MECHANICAL ENGINEERING
SCIENCE[STATICS]
COURSE CODE: MEC111
YEAR I- SE MESTER I
THEORY
Version 1: December 2008
1
MECHANICAL ENGINEERING SCIENCE [STATICS]
TABLE OF CONTENTS
Week 1
1.0 Scalar and Vector quantities
1.1 Scalar and vector quantities
1.1.1Vectors and Direction
1.1.2 Conventions for Describing Directions of Vectors
1.1.3 Representing the Magnitude of a Vector
1.1.4 Vector Resolution
1.1.5 Parallelogram Method of Vector Resolution
1.1.6 Trigonometric Method of Vector Resolution
1.1.7 Vector Components
1.1.8 Vector Addition
1.2 The Pythagorean Theorem
1.2.1 Using Trigonometry to Determine a Vector's Direction
1.2.3 Use of Scaled Vector Diagrams to Determine a Resultant
Week 2
2.0 Torque and static equilibrium
2.1 Torque
2.2 By friction and sliding.
2.3 Force at an angle
2.4 Static equilibrium
2.5 Torque and angular acceleration
2.6 Relationship between torque, power and energy
2.7 Conversion to other units
2.8 Objects in static equilibrium
2.9 Net forces and torques
2.10 Sum of forces is zero
2.11 Net torque around ANY axis must be zero.
2.12 Properties of Solids: Density, elasticity, and strength of materials.
2.13 Elasticity and Young's Modulus
2
Week 3
3.1 Forces
3.1.1 Forces in Two Dimensions
3.1.2 Equilibrium and Statics
3.2 Forces in Two Dimensions
3.2.1Inclined Planes
3.2.2 Addition of Forces
Week 4
4.0 Levers and Moments
4.1 Levers
4.2 Three classes of Lever
4.3 Moments
4.3.1 Moment of a force about a pivot
4.3.2 The Law of the Lever
Week 5
5.0 Force resolution
5.1 The Parallelogram of forces
5.2 Forces in a plane
5.2.1 Force on a par&cle. Resultant of two forces
5.3 Derivation of the area formula
3
Week 6
6.1 Resultant force
6.2 Vectors - Fundamentals and Operations
6.3 Resultants
6.4 Vectors - Fundamentals and Operations
6.5 Relative Velocity and Riverboat Problems
6.6 Analysis of a Riverboat's Motion
6.7 Check Your Understanding
Week 7
7.1 Polygon of Forces
7.2 Principle of polygon of Forces
Week 8
8.1 Lami’s Theorem
8.2 Resolving a Force
8.3 Independence of Perpendicular Components of Motion
8.4 Check Your Understanding
Week 9
9.0 Equilibrium of bodies
9.1 Equilibrium
9.2 Balance of Moments
9.3 Clockwise Moment.
9.4 Anticlockwise Moment.
9.5 Varignon’s Principle of Moments (or Law of Moments)
9.6 Principle of Moments
4
Week 10
10.0 Friction
10.1 Fundamentals and Application of Friction
10.2 Fric&on
10.3 Fric&on and Surface Roughness
10.4 Sta&c Fric&on
10.5 Kine&c Fric&on
10.6 Fric&on Plot
10.7 Rolling Fric&on
10.8 Coulomb Friction
10.9 Other types of friction
10.9.1 Triboelectric effect
10.10 Reducing friction
10.11 Devices related to Friction
10.12 Lubricants
10.13 Energy of friction
10.14 Work of friction
Week 11
11.1 Fundamentals and application of friction (continues)
11.2 Coefficient of static friction
11.3 The normal force
11.4 Frequent mistake
11.5 Engineering coefficients of friction
11.6 Coefficients of Fric&on
11.7 Normal Force
11.8 The microscopic origin of friction
5
Week 12
12.1 Fundamentals and application of friction (continues)
12.2 Angle of repose
12.3 Applications of theory
12.4 Measurement
12.5 Exploitation by antlion larvae
Week 13
13.0 Frames, Struts and Ties
13.1 Frame and structures
13.2 Simple Frames
13.3 Struts and Ties
13.4 Example: Frame (non-truss, rigid body) Problems
Week 14
14.1 Structural Forces
14.2 More forces in action
14.3 Frames (non-truss, rigid body structures)
14.4 Additional Examples
14.5 Problems Assignment - Frames 1
14.6 Problems Assignment - Frames 2
14.7 Problems Assignment - Frames 3
Week 15
15.0 Forces - More Questions
15.1 Statically Determinate - Example 2
15.2 Statically Determinate - Example 3
15.3 Trusses - Example 1
15.4 Trusses - Example 2
15.5 Trusses -Example 3
15.6 Problem Assignment - Trusses 1
15.7 Problem Assignment - Trusses 2
15.8 Problem Assignment - Trusses 3
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MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 1 1.0 Scalar and Vector quantities
1.1Scalar and vector quantities
Two kinds of quantities are distinguished in mechanics. SCALAR quantities have only
magnitude. Examples are: density, energy; mass; speed; temperature; time, and volume
e.t.c.
VECTOR quantities have magnitude and direction. Examples are acceleration;
displacement;
force; moment; momentum; and velocity e.t.c. Vectors are commonly represented by
arrows in
drawings, with letters or numbers by the arrows indicating the quantity of them.
F

→
OR
15kN
  
→
7
•
•
A vector is represented by a directed line as shown below,
 A
O →
V
•
It may be noted that the lenght OA represents the magnitude of the vector OA i.e. from O ( i.e. starting point)
to A (i.e. end point). It is also known as vector V.
•
(ii) Unit vector.
•
A vector, whose magnitude is unity, is known as a unit vector.
•
→ A
O
•
(iii) Equal vectors
•
The vectors , which are parallel to each other , have same direction, (i.e. same sense ). and equal magnitudes,
are known as equal vectors.
1
•
O 7N

→ A
O 7N

→ B
•
(iv) Like vector
•
The vectors, which are parallel to each other and have same sense but unequal magnitude, are known as like
vectors.
•
O 7N

→ A
10N
O  → B
•
(v) Addition of vectors
•
Consider two vectors PQ=4N and RS=5N, which are required to be added as shown below:
•
 
→  → =  →
•
 
→ ←   =
4N
4N
5N
5N
9N
←
1N
•
8
1.1.1 Vectors and Direction
A study of motion will involve the introduction of a variety of quantities which are used
to describe the physical world. Examples of such quantities include distance,
displacement, speed, velocity, acceleration, force, mass, momentum, energy, work,
power, etc. All these quantities can by divided into two categories - vectors and scalars. A
vector quantity is a quantity which is fully described by both magnitude and direction. On
the other hand, a scalar quantity is a quantity which is fully described by its magnitude.
The emphasis of this unit is to understand some fundamentals about vectors and to apply
the fundamentals in order to understand motion and forces which occur in two
dimensions.
Examples of vector quantities which have been
previously discussed include displacement, velocity,
acceleration, and force. Each of these quantities are
unique in that a full description of the quantity
demands that both a magnitude and a direction are
Fig. 1
listed. For example, suppose your teacher tells you
"A bag of gold is located outside the classroom. To find it, displace yourself 20 meters."
This statement may provide yourself enough information to pique your interest; yet, there
is not enough information included in the statement to find the bag of gold. The
displacement required to find the bag of gold has not been fully described. On the other
hand, suppose your teacher tells you "A bag of gold is located outside the classroom. To
find it, displace yourself from the center of the classroom door 20 meters in a direction 30
degrees to the west of north." This statement now provides a complete description of the
displacement vector - it lists both magnitude (20 meters) and direction (30 degrees to the
west of north) relative to a reference or starting position (the center of the classroom
door). Vector quantities are not fully described unless both magnitude and direction are
listed.
Vector quantities are often represented by scaled
vector diagrams. Vector diagrams depict a vector by
use of an arrow drawn to scale in a specific direction.
Vector diagrams were introduced and used in earlier
units to depict the forces acting upon an object. Such
diagrams are commonly called as free-body diagrams.
An example of a scaled vector diagram is shown in the
diagram at the right. The vector diagram depicts a
displacement vector. Observe that there are several
characteristics of this diagram which make it an
appropriately drawn vector diagram.
•
•
•
a scale is clearly listed
2
a vector arrow (with arrowhead) is drawn in a specified direction. TheFig.
vector
arrow has a head and a tail.
the magnitude and direction of the vector is clearly labeled. In this case, the
diagram shows the magnitude is 20 m and the direction is (30 degrees West of
North).
9
Fig. 3
1.1.2 Conventions for Describing Directions of Vectors
Vectors can be directed due East, due West, due South, and due North. But
some vectors are directed northeast (at a 45 degree angle); and some vectors
are even directed northeast, yet more north than east. Thus, there is a clear
need for some form of a convention for identifying the direction of a vector
which is not due East, due West, due South, or due North. There are a variety
of conventions for describing the direction of any vector. The two conventions which will
be discussed and used in this unit are described below:
1. The direction of a vector is often expressed as an angle of rotation of the vector
about its "tail" from either east, west, north, or south. For example, a vector can be
said to have a direction of 40 degrees North of West (meaning a vector pointing
West has been rotated 40 degrees towards the northerly direction) of 65 degrees
East of South (meaning a vector pointing South has been rotated 65 degrees
towards the easterly direction).
2. The direction of a vector is often expressed as an counterclockwise angle of
rotation of the vector about its "tail" from due East. Using this convention, a
vector with a direction of 30 degrees is a vector which has been rotated 30 degrees
in a counterclockwise direction relative to due east. A vector with a direction of
160 degrees is a vector which has been rotated 160 degrees in a counterclockwise
direction relative to due east. A vector with a direction of 270 degrees is a vector
which has been rotated 270 degrees in a counterclockwise direction relative to due
east. This is one of the most common conventions for the direction of a vector and
will be utilized throughout this unit.
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Fig. 4
Two illustrations of the second convention (discussed above) for identifying the direction
of a vector are shown below.
(a)
Fig. 5
(b)
Observe in the first example that the vector is said to have a direction of 40 degrees. You
can think of this direction as follows: suppose a vector pointing East had its tail pinned
down and then the vector was rotated an angle of 40 degrees in the clockwise direction.
Observe in the second example that the vector is said to have a direction of 240 degrees.
This means that the tail of the vector was pinned down and the vector was rotated an
angle of 240 degrees in the counterclockwise direction beginning from due east. A
rotation of 240 degrees is equivalent to rotating the vector through two quadrants (180
degrees) and then an additional 60 degrees into the third quadrant.
1.1.3Representing the Magnitude of a Vector
The magnitude of a vector in a scaled vector diagram is
depicted by the length of the arrow. The arrow is drawn a
precise length in accordance with a chosen scale. For
example, the diagram at the right shows a vector with a
magnitude of 20 miles. Since the scale used for constructing
the diagram is 1 cm = 5 miles, the vector arrow is drawn with
a length of 4 cm. That is, 4 cm x (5 miles/1 cm) = 20 miles.
Fig. 6
Using the same scale (1 cm = 5 miles), a displacement vector which is 15 miles will be
represented by a vector arrow which is 3 cm in length. Similarly, a 25 mile displacement
vector is represented by a 5-cm long vector arrow. And finally, an 18 mile displacement
vector is represented by a 3.6-cm long arrow. See the examples shown below.
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(a)
(c)
(b)
Fig. 7
In conclusion, vectors can be represented by use of a scaled vector diagram. On such a
diagram, a vector arrow is drawn to represent the vector. The arrow has an obvious tail
and arrowhead. The magnitude of a vector is represented by the length of the arrow. A
scale is indicated (such as, 1 cm = 5 miles) and the arrow is drawn the proper length
according to the chosen scale. The arrow points in the precise direction. Directions are
described by the use of some convention. The most common convention is that the
direction of a vector is the counterclockwise angle of rotation which that vector makes
with respect to due East.
In the remainder of this lesson, in the entire unit, and in future units, scaled vector
diagrams and the above convention for the direction of a vector will be frequently used to
describe motion and solve problems concerning motion. For this reason, it is critical that
you have a comfortable understanding of the means of representing and describing vector
quantities.
1.1.4 Vector Resolution
As mentioned earlier in this lesson, any vector directed at an angle to the horizontal (or
the vertical) can be thought of as having two parts (or components). That is, any vector
directed in two dimensions can be thought of as having two components. For example, if
a chain pulls upward at an angle on the collar of a dog, then there is a tension force
directed in two dimensions. This tension force has two components: an upward
component and a rightward component. As another example, consider an airplane which
is displaced northwest from O'Hare International Airport (in Chicago) to a destination in
Canada. The displacement vector of the plane is in two dimensions (northwest). Thus, this
displacement vector has two components: a northward component and a westward
component.
In this unit, we learn two basic methods for determining the magnitudes of the
components of a vector directed in two dimensions. The process of determining the
12
magnitude of a vector is known as vector resolution. The two methods of vector
resolution which we will examine are
•
•
The parallelogram method
The trigonometric method
1.1.5 Parallelogram Method of Vector Resolution
The parallelogram method of vector resolution involves using an accurately drawn, scaled
vector diagram to determine the components of the vector. Briefly put, the method
involves drawing the vector to scale in the indicated direction, sketching a parallelogram
around the vector such that the vector is the diagonal of the parallelogram, and
determining the magnitude of the components (the sides of the parallelogram) using the
scale. If one desires to determine the components as directed along the traditional x- and
y-coordinate axes, then the parallelogram is a rectangle with sides which stretch vertically
and horizontally. A step-by-step procedure for using the parallelogram method of vector
resolution is:
1. Select a scale and accurately draw the vector to scale in the indicated direction.
2. Sketch a parallelogram around the vector: beginning at the tail of the vector,
sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the
head of the vector; the sketched lines will meet to form a rectangle (a special case
of a parallelogram).
3. Draw the components of the vector. The components are the sides of the
parallelogram. The tail of the components starts at the tail of the vector and
stretches along the axes to the nearest corner of the parallelogram. Be sure to place
arrowheads on these components to indicate their direction (up, down, left, right).
4. Meaningfully label the components of the vectors with symbols to indicate which
component is being represented by which side. A northward force component
might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
5. Measure the length of the sides of the parallelogram and use the scale to determine
the magnitude of the components in real units. Label the magnitude on the
diagram.
The step-by-step procedure above is illustrated in the diagram below to show how a
velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the
horizontal may be resolved into two components. The diagram shows that the vector is
first drawn to scale in the indicated direction; a parallelogram is sketched about the
vector; the components are labeled on the diagram; and the result of measuring the length
of the vector components and converting to m/s using the scale. (NOTE: because different
computer monitors have different resolutions, the actual length of the vector on your
monitor may not be 5 cm.)
13
(a)
(b)
(c)
Fig. 8
1.1.6 Trigonometric Method of Vector Resolution
The trigonometric method of vector resolution involves using trigonometric functions to
determine the components of the vector. The use of trigonometric functions to determine
the direction of a vector was described. Now in this part of lesson 1, trigonometric
functions will be used to determine the components of a single vector. Recall from the
earlier discussion that trigonometric functions relate the ratio of the lengths of the sides of
a right triangle to the measure of an acute angle within the right triangle. As such,
trigonometric functions can be used to determine the length of the sides of a right triangle
if an angle measure and the length of one side are known.
The method of employing trigonometric functions to determine the components of a
vector are as follows:
1. Construct a rough sketch (no scale needed) of the vector in the indicated direction.
Label its magnitude and the angle which it makes with the horizontal.
2. Draw a rectangle about the vector such that the vector is the diagonal of the
rectangle. Beginning at the tail of the vector, sketch vertical and horizontal lines.
Then sketch horizontal and vertical lines at the head of the vector. The sketched
lines will meet to form a rectangle.
3. Draw the components of the vector. The components are the sides of the rectangle.
The tail of each component begins at the tail of the vector and stretches along the
axes to the nearest corner of the rectangle. Be sure to place arrowheads on these
components to indicate their direction (up, down, left, right).
4. Meaningfully label the components of the vectors with symbols to indicate which
component is being represented by which side. A northward force component
might be labeled Fnorth. A rightward force velocity component might be labeled vx;
etc.
5. To determine the length of the side opposite the indicated angle, use the sine
function. Substitute the magnitude of the vector for the length of the hypotenuse.
Use some algebra to solve the equation for the length of the side opposite the
indicated angle.
14
6. Repeat the above step using the cosine function to determine the length of the side
adjacent to the indicated angle.
Example 1
The above method is illustrated below for determining the components of the force acting
upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle
of 40 degrees, the components of this force can be determined using trigonometric
functions.
Fig. 9
In conclusion, a vector directed in two dimensions has two components - that is, an
influence in two separate directions. The amount of influence in a given direction can be
determined using methods of vector resolution. Two methods of vector resolution have
been described here - a graphical method (parallelogram method) and a trigonometric
method.
1.1.7 Vector Components
A vector is a quantity which has both magnitude and direction. Displacement, velocity,
acceleration, and force are the vector quantities which we have discussed thus far in the
Physics Classroom Tutorial. In the first couple of units, all vectors which we discussed
were simply directed up, down, left or right. When there was a free-body diagram
depicting the forces acting upon an object, each individual force was directed in one
dimension - either up or down or left or right. When an object had an acceleration and we
described its direction, it was directed in one dimension - either up or down or left or
right. Now in this unit, we begin to see examples of vectors which are directed in two
15
dimensions - upward and rightward, northward and westward, eastward and southward,
etc.
Fig. 10
In situations in which vectors are directed at angles to the customary coordinate
axes, a useful mathematical trick will be employed to transform the vector into two
parts with each part being directed along the coordinate axes. For example, a
vector which is directed northwest can be thought of as having two parts - a
northward part and a westward part. A vector which is directed upward and
rightward can be thought of as having two parts - an upward part and a rightward
part.
Fig. 11
Any vector directed in two dimensions can be
thought of as having an influence in two different
directions. That is, it can be thought of as having
two parts. Each part of a two-dimensional vector is
known as a component. The components of a
vector depict the influence of that vector
Fig. 12
in a given direction. The combined influence of the two components is equivalent to the
influence of the single two-dimensional vector. The single two-dimensional vector could
be replaced by the two components.
16
If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then
the tension force in the chain has two components - an upward component and a
rightward component. To Fido, the influence of the chain on his body is equivalent to the
influence of two chains on his body - one pulling upward and the other pulling rightward.
If the single chain were replaced by two chains. with each chain having the magnitude
and direction of the components, then Fido would not know the difference. This is not
because Fido is dumb (a quick glance at his picture reveals that he is certainly not that),
but rather because the combined influence of the two components is equivalent to the
influence of the single two-dimensional vector.
(a)
(b)
Fig. 13
Consider a picture which is hung to a wall by
means of two wires which are stretched vertically
and horizontally. Each wire exerts a tension force
upon the picture to support its weight. Since each
wire is stretched in two dimensions (both vertically
and horizontally), the tension force of each wire has
two components - a vertical component and a
horizontal component. Focusing on the wire on the
left, we could say that the wire has a leftward and
an upward component. This is to say that the wire
on the left could be replaced by two wires, one
pulling leftward and the other pulling upward. If the
single wire were replaced by two wires (each one
having the magnitude and direction of the components), then there
17
Fig. 14
would be no affect upon the stability of the picture. The combined influence of the two
components is equivalent to the influence of the single two-dimensional vector.
Fig. 15
Consider an airplane which is flying from Chicago's O'Hare International Airport to a
destination in Canada. Suppose that the plane is flying in such a manner that its resulting
displacement vector is northwest. If this is the case, then the displacement of the plane has
two components - a component in the northward direction and a component in the
westward direction. This is to say that the plane would have the same displacement if it
were to take the trip into Canada in two segments - one directed due North and the other
directed due West. If the single displacement vector were replaced by these two
individual displacement vectors, then the passengers in the plane would end up in the
same final position. The combined influence of the two components is equivalent to the
influence of the single two-dimensional displacement.
18
Fig. 16
Any vector directed in two dimensions can be thought of as having two different
components. The component of a single vector describes the influence of that
vector in a given direction. In the next part of this lesson, we will investigate two
methods for determining the magnitude of the components. That is, we will
investigate how much influence a vector exerts in a given direction.
1.1.8 Vector Addition
A variety of mathematical operations can be performed with and upon vectors. One such
operation is the addition of vectors. Two vectors can be added together to determine the
result (or resultant). This process of adding two or more vectors has already been
discussed in an earlier unit. Recall in our discussion of Newton's laws of motion, that the
net force experienced by an object was determined by computing the vector sum of all the
individual forces acting upon that object. That is the net force was the result (or resultant)
of adding up all the force vectors. During that unit, the rules for summing vectors (such as
force vectors) were kept relatively simple.
19
Example 2
Observe the following summations of two force vectors:
Fig. 17
These rules for summing vectors were applied to free-body diagrams in order to
determine the net force (i.e., the vector sum of all the individual forces). Sample
applications are shown in the diagram below.
(a)
(c)
(b)
Fig. 18
In this unit, the task of summing vectors will be extended to more complicated cases in
which the vectors are directed in directions other than purely vertical and horizontal
directions. For example, a vector directed up and to the right will be added to a vector
directed up and to the left.
20
The vector sum will be determined for the more complicated cases shown in the diagrams
below.
Fig. 19
There are a variety of methods for determining the magnitude and direction of the result
of adding two or more vectors. The two methods which will be discussed in this lesson
and used throughout the entire unit are:
•
•
the Pythagorean theorem and trigonometric methods
the head-to-tail method using a scaled vector diagram
1.2 The Pythagorean Theorem
The Pythagorean theorem is a useful method for determining the result of adding two (and
only two) vectors which make a right angle to each other. The method is not applicable
for adding more than two vectors or for adding vectors which are not at 90-degrees to
each other. The Pythagorean theorem is a mathematical equation which relates the length
of the sides of a right triangle to the length of the hypotenuse of a right triangle.
Fig. 20
21
To see how the method works, consider the following problem:
Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine
Eric's resulting displacement.
Example 3
This problem asks to determine the result of adding two displacement vectors which are at
right angles to each other. The result (or resultant) of walking 11 km north and 11 km east
is a vector directed northeast as shown in the diagram to the right. Since the northward
displacement and the eastward displacement are at right angles to each other, the
Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the
right triangle).
Fig. 21
The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6
km. Later, the method of determining the direction of the vector will be discussed.
Exercise 1
Let's test your understanding with the following two practice problems. In each
22
case, use the Pythagorean theorem to determine the magnitude of the vector sum.
Ans. (A) R = 11.2 km
(B) R = 50 km
Fig. 22
1.2.1 Using Trigonometry to Determine a Vector's Direction
The direction of a resultant vector can often be determined by use of trigonometric
functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA
from their course in trigonometry. SOH CAH TOA is a mnemonic which helps one
remember the meaning of the three common trigonometric functions - sine, cosine, and
tangent functions. These three functions relate an acute angle in a right triangle to the
ratio of the lengths of two of the sides of the right triangle. The sine function relates the
measure of an acute angle to the ratio of the length of the side opposite the angle to the
length of the hypotenuse. The cosine function relates the measure of an acute angle to the
ratio of the length of the side adjacent the angle to the length of the hypotenuse. The
tangent function relates the measure of an angle to the ratio of the length of the side
opposite the angle to the length of the side adjacent to the angle.
23
The three equations below summarize these three functions in equation form.
Fig. 23
These three trigonometric functions can be applied to the hiker problem in order to
determine the direction of the hiker's overall displacement. The process begins by the
selection of one of the two angles (other than the right angle) of the triangle. Once the
angle is selected, any of the three functions can be used to find the measure of the angle.
Write the function and proceed with the proper algebraic steps to solve for the measure of
the angle. The work is shown below.
Fig. 24
Once the measure of the angle is determined, the direction of the vector can be found. In
this case the vector makes an angle of 45 degrees with due East. Thus, the direction of
this vector is written as 45 degrees. (Recall from earlier in this lesson that the direction of
a vector is the counterclockwise angle of rotation which the vector makes with due East.)
The measure of an angle as determined through use of SOH CAH TOA is not always the
direction of the vector. The following vector addition diagram is an example of such a
situation. Observe that the angle within the triangle is determined to be 26.6 degrees using
24
SOH CAH TOA. This angle is the southward angle of rotation which the vector R makes
with respect to West. Yet the direction of the vector as expressed with the CCW
(counterclockwise from East) convention is 206.6 degrees.
Fig. 25
25
In the above problems, the magnitude and direction of the sum of two vectors is
determined using the Pythagorean theorem and trigonometric methods (SOH CAH TOA).
The procedure is restricted to the addition of two vectors which make right angles to each
other. When the two vectors which are to be added do not make right angles to one
another, or when there are more than two vectors to add together, we will employ a
method known as the head-to-tail vector addition method. This method is described
below.
1.2.3 Use of Scaled Vector Diagrams to Determine a Resultant
The magnitude and direction of the sum of two or more vectors can also be determined by
use of an accurately drawn scaled vector diagram. Using a scaled diagram, the head-totail method is employed to determine the vector sum or resultant. A common Physics lab
involves a vector walk. Either using centimeter-sized displacements upon a map or metersized displacements in a large open area, a student makes several consecutive
displacements beginning from a designated starting position. Suppose that you were given
a map of your local area and a set of 18 directions to follow. Starting at home base, these
18 displacement vectors could be added together in consecutive fashion to determine the
result of adding the set of 18 directions. Perhaps the first vector is measured 5 cm, East.
Where this measurement ended, the next measurement would begin. The process would
be repeated for all 18 directions. Each time one measurement ended, the next
measurement would begin. In essence, you would be using the head-to-tail method of
vector addition.
Fig. 27
The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning
at a designated starting position. Where the head of this first vector ends, the tail of the
second vector begins (thus, head-to-tail method). The process is repeated for all vectors
which are being added. Once all the vectors have been added head-to-tail, the resultant is
then drawn from the tail of the first vector to the head of the last vector; i.e., from start to
finish. Once the resultant is drawn, its length can be measured and converted to real units
using the given scale. The direction of the resultant can be determined by using a
protractor and measuring its counterclockwise angle of rotation from due East.
26
Exercise 2
A step-by-step method for applying the head-to-tail method to determine the sum of two
or more vectors is given below.
1. Choose a scale and indicate it on a sheet of paper. The best choice of scale is one
which will result in a diagram which is as large as possible, yet fits on the sheet of
paper.
2. Pick a starting location and draw the first vector to scale in the indicated direction.
Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm
= 20 m).
3. Starting from where the head of the first vector ends, draw the second vector to
scale in the indicated direction. Label the magnitude and direction of this vector
on the diagram.
4. Repeat steps 2 and 3 for all vectors which are to be added
5. Draw the resultant from the tail of the first vector to the head of the last vector.
Label this vector as Resultant or simply R.
6. Using a ruler, measure the length of the resultant and determine its magnitude by
converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
7. Measure the direction of the resultant using the counterclockwise convention
discussed earlier in this lesson.
An example of the use of the head-to-tail method is illustrated below. The problem
involves the addition of three vectors:
20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.
SCALE: 1 cm = 5 m
Fig. 28
27
Exercise 3
The head-to-tail method is employed as described above and the resultant is determined
(drawn in red). Its magnitude and direction is labeled on the diagram.
SCALE: 1 cm = 5 m
Fig. 29
Interestingly enough, the order in which three vectors are added has no affect upon either
the magnitude nor the direction of the resultant. The resultant will still have the same
magnitude and direction.
28
For example, consider the addition of the same three vectors in a different order.
15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.
SCALE: 1 cm = 5 m
Fig. 30
When added together in this different order, these same three vectors still produce a
resultant with the same magnitude and direction as before (22 m, 310 degrees). The order
in which vectors are added using the head-to-tail method is insignificant.
SCALE: 1 cm = 5 m
Fig. 31
29
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 2 2.0
Torque and static equilibrium
2.1Torque
What is torque?
Torque is a measure of how much a force acting
on an object causes that object to rotate. The object
rotates about an axis, which we will call the pivot
point, and will label 'O'. We will call the force 'F'. The
distance from the pivot point to the point where the
force acts is called the moment arm, and is denoted by
'r'. Note that this distance, 'r', is also a vector, and
points from the axis of rotation to the point where the
force acts. (Refer to Figure 1 for a pictoral
representation of these definitions.)
Figure 1 Definitions
Torque is defined as
= r x F = r F sin(θ).
In other words, torque is the cross product between the distance vector (the distance
from the pivot point to the point where force is applied) and the force vector, 'a' being the
angle between r and F.
Using the right hand rule, we can find the direction of the torque vector. If we put our
fingers in the direction of r, and curl them to the direction of F, then the thumb points in
the direction of the torque vector.
Imagine pushing a door to open it. The force of your push (F) causes the door to rotate
about its hinges (the pivot point, O). How hard you need to push depends on the distance
you are from the hinges (r) (and several other things, but let's ignore them now). The
closer you are to the hinges (i.e. the smaller r is), the harder it is to push. This is what
happens when you try to push open a door on the wrong side. The torque you created on
the door is smaller than it would have been had you pushed the correct side (away from
its hinges).
Note that the force applied, F, and the moment arm, r, are independent of the object.
Furthermore, a force applied at the pivot point will cause no torque since the moment arm
would be zero (r = 0).
30
Another way of expressing the above
equation is that torque is the product of the
magnitude of the force and the
perpendicular distance from the force to
the axis of rotation (i.e. the pivot point).
Let the force acting on an object be broken
up into its tangential (Ftan) and radial
(Frad) components (see Figure 2). (Note
Figure 2 Tangential and radial components
that the tangential component is
of force F
perpendicular to the moment arm, while
the radial component is parallel to the
moment arm.) The radial component of the
force has no contribution to the torque
because it passes through the pivot point.
So, it is only the tangential component of
the force which affects torque (since it is
perpendicular to the line between the point
of action of the force and the pivot point).
There may be more than one force acting on an object, and each of these forces may act
on different point on the object. Then, each force will cause a torque. The net torque is
the sum of the individual torques.
Rotational Equilibrium is analogous to translational equilibrium, where the sum of the
forces are equal to zero. In rotational equilibrium, the sum of the torques is equal to
zero. In other words, there is no net torque on the object.
Note that the SI units of torque is a Newton-metre, which is also a way of expressing
a Joule (the unit for energy). However, torque is not energy. So, to avoid confusion, we
will use the units N.m, and not J. The distinction arises because energy is a scalar
quanitity, whereas torque is a vector.
2.2 By friction and sliding.
For this reason, when measuring the effective power produced by a rotating engine and
the energy spent in the system to generate a movement, you will often need to take into
account the angle of rotation, and then, adding the radian in the unit system is necessary
as well as making a difference between the measurement of arcs (in radian meter) and the
measurement of straight segment distances (in meters), as a way to effectively compute
the efficiency of the mobile system and the capacity of a motor engine to convert between
31
rotational power (in radian watt) and linear power (in watts): in a friction-free ideal
system, the two measurements would have equal value, but this does not happen in
practice, each conversion losing energy in friction (it's easier to limit all losses of energy
caused by sliding, by introducing mechanical constraints of forms on the surfaces of
contacts).
Depending on works, the extended units including radians as a fundamental dimension
may or may not be used.
2.3 Force at an angle
If a force of magnitude F is at an angle θ from the displacement arm of length r (and
within the plane perpendicular to the rotation axis), then from the definition of cross
product, the magnitude of the torque arising is:
τ = r F sinθ
2.4 Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but
also the sum of the torques (moments) about any point. For a two-dimensional situation
with horizontal and vertical forces, the sum of the forces requirement is two equations:
ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically
determinate equilibrium problems in two-dimensions, we use three equations.
2.5 Torque and angular acceleration
In this section, we will develop the relationship between torque and angular acceleration.
You will need to have a basic understanding of moments of inertia for this section.
Imagine a force F acting on some object at a
distance r from its axis of rotation. We can break
up the force into tangential (Ftan), radial (Frad)
(see Figure 1). (This is assuming a twodimensional scenario. For three dimensions -- a
more realistic, but also more complicated situation
-- we have three components of force: the
tangential component Ftan, the radial component
Frad and the z-component Fz. All components of
force are mutually perpendicular, or normal.)
Figure 3 Radial and Tangential
Components of Force, two
From Newton's Second Law,
dimensions
32
Ftan = m atan
However, we know that angular acceleration, ,
and the tangential acceleration atan are related by:
atan = r α
Then,
Ftan = m r α
If we multiply both sides by r (the moment arm),
the equation becomes
Ftan r = m r α
Note that the radial component of the force goes
through the axis of rotation, and so has no
contribution to torque. The left hand side of the
equation is torque. For a whole object, there may
be many torques. So the sum of the torques is
equal to the moment of inertia (of a particle mass,
which is the assumption in this derivation),
I=mr
Figure 4 Radial, Tangential and zComponents of Force, three
dimensions
multiplied by the angular acceleration, α.
If we make an analogy between translational and rotational motion, then this relation
between torque and angular acceleration is analogous to the Newton's Second Law.
Namely, taking torque to be analogous to force, moment of inertia analogous to mass, and
angular acceleration analogous to acceleration, then we have an equation very much like
the Second Law.
Torque is part of the basic specification of an engine: the power output of an engine is
expressed as its torque multiplied by its rotational speed of the axis. Internal-combustion
engines produce useful torque only over a limited range of rotational speeds (typically
from around 1,000–6,000 rpm for a small car). The varying torque output over that range
can be measured with a dynamometer, and shown as a torque curve. The peak of that
torque curve usually occurs somewhat below the overall power peak. The torque peak
cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in
automotive engineering, concerned as it is with transmitting power from the engine
through the drive train to the wheels. Power is typically a function of torque and engine
33
speed. The gearing of the drive train must be chosen appropriately to make the most of
the motor's torque characteristics.
Steam engines and electric motors tend to produce maximum torque close to zero rpm,
with the torque diminishing as rotational speed rises (due to increasing friction and other
constraints). Therefore, these types of engines usually have quite different types of drive
trains from internal combustion engines.
Torque is also the easiest way to explain mechanical advantage in just about every simple
machine.
2.6 Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if
torque is allowed to act through a rotational distance, it is doing work. Power is the work
per unit time. However, time and rotational distance are related by the angular speed
where each revolution results in the circumference of the circle being travelled by the
force that is generating the torque. The power injected by the applied torque may be
calculated as:
On the right hand side, this is a scalar product of two vectors, giving a scalar on the left
hand side of the equation. Mathematically, the equation may be rearranged to compute
torque for a given power output. Note that the power injected by the torque depends only
on the instantaneous angular speed - not on whether the angular speed increases,
decreases, or remains constant while the torque is being applied (this is equivalent to the
linear case where the power injected by a force depends only on the instantaneous speed not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a
large electrical power grid. In such an arrangement, the generator’s angular speed is fixed
by the grid's frequency, and the power output of the plant is determined by the torque
applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton meters
and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton meter is dimensionally equivalent to the joule, which is the unit of
energy. However, in the case of torque, the unit is assigned to a vector, whereas for
energy, it is assigned to a scalar.
2.7 Conversion to other units
For different units of power, torque, or angular speed, a conversion factor must be
inserted into the equation. Also, if rotational speed (revolutions per time) is used in place
of angular speed (radians per time), a conversion factor of 2π must be added because there
are 2π radians in a revolution:
34
,
where rotational speed is in revolutions per unit time.
Useful formula in SI units:
Where 60,000 comes from 60 seconds per minute times 1000 watts per kilowatt.
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical)
for power, foot-pounds (lbf· ft) for torque and rpm (revolutions per minute) for angular
speed. This results in the formula changing to:
The constant below in, ft·lbf./min, changes with the definition of the horsepower; for
example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion
factor.
2.8 Objects in static equilibrium
An object is described as being in static equilibrium when it neither translates, nor rotates.
This is the aim in structures. Such objects may have substantial applied forces and
torques, but the idea is that there will be no NET force, and there will be no NET torque.
Thus, in calculating the behavior, we need to use all the previous information about free
body diagrams to determine all the forces. This type of calculation is a classic in building
design or in the design of any type of object that SHOULD NOT MOVE.
2.9 Net forces and torques
Equilibrium calculations come in two flavors:
1. Take a collection of known forces and torques, and figure out if the object is going
to move.
2. Take an object that you insist MUST NOT MOVE, and calculate the forces and
torques needed to be sure that it does not.
There are three things that we require of any equilibrium static object:
1. The net force MUST BE ZERO. Otherwise, the object will accelerate.
35
•
on the force required from a door hinge to hold the door.
2. The net torque MUST BE ZERO. Otherwise, the object will angularly accelerate.
•
•
•
on the force a wall places on a ladder as a person climbs.
on the friction force required on a ladder as a person climbs.
on when a ladder is most likely to slip as a person climbs.
3. The object must be strong enough to withstand all the forces and torques.
Let's do an example to show how these ideas are actually used:
Example: A bridge of 10,000kg mass and 30 meter length has a toll booth of 5000kg mass
located 10 meters from one end. What forces must be applied on the ends of the bridge to
support the structure?
As usual, let's start with a picture of the forces:
Fig. 5
In this picture, we show the forces on the two bridge ends, AND we show the gravity
forces of the weight of the bridge and the weight of the toll booth. Notice that gravity
forces always act as though they were located at a central point, referred to as the center
of gravity. The center of gravity is found by calculating (it is essentially the same idea as)
the position of the center of mass. Recall that for an object made of many masses, the
center of mass is given by:
For most of the object that we will consider, for example, the bridge, the center of mass is
located at the center of the object. Therefore, gravity acts on the bridge through its center.
Similarly, gravity acts on the toll booth through its center.
OK, so in working these problems, your first effort should be to require:
2.10 Sum of forces is zero
36
If the object is not going to accelerate, the net force is zero.
For this case, we add up the forces and insist that they add to zero net force:
Therefore, we find the fairly obvious result that the total upward force must cancel
gravity:
Therefore, we have an equation that relates the two support forces, but it is not enough to
determine the forces individually. We need another equation. You usually get it by
looking at the torque:
37
2.11 Net torque around ANY axis must be zero.
If the object is not going to angularly accelerate, the net torque is zero.
We can state that the net torque is zero with respect to ANY axis of rotation. It is usually
helpful to choose the axis to simplify the problem. In particular, if you choose an axis so
that one or more forces have no lever arm, then they drop out of the problem.
For example, we could take the axis to be at the place where F1 is applied. Then, the
torques are:
Therefore, we can solve immediately for the second support force:
Therefore, the value of F1 is:
2.12 Properties of Solids: Density, elasticity, and strength of materials.
Balance of force and torque is not enough in practical situations. You need materials that
can tolerate the forces.
Most static structures are constructed from solid materials. Therefore, we are headed
toward the study of the properties of solids. These are the basic pieces of information that
you must know about solid materials if you try to build static structures.
Density
One of the most basic properties of solids is their density or mass per unit volume:
For most materials, while the mass depends upon how large the object is, the density does
not. Therefore, if you know the material you intend to use, its density (say around 2800
kg/m^3 for concrete), and the volume of material you will use, then you can determine the
total mass.
The mass is then useful for determining the forces that gravity will place on the structure.
As you can tell, one of the major design goals in a building or bridge is the use of strong,
but low density materials so that the forces from gravity are smaller.
38
2.13 Elasticity and Young's Modulus
Next, let's look at how materials respond to applied forces. For example, imagine that a
bar of steel is pulled on or pushed on by some force. The picture looks like this:
Fig. 6
If the force is a pull, we say that the bar is in tension. If it is a push, we call it
compression. Solid objects respond to forces by stretching (for tension) or compressing
(for compression). In other words, the length of the bar changes by some amount. Robert
Hooke (a contemporary of Newton's) pointed out that for smallish forces, the change in
length is in direct proportion to the force:
This relation is called Hooke's Law, where k is the 'spring constant'. We have seen exactly
this law previously when we talked about forces and energy in springs. Hooke's Law says
that all materials act like springs for small forces.
While this result is useful, it is found that the spring constant depends upon the length of
the bar and the cross sectional area of the bar. By experiment, we observe that the change
in length is proportional to the total length, the force, inversely proportional to the area,
and depends THEN upon a factor that depends ONLY on the material. This factor is
called Young's modulus, E:
so that
For any bar of material, if you know the length, area, and Young's modulus of the
material, you can design how flexible it will be. Giancoli also discusses the cases of shear
and bulk modulus.
While it is satisfying to be able to predict spring constants for bars, etc., again, we have an
opportunity to define properties that are independent of the shape if we look at force per
unit area, or stress. Then, in terms of stress, the change in length divided by the total
length, is just the stress divided by Young's modulus. Emphasis on the stress, rather than
39
the total force, allows us to discuss the breaking strength of objects, without requiring that
we worry about shape.
Breaking strength.
If you extend your experiments to larger forces, a plot of the length change versus the
force looks like this:
Fig. 7
Eventually, ANY MATERIAL CAN BE BROKEN. If the forces lead to impossible
stresses in the material, it will fail. We call the maximum stress that a material can
tolerate, the ultimate strength or breaking strength. Giancoli offers a list of typical
strengths for compressive and tensile stress for several materials.
•
on the stress on a hanging steel cable at different cable diameters.
Important points
•
•
For an object to remain static, the net force and net torque on the object must be
zero. Otherwise, the object will translationally or rotationally accelerate. Very bad
for buildings.
The object must also have the mechanical strength to withstand the forces. We
characterize mechanical strength via Young's modulus, the shear and bulk
modulus, and by the ultimate breaking strength.
40
Example problem on torque: the swinging door
Question
In a hurry to catch a cab, you rush through a frictionless swinging door and onto the
sidewalk. The force you exerted on the door was 50N, applied perpendicular to the plane
of the door. The door is 1.0m wide. Assuming that you pushed the door at its edge, what
was the torque on the swinging door (taking the hinge as the pivot point)?
Hints
1.
2.
3.
4.
Where is the pivot point?
What was the force applied?
How far from the pivot point was the force applied?
What was the angle between the door and the direction of force?
Solution
The pivot point is at the hinges of the
door, opposite to where you were
pushing the door. The force you used
was 50N, at a distance 1.0m from the
pivot point. You hit the door
perpendicular to its plane, so the angle
between the door and the direction of
force was 90 degrees. Since
Figure 8 Diagram of Example Problem 1
= r x F = r F sin(θ)
then the torque on the door was:
= (1.0m) (50N) sin(90)
= 50 N m
Note that this is only the magnitude of the torque; to complete the answer, we need to find
the direction of torque. Using the right hand rule, we see that the direction of torque is out
of the screen.
41
Example on the Right Hand Rule
Question
Which direction is the torque in this diagram, with respect to the pivot point labelled O?
(a) Diagram of the problem
(b) Diagram of the problem, force has been
translated in order to simplify the use of the
right hand rule
Fig. 9
Solution
Here we are assuming that the force, F, and moment arm, r vectors were originally placed
'head-to-head' (that is, F was pointing to the arrowhead of r, not at its pivot point). This is
shown in Figure 1. However, by translating the force vector to its position in Figure 2, the
use of the Right Hand Rule becomes more obvious.
Without this clarification, it is possible to interpret Figure 2 as having the force vector
going through the pivot point, in which case there would be no torque (see What is
Torque?). This is due to the definition of the moment arm, which is the distance between
the pivot point and the point where the force acts. If the force acts right on the pivot point,
then r = 0, so there would be no torque. (Having a moment arm of zero would be like
trying to open a door by pushing on its hinges; nothing happens because no torque has
resulted from the applied force.)
Recall the use of the Right Hand Rule in torque
calculations. Fingers are to point in the direction of the
first vector, and are curled towards the second vector. In
this case, torque is the cross product of the moment arm
and torque. So the fingers would point to in the same
direction as the moment arm, and are curled to the
direction of the force (clockwise). The direction of your
thumb is the direction of torque; in this case, torque is into
Figure 10 Diagram of the solved problem (resulting torque is into the
the screen.
42
We can represent "into" and "out of" using symbols when
drawing three-dimensional diagrams. The symbol for
"into" is
(it's supposed to be the tail of an arrow), and for "out of"
is (this is the tip of the arrowhead).
Figure 10 indicates the direction of torque using the "into"
symbol (shown in blue).
Exercises 1.
A constant torque of 2 kN-m is exerted on a crankshaft to start the engine. The flywheel
has a mass of 1800 kg and a radius of gyration 1m. If there is a resisting torque of 1 kNm, find the speed of the engine after 1 minute. (ANS. 320.9 r.p.m.)
Exercises 2.
A flywheel of mass 400 kg and a radius of gyration 1m losses its speed from 300 r.p.m. to
120 seconds. Determine the retarding torque acting on it. (Ans. 20.8 N-m)
43
MECHANICAL ENG’G. SCIENCE (STATICS) – MEC111
WEEK 3
44
3.1 Forces
There are different types of forces that act in different ways on structures such as bridges, chairs,
buildings, in fact any structure. The main examples of forces are shown below. Study the diagram
and text and then draw a diagram/pictogram to represent each of these forces.
(a)STATIC LOAD (standing
still)
A Static Load: A good example
of this is a person seen on the
left. He is holding a stack of
books on his hand but he is not
moving. The force downwards
is STATIC.
A Dynamic Load: A good
example of a dynamic load is
the person on the right. He is
carrying a weight of books but
walking. The force is moving or
DYNAMIC.
(d)DYNAMIC LOAD
(moving)
Internal Resistance: The person
in the diagram is sat on the
mono-bicycle and the air filled
tyre is under great pressure. The
air pressure inside it pushes
back against his/her weight.
(b)INTERNAL RESISTANCE
Tension: The rope is in
“tension” as the two people
pull on it. This stretching puts
the rope in tension.
Compression: The weight
lifter finds that his body is
compressed by the weights he
is holding above his head.
Shear Force: A good
example of shear force is seen
with a simple scissors. The
two handles put force in
different directions on the pin
that holds the two parts
together. The force applied to
the pin is called45
shear force.
(e)TENSION
(f)SHEAR FORCE
(c)COMPRESSION
Torsion: The plastic ruler is twisted
between both hands. The ruler is said to
be in a state of torsion.
(g)TORSION
Fig. 3.1
SYSTEM OF FORCES
COPLANAR FORCES
When two or more forces act in the same plane, they are said to be COPLANAR. The figure below
shows an example of two coplanar forces, the forces being in the plane , or on the surface, of the
paper. The line of action of the forces passes through the concurrent point.
F2
o
F1
Concurrent point
CONCURRENT FORCES
These are forces whose lines of action lie on the same point. They are said to be CONCURRENT. As
shown in the figure above. point O is the concurrent point.
46
COPLANAR-CONCURRENT FORCES
These are forces whose lines of action meet at, or originate from the same point and also lie on the
same plane.
RESULTANT FORCE
Referring to the Fig ( ), it is often an advantage to replace the forces F1 ands F2 by a single force,
known as the resultant force or simply the resultant, which must pass through the concurrent
point. The resultant will have the same effect as the two forces.
Example1
Determine the resultant of the two coplanar forces shown in the figure below.
b
10N
8N
8N
120o
Resultant
a
(a) space diagram
scale: 10mm = 1N
10N
(b) Vector diagram
O
(1) Draw vector oa 100mm long parallel to the 10Nforce in Fig.(a).
Note that oa means the direction of the vector is from o towards a. Compare this with the
direction of the 10N force in the space diagram.
(2) Draw vector ab 80mm long parallel to the 8Nforce, the direction from a towards b being the
same as the direction of the 8N force in the space diagram.
(3) Join ab, the vector ab represents the resultant force and is 92mm long.
Therefore: the resultant is 9.2N
It is important to note that the direction of the resultant vector, ab, is always in opposition to
the direction of the other vectors.
Example 2
Determine the magnitude and direction of the resultant of the two coplanar forces shown in the figure
below.
Referring to the vector diagram, Fig,(b). let 5mm represent 1N
47
(1) Draw vector oa 80mm long, parallel to and in the same direction as the 16Nforce in Fig.(a).in
the space diagram.
(2) Draw vector ab 50mm long, parallel to the and in the same direction as the 10Nforce in the
space diagram.
(3) Join ab, scaling the vector diagram, Fig.(b), the resultant force ab is 86.5mm long.
Therefore: the resultant is 17.3N
To find the direction of the resultant, measure the angle θ and refer this back to the space diagram as
shown in Fig,( c)
b
16N
R
100O
10N
10N
θ
(a)Space diagram
a
o
16N
(b) Vector diagram
Scale: 5mm = 1N
145.4o
R
θ= 34.6o
16N
100O
10N
(c)
Exercises
Determine the magnitude and direction of the resultant force for each of the system of concurrent
coplanar forces shown in fig.( ). State the direction relative to the larger force in each case.
10N
90o
10N
5N
30o
120o
15N
1
2
48
15N
10N
3
THE TRIANGLE OF FORCES
If three coplanar forces, acting at a point, are in equilibrium; and straight lines be drawn parallel to
the directions of the forces; then the lengths of the sides of the triangle so formed are proportional to
the magnitudes of the forces which they represent.
Let the three straight lines which are concurrent at O represent the directions of the three forces F1,
F2, F3 whose magnitude are known.
A
F2
49
F1
B
o
F3
F2
C
(a)Space diagram
F3
(b)Force diagram
F1
Now:
(i) Measure the lengths of the sides of the triangle ABC.
(ii) Find the ratios of the lengths of the sides to the forces to which they are parallel. i.e
(iii) It will be found that; these ratios are equal.
F
1
AB
=
F
2
BC
=
F
3
CA
From the Fig.(b) , the force F3 maintains equilibrium in conjunction with F1 and F2 and is called
The Equilibrant. However the
force F3 , with reversed
direction is called the
Resultant of F1andF2.
A
F2
F1
C
F3
50
B
51
3.1.1 Forces in Two Dimensions
3.1.2 Equilibrium and Statics
When all the forces which act upon an object are
balanced, then the object is said to be in a state of
equilibrium. The forces are considered to be
balanced if the rightward forces are balanced by
the leftward forces and the upward forces are
balanced by the
Fig. 3.2
downward forces. This however does not necessarily mean that all the forces are equal to
each other. Consider the two objects pictured in the force diagram shown below. Note that
the two objects are at equilibrium because the forces which act upon them are balanced;
however, the individual forces are not equal to each other. The 50 N force is not equal to
the 30 N force.
Fig. 3.3 These two objects are at equilibrium since the forces are balanced.
However, the forces are not equal.
If an object is at equilibrium, then the forces are balanced. Balanced is the key word
which is used to describe equilibrium situations. Thus, the net force is zero and the
acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s. This
extends from Newton's first law of motion. But having an acceleration of 0 m/s/s does not
mean the object is at rest. An object at equilibrium is either ...
•
•
at rest and staying at rest , or
in motion and continuing in motion with the same
speed and direction.
This too extends from Newton's first law of motion.
If an object is at rest and is in a state of equilibrium, then we
would say that the object is at "static equilibrium." "Static"
means stationary or at rest. A common physics lab is to hand
an object by two or more strings and to measure the forces
which are exerted at angles upon the object to support its
Fig. 3.4
52
weight. The state of the object is analyzed in terms of the forces acting upon the object.
The object is a point on a string upon which three forces were acting. See diagram at
right. If the object is at equilibrium, then the net force acting upon the object should be 0
Newtons. Thus, if all the forces are added together as vectors, then the resultant force (the
vector sum) should be 0 Newtons. (Recall that the net force is "the vector sum of all the
forces" or the resultant of adding all the individual forces head-to-tail.) Thus, an
accurately drawn vector addition diagram can be constructed to determine the resultant.
Sample data for such a lab are shown below.
Magnitude
Direction
Force A
Force B
Force C
3.4 N
9.2 N
9.8 N
161 deg.
70 deg.
270 deg
Fig. 3.5
For most students, the resultant was 0 Newtons (or at least very close to 0 N). This is what
we expected - since the object was at equilibrium, the net force (vector sum of all the
forces) should be 0 N.
Fig. 3.6
Another way of determining the net force (vector sum of all the forces) involves using the
trigonometric functions to resolve each force into its horizontal and vertical components.
Once the components are known, they can be compared to see if the vertical forces are
balanced and if the horizontal forces are balanced. The diagram below shows vectors A,
B, and C and their respective components. For vectors A and B, the vertical components
53
can be determined using the sine of the angle and the horizontal components can be
analyzed using the cosine of the angle. The magnitude and direction of each component
for the sample data are shown in the table below the diagram.
Fig. 3.7
The data in the table above show that the forces nearly balance. An analysis of the
horizontal components show that the leftward component of A nearly balances the
rightward component of B. An analysis of the vertical components show that the sum of
the upward components of A + B nearly balance the downward component of C. The
vector sum of all the forces is (nearly) equal to 0 Newtons. But what about the 0.1 N
difference between rightward and leftward forces and the 0.2 N difference between the
upward and downward forces? Why do the components of force only nearly balance? The
sample data used in this analysis are the result of measured data from an actual
experimental setup. The difference between the actual results and the expected results is
due to the error incurred when measuring force A and force B. We would have to
conclude that this low margin of experimental error reflects an experiment with excellent
results. We could say it's "close enough for government
work."
The above analysis of the forces acting upon an object in
equilibrium is commonly used to analyze situations
involving objects at static equilibrium. The most common
54
application involves the analysis of the forces acting upon a sign that is at rest. For
example, consider the picture at the right which hangs on a wall. The picture is in a state
of equilibrium, and thus all the forces acting upon the picture must be balanced. That is,
all horizontal components must add to 0 Newtons and all vertical components must add to
0 Newtons. The leftward pull of cable A must balance the rightward pull of cable B and
the sum of the upward pull of cable A and cable B must balance the weight of the sign.
Suppose the tension in both of the cables is measured to be 50 N and that the angle which
each cable makes with the horizontal is known to be 30 degrees. What is the weight of the
sign? This question can be answered by conducting a force analysis using trigonometric
functions. The weight of the sign is equal to the sum of the upward components of the
tension in the two cables. Thus, a trigonometric function can be used to determine this
vertical component. A diagram and accompanying work is shown below.
Fig. 3.9
Since each cable pulls upwards with a force of 25 N, the total upward pull of the sign is
50 N. Therefore, the force of gravity (also known as weight) is 50 N, down. The sign
weighs 50 N.
In the above problem, the tension in the cable and the angle which the cable makes with
the horizontal are used to determine the weight of the sign. The idea is that the tension,
the angle, and the weight are related. If the any two of these three are known, then the
third quantity can be determined using trigonometric functions.
As another example which illustrates this idea, consider the
symmetrical hanging of a sign as shown at the right. If the sign is
known to have a mass of 5 kg and if the angle between the two
cables is 100 degrees, then the tension in the cable can be
determined. Assuming that the sign is at equilibrium (a good
55
Fig. 3.10
assumption if it is remaining at rest), the two cables must supply enough upward force to
balance the downward force of gravity. The force of gravity (also known as weight) is 49
N (Fgrav = m*g), so each of the two cables must pull upwards with 24.5 N of force. Since
the angle between the cables is 100 degrees, then each cable must make a 50 degree angle
with the vertical and a 40 degree angle with the horizontal. A sketch of this situation (see
diagram below) reveals that the tension in the cable can be found using the sine function.
The triangle below illustrates these relationships.
Fig. 3.11
There is an important principle which emanates from some of the trigonometric
calculations performed above. The principle is that as the angle with the horizontal
increases, the amount of tensional force required to hold the sign at equilibrium decreases.
To illustrate this, consider a 10-Newton picture held by three different wire orientations as
shown in the diagrams below. In each case, two wires are used to support the picture;
each wire must support one-half of the sign's weight (5 N). The angle which the wires
make with the horizontal is varied from 60 degrees to 15 degrees. Use this information
and the diagram below to determine the tension in the wire for each orientation.
(a)
(b)
Fig. 3.12
56
(c)
In conclusion, equilibrium is the state of an object in which all the forces acting upon it
are balanced. In such cases, the net force is 0 Newtons. Knowing the forces acting upon
an object, trigonometric functions can be utilized to determine the horizontal and vertical
components of each force. If at equilibrium, then all the vertical components must balance
and all the horizontal components must balance.
57
Check Your Understanding
The following questions are meant to test your understanding of equilibrium situations.
Click the button to view the answers to these questions.
1. The following picture is hanging on a wall. Use trigonometric functions to determine
the weight of the picture.
Fig. 3.13
2. The sign below hangs outside the physics classroom, advertising the most important
truth to be found inside. The sign is supported by a diagonal cable and a rigid horizontal
bar. If the sign has a mass of 50 kg, then determine the tension in the diagonal cable
which supports its weight.
Fig. 3.14
3. The following sign can be found in Glenview. The sign has a mass of 50 kg. Determine
the tension in the cables.
58
Fig. 3.15
4. After its most recent delivery, the infamous stork announces the good news. If the sign
has a mass of 10 kg, then what is the tensional force in each cable? Use trigonometric
functions and a sketch to assist in the solution.
Fig. 3.16
3.2 Forces in Two Dimensions
3.2.1Inclined Planes
An object placed on a tilted surface will often slide down the
surface. The rate at which the object slides down the surface is
dependent upon how tilted the surface is; the greater the tilt of the
surface, the faster the rate at which the object will slide down it. In
physics, a tilted surface is called an inclined plane. Objects are
known to accelerate down inclined planes because of an unbalanced
force. To understand this type of motion, it is important to analyze
the forces acting upon an object on an inclined plane. The diagram
at the right depicts the two forces acting upon a crate which is
positioned on an inclined plane (assumed to be friction-free). As
shown in the diagram, there are always at least two forces acting
upon any object that is positioned on an inclined plane - the force of gravity and theFig. 3.17
normal force. The force of gravity (also known as weight) acts in a downward direction;
59
yet the normal force acts in a direction perpendicular to the surface (in fact, normal means
"perpendicular").
The first peculiarity of inclined plane problems is that the normal force is not directed in
the direction which we are accustomed to. Up to this point in the course, we have always
seen normal forces acting in an upward direction, opposite the direction of the force of
gravity. But this is only because the objects were always on horizontal surfaces and never
upon inclined planes. The truth about normal forces is not that they are always upwards,
but rather that they are always directed perpendicular to the surface that the object is on.
Fig. 3.18
The task of determining the net force acting upon an object on an inclined plane is a
difficult manner since the two (or more) forces are not directed in opposite directions.
Thus, one (or more) of the forces will have to be resolved into perpendicular components
so that they can be easily added to the other forces acting upon the object. Usually, any
force directed at an angle to the horizontal is resolved into horizontal and vertical
components. However, this is not the process that we will pursue with inclined planes.
Instead, the process of analyzing the forces acting upon objects on inclined planes will
involve resolving the weight vector (Fgrav) into two perpendicular components. This is the
second peculiarity of inclined plane problems. The force of gravity will be resolved into
two components of force - one directed parallel to the inclined surface and the other
directed perpendicular to the inclined surface.
The diagram below shows how the force of gravity has been replaced by two components
- a parallel and a perpendicular component of force.
Fig. 3.19
60
The perpendicular component of the force of gravity is directed opposite the normal force
and as such balances the normal force. The parallel component of the force of gravity is
not balanced by any other force. This object will subsequently accelerate down the
inclined plane due to the presence of an unbalanced force. It is the parallel component of
the force of gravity which causes this acceleration. The parallel component of the force of
gravity is the net force.
Fig. 3.20
The task of determining the magnitude of the two components of the force of gravity is a
mere manner of using the equations. The equations for the parallel and perpendicular
components are:
In the absence of friction and other forces (tension, applied, etc.), the acceleration of an
object on an incline is the value of the parallel component (m*g*sine of angle) divided by
the mass (m). This yields the equation
(in the absence of friction and other forces)
In the presence of friction or other forces (applied force,
tensional forces, etc.), the situation is slightly more
complicated. Consider the diagram shown at the right.
The perpendicular component of force still balances the
normal force since objects do not accelerate
perpendicular to the incline. Yet the frictional force
must also be considered when determining the net force.
As in all net force problems, the net force is the vector
sum of all the forces. That is, all the individual forces
are added together as vectors. The perpendicular
component and the normal force add to 0 N. The
parallel component and the friction force add together to
Fig. 3.21
61
yield 5 N. The net force is 5 N, directed along the incline towards the floor.
The above problem (and all inclined plane problems) can be simplified through a useful
trick known as "tilting the head." An inclined plane problem is in every way like any
other net force problem with the sole exception that the surface has been tilted. Thus, to
transform the problem back into the form with which you are more comfortable, merely
tilt your head in the same direction that the incline was tilted. Or better yet, merely tilt the
page of paper (a sure remedy for TNS - "tilted neck syndrome" or "taco neck syndrome")
so that the surface no longer appears level. This is illustrated below.
Fig. 3.21
Fig. 3.22
Once the force of gravity has been resolved into its two components and the inclined
plane has been tilted, the problem should look very familiar. Merely ignore the force of
gravity (since it has been replaced by its two components) and solve for the net force and
acceleration.
As an example consider the situation depicted in the
diagram at the right. The free-body diagram shows the
forces acting upon a 100-kg crate which is sliding
down an inclined plane. The plane is inclined at an
angle of 30 degrees. The coefficient of friction
between the crate and the incline is 0.3. Determine the
net force and acceleration of the crate.
Begin the above problem by finding the force of
gravity acting upon the crate and the components of
this force parallel and perpendicular to the incline. The force of gravity is 980 N
Fig. 3.23
and the components of this force are Fparallel = 490 N (980 N • sin 30 degrees) and
Fperpendicular = 849 N (980 N • cos30 degrees). Now the normal force can be determined to
be 849 N (it must balance the perpendicular component of the weight vector). The force
of friction can be determined from the value of the normal force and the coefficient of
friction; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N). The net force is the vector sum
of all the forces. The forces directed perpendicular to the incline balance; the forces
directed parallel to the incline do not balance. The net force is 235 N (490 N - 255 N).
The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100 kg).
62
Practice
The two diagrams below depict the free-body diagram for a 1000-kg roller coaster on the
first drop of two different roller coaster rides. Use the above principles of vector
resolution to determine the net force and acceleration of the roller coaster cars. Assume a
negligible affect of friction and air resistance. When done, click the button to view the
answers.
Fig. 3.24
The affects of the incline angle on the acceleration of a roller coaster (or any object on an
incline) can be observed in the two practice problems above. As the angle is increased,
the acceleration of the object is increased. The explanation of this relates to the
components which we have been drawing. As the angle increases, the component of force
parallel to the incline increases and the component of force perpendicular to the incline
decreases. It is the parallel component of the weight vector which causes the acceleration.
Thus, accelerations are greater at greater angles of incline. The diagram below depicts this
relationship for three different angles of increasing magnitude.
63
Fig. 3.25
Roller coasters produce two thrills associated
with the initial drop down a steep incline. The
thrill of acceleration is produced by using large
angles of incline on the first drop; such large
angles increase the value of the parallel
component of the weight vector
Fig. 3.26
(the component which causes acceleration). The thrill of weightlessness is produced by
reducing the magnitude of the normal force to values less than their usual values. It is
important to recognize that the thrill of weightlessness is a feeling associated with a lower
than usual normal force. Typically, a person weighing 700 N will experience a 700 N
normal force when sitting in a chair. However, if the chair is accelerating down a 60degrees incline, then the person will experience a 350 Newton normal force. This value is
less than normal and contributes to the feeling of weighing less than one's normal weight i.e., weightlessness.
Exercise 1
The following questions are intended to test your understanding of the mathematics and
concepts of inclined planes.
1. Two boys are playing ice hockey on a neighborhood street. A stray puck travels across
the friction-free ice and then up the friction-free incline of a driveway. Which one of the
following ticker tapes (A, B, or C) accurately portrays the motion of the puck as it travels
across the level street and then up the driveway?
Explain your answer.
64
Exercise 2
2. Little Johnny stands at the bottom of the driveway and kicks a soccer ball. The ball
rolls northward up the driveway and then rolls back to Johnny. Which one of the
following velocity-time graphs (A, B, C, or D) most accurately portrays the motion of the
ball as it rolls up the driveway and back down?
Explain your answer.
Exercise 3
3. A golf ball is rolling across a horizontal section of the green on the 18th hole. It then
encounters a steep downward incline (see diagram). Friction is involved. Which of the
following ticker tape patterns (A, B, or C) might be an appropriate representation of the
ball's motion?
Explain why the inappropriate patterns are inappropriate.
Exercise 4
4. Missy dePenn's eighth frame in the Wednesday night bowling
league was a disaster. The ball rolled off the lane, passed through
the freight door in the building's rear, and then down the
driveway. Millie Meater (Missy's teammate), who was spending
every free moment studying for her physics test, began visualizing
the velocity-time graph for the ball's motion. Which one of the velocity-time graphs (A,
B, C, or D) would be an appropriate representation of the ball's motion as it rolls across
the horizontal surface and then down the incline? Consider frictional forces.
65
Exercise 5
5. Three lab partners - Olive N. Glenveau, Glen Brook, and Warren Peace - are discussing
an incline problem (see diagram). They are debating the value of the normal force. Olive
claims that the normal force is 250 N; Glen claims that the normal force is 433 N; and
Warren claims that the normal force is 500 N. While all three answers seem reasonable,
only one is correct. Indicate which two answers are wrong and explain why they are
wrong.
Exercise 6
6. Lon Scaper is doing some lawn work when a 2-kg tire
escapes from his wheelbarrow and begins rolling down a
steep hill (a 30° incline) in San Francisco. Sketch the parallel
and perpendicular components of this weight vector.
Determine the magnitude of the components using
trigonometric functions. Then determine the acceleration of
the tire. Ignore resistance force.
Finally, determine which one of the velocity-time graph would represent the motion of the
tire as it rolls down the incline.
66
Explain your answer.
Exercise 7
7. In each of the following diagrams, a 100-kg box is sliding down a frictional surface at a
constant speed of 0.2 m/s. The incline angle is different in each situation. Analyze each
diagram and fill in the blanks.
Diagram A
Diagram B
67
3.2.2 Addition of Forces
In Unit 2 we studied the use of Newton's second law and free-body diagrams to determine
the net force and acceleration of objects. In that unit, the forces acting upon objects were
always directed in one dimension. There may have been both horizontal and vertical
forces acting upon objects; yet there were never individual forces which were directed
both horizontally and vertically. Furthermore, when a free-body diagram analysis was
performed, the net force was either horizontal or vertical; the net force (and corresponding
acceleration) was never both horizontal and vertical. Now times have changed and you
are ready for situations involving forces in two dimensions. In this unit, we will examine
the affect of forces acting at angles to the horizontal, such that the force has an influence
in two dimensions - horizontally and vertically. For such situations, Newton's second law
applies as it always did for situations involving one-dimensional net forces. However, to
use Newton's laws, common vector operations such as vector addition and vector
resolution will have to be applied. In this part of Lesson 3, the rules for adding vectors
will be reviewed and applied to the addition of force vectors.
Methods of adding vectors were discussed earlier in Lesson 1 of this unit. During that
discussion, the head to tail method of vector addition was introduced as a useful method
of adding vectors which are not at right angles to each other. Now we will see how that
method applies to situations involving the addition of force
vectors.
A force board (or force table) is a common physics lab apparatus
that has three (or more) chains or cables attached to a center
ring. The chains or cables exert forces upon the center ring in
three different directions. Typically the experimenter adjusts the
direction of the three forces, makes measurements of the amount
of force in each direction, and determines the vector sum of three forces. Forces
perpendicular to the plane of the force board are typically ignored in the analysis.
Suppose that a force board or a force table is used such that there are three forces acting
upon an object. (The object is the ring in the center of the force board or force table.) In
this situation, two of the forces are acting in two-dimensions. A top view of these three
forces could be represented by the following diagram.
68
The goal of a force analysis is to determine the net force and the corresponding
acceleration. The net force is the vector sum of all the forces. That is, the net force is the
resultant of all the forces; it is the result of adding all the forces together as vectors. For
the situation of the three forces on the force board, the net force is the sum of force
vectors A + B + C.
One method of determining the vector sum of these three forces (i.e., the net force") is to
employ the method of head-to-tail addition. In this method, an accurately drawn scaled
diagram is used and each individual vector is drawn to scale. Where the head of one
vector ends, the tail of the next vector begins. Once all vectors are added, the resultant
(i.e., the vector sum) can be determined by drawing a vector from the tail of the first
vector to the head of the last vector. This procedure is shown below. The three vectors are
added using the head-to-tail method. Incidentally, the vector sum of the three vectors is 0
Newtons - the three vectors add up to 0 Newtons. The last vector ends where the first
vector began such that there is no resultant vector.
The purpose of adding force vectors is to determine the net force acting upon an object. In
the above case, the net force (vector sum of all the forces) is 0 Newtons. This would be
expected for the situation since the object (the ring in the center of the force table) is at
rest and staying at rest. We would say that the object is at equilibrium. Any object upon
which all the forces are balanced (Fnet = 0 N) is said to be at equilibrium.
69
Quite obviously, the net force is not always 0 Newtons. In fact, whenever objects are
accelerating, the forces will not balance and the net force will be nonzero. This is
consistent with Newton's first law of motion. For example consider the situation described
below.
An Example to Test Your Understanding
A pack of five Artic wolves are exerting five
different forces upon the carcass of a 500-kg dead
polar bear. A top view showing the magnitude and
direction of each of the five individual forces is
shown in the diagram at the right. The
counterclockwise convention is used to indicate the
direction of each force vector. Remember that this is
a top view of the situation and as such does not depict
the gravitational and normal forces (since they would
be perpendicular to the plane of your computer
monitor); it can be assumed that the gravitational and
normal forces balance each other. Use a scaled vector
diagram to determine the net force acting upon the polar bear. Then compute the
acceleration of the polar bear (both magnitude and direction).
The task of determining the vector sum of all the forces for the polar bear problem
involves constructing an accurately drawn scaled vector diagram in which all five forces
are added head-to-tail.
The following five forces must be added.
70
The scaled vector diagram for this problem would look like the following:
The above two problems (the force table problem and the polar bear problem) illustrate
the use of the head-to-tail method for determining the vector sum of all the forces. The
resultant in each of the above diagrams represent the net force acting upon the object.
This net force is related to the acceleration of the object. Thus, to put the contents of this
page in perspective with other material studied in this course, vector addition methods can
be utilized to determine the sum of all the forces acting upon an object and subsequently
the acceleration of that object. And the acceleration of an object can be combined with
kinematic equations to determine motion information (i.e., the final velocity, the distance
traveled, etc.) for a given object.
71
In addition to knowing graphical methods of adding the forces acting upon an object, it is
also important to have a conceptual grasp of the principles of adding forces. Let's begin
by considering the addition of two forces, both having a magnitude of 10 Newtons.
Suppose the question is posed:
10 Newtons + 10 Newtons = ???
How would you answer such a question? Would you quickly conclude 20 Newtons,
thinking that two force vectors can be added like any two numerical quantities? Would
you pause for a moment and think that the quantities to be added are vectors (force
vectors) and the addition of vectors follow a different set of rules than the addition of
scalars? Would you pause for a moment, pondering the possible ways of adding 10
Newtons and 10 Newtons and conclude "it depends upon their direction?" In fact, 10
Newtons + 10 Newtons could give almost any resultant, provided that it has a magnitude
between 0 Newtons and 20 Newtons. Study the diagram below in which 10 Newtons and
10 Newtons are added to give a variety of answers; each answer is dependent upon the
direction of the two vectors which are to be added. For this example, the minimum
magnitude for the resultant is 0 Newtons (occurring when 10 N and 10 N are in the
opposite direction); and the maximum magnitude for the resultant is 20 N (occurring
when 10 N and 10 N are in the same direction).
72
The above diagram shows what is occasionally a difficult concept to believe. Many
students find it difficult to see how 10 N + 10 N could ever be equal to 10 N. For reasons
to be discussed in the next section of this lesson, 10 N + 10 N would equal 10 N
whenever the two forces to be added are at 30 degrees to the horizontal. For now, it ought
to be sufficient to merely show a simple vector addition diagram for the addition of the
two forces (see diagram below).
Exercise
Answer the following questions and then view the answers by clicking on the button.
1. Barb Dwyer recently submitted her vector addition homework assignment. As seen
below, Barb added two vectors and drew the resultant. However, Barb Dwyer failed to
label the resultant on the diagram. For each case, which is the resultant (A, B, or C)?
Explain.
73
2. Consider the following five force vectors.
Sketch the following and draw the resultant (R). Do not draw a scaled vector diagram;
merely make a sketch. Label each vector. Clearly label the resultant (R).
A+C+D
B+E+D
3. On two different occasions during a high school soccer game, the ball was kicked
simultaneously by players on opposing teams. In which case (Case 1 or Case 2) does the
ball undergo the greatest acceleration? Explain your answer.
4. Billie Budten and Mia Neezhirt are having an intense argument at the lunch table.
They are adding two force vectors together to determine the resultant force. The
magnitude of the two forces is 3 N and 4 N. Billie is arguing that the sum of the two
forces is 7 N. Mia argues that the two forces add together to equal 5 N. Who is right?
Explain.
5. Matt Erznott entered the classroom for his physics class. He quickly became amazed
by the remains of some of teacher's whiteboard scribbling. Evidently, the teacher had
taught his class on that day that
74
Explain why the equalities are indeed equalities and the inequality must definitely be an
inequality.
75
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 4
4.0 Levers and Moments
4.1 Levers
Levers are used to lift heavy weights with the
least amount of effort. In the example opposite,
the heavy weight on the left hand side is been
lifted by the person because of the lever. The
longer the 'rod' the easier it is to lift the weight.
Under normal circumstances the person would
not be able to lift the weight at all. The fulcrum is
the place where the rod pivots (or rotates).
Fig. 1
The load is the scientific name for the weight. The effort is quite simply the
amount of effort used to push down on the rod in order to move the weight.
We use levers in every day life.
Bicycle brakes work due to the fact
that they are based on a lever. The
diagram opposite shows the fulcrum
and the effort.
Another good example of a lever is a
simple door handle or a wheel barrow.
Fig. 2
Draw three examples of levers that are used in everyday life.
76
4.2 Three classes of Lever
There are three classes of lever and each class has fulcrum, load and effort which together can
move a heavy weight.
(a)
CLASS 1
The workman uses a trolley to move the large
packing case. The fulcrum is the wheel.
(b)
CLASS 2
The gardener uses a wheel barrow to lift tools
and garden waste. The load is in the centre of the
barrow
(c)
CLASS 3
The fisherman catches the fish which becomes the
load at the end of the lever.
Fig. 3
77
Fig. 4
Draw your own examples of the three classes of lever. Think in terms of examples that you have
used at home, work or school.
How did the Egyptians use levers to build the pyramids?
78
Machines allow a force called the effort to overcome another force called the load. If an
effort of 10 N applied to a machine can move a load of 25 N, the mechanical advantage,
MA, of that machine = 25/10 = 2.5. If MA > 1, i.e. heavier loads are moved by smaller
efforts, then the effort must move further than the load. Velocity ratio, VR, = distance
travelled by the effort / distance travelled by the load. Machines lets us overcome a
resistance at one place with an effort by applying a force at another place to move a load.
Most people cannot crush an empty match box between their thumb and fingers however
crushing the matchbox with pliers needs little effort. The effort force is multiplied
because the distance to where you grip the handles of the pliers is much greater than the
distance to the end of the jaws of the pliers. A big stone may be too heavy to carry but
you can move it on a wheelbarrow. Our fingers may be too big to pick up very small
objects or pick up objects inside small spaces but you can use tweezers to help us. Lever,
pulley and axle are simple machines developed according to the lever principle. Inclined
plane, wedge and screw are simple machines developed according to the principle of
inclined plane. The machines may save labour but do not save work because force (effort)
X distance (effort) = force (load) X distance (load), neglecting the force of friction.
Mechanical advantage (MA) is the number of times the load moved by a machine is
greater than the effort applied to that machine, i.e. MA = load / effort. MA has no unit, as
it is a ratio. If MA > 1, load > effort, i.e. you an use a smaller effort to move a bigger
load. However the effort must move further than the load. Distance moved by the effort /
distance moved by the load = velocity ratio, VR. Efficiency of a machine = energy output
/ energy input, as percentage. No machine is 100% efficient because always some energy
is lost due to friction. Work = force X distance. Efficiency = work done on the load /
work done by the effort = load X distance moved by the load / effort X distance moved by
the effort = MA X 1/VR = MA/VR X 100%.
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Fig. 5 types of levers
3 types of levers: The distance of the effort from the fulcrum is called the effort: arm and
the distance of the load from the fulcrum, the load arm. To arrange the lever so that a
small effort would lift a big load the effort arm must be as long as possible and the load
arm as short as possible. A fishing rod gives a loss in force, but gives a gain in distance
and speed instead. Examples of levers: scissors, wheelbarrow, forearm, claw hammer to
draw a nail, sugar tongs, boat oars, nut-crackers, pliers, can opener, bottle opener, crow
bar.
Lever is a simple machine consisting of a rigid rod pivoted at a fixed point called the
fulcrum, used for shifting or raising a heavy load or applying force. The lever principle
states that motive force × the arm of the motive force = resistance × the arm of resistance.
Classify levers into 3 orders or types according to where the effort is applied, and the load
moving force developed, in relation to the position of the fulcrum. A first order or type 1
lever has the load and effort on opposite sides of the fulcrum, e.g. seesaw, beam balance,
pair of scissors (two first order levers!), tin snips, bolt cutters. A second order or type 2
lever has the load and effort on the same sides with the load nearer the fulcrum, e.g.
nutcrackers, wheelbarrow, fishing rod, broomstick, biceps muscle on upper arm. A third
order or type 3 lever has the effort nearer the fulcrum than the load with both on the same
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side of it, e.g. tweezers, tongs, chopsticks. Machines are used to change the direction of
forces. Usually they allows a smaller applied force, the effort, to overcome a larger
resistance force, the load. Levers have a rigid beam supported a one point, the fulcrum (F)
with a load force (L) applied at one point and an effort force (F) applied at another point.
The Lever Principle can be stated as: Load x Length of load arm = Force x Length of
force arm. Each side of this equation is a moment, i.e. Force x Perpendicular distance to
pivot. Hence moments clockwise = moments anti-clockwise. The three types of lever
depend upon the relative positions of F, L, and E: Type 1 Fulcrum between load and
effort (E F L); Type 2 Load between effort and fulcrum (F L E); Type 3 Effort between
fulcrum and load (L E F). The mechanical advantage of a machine, M. A., is the ratio of
the load to the effort, L/E. The velocity ratio of a machine, V. R., is the ratio of the
distance moved by the effort to the distance moved by the load. The efficiency of the
machine is the ratio M. A. /V. R. Efficiency is always less than 100% because when a
machine is used there is always some energy loss.
Application of levers in everyday life
See diagram 21.1.1
Look for levers used at everyday life and classify them according to the type of lever, e.g.
Chopsticks belongs to the third order lever. Using it needs greater effort but can prevent
food from slipping away.
4.3 Moments
4.3.1 Moment of a force about a pivot
The moment of a force P about a pivot is the product of the force
and the perpendicular distance x that the force is away from the
pivot.
Why not try out a calculation yourself? If P is a number of newtons
and x is a number of meters what how many newton meters (Nm)
will the moment be?
You can try it out as many times as you like.
81
4.3.2 The Law of the Lever
The law of the lever is that if
you want to counter act a
moment caused by a force F1
and a lever arm x1 you will need
a force and lever arm in the
opposite direction. If the force
and lever arm multiplied
together (F2, x2) give you the
same moment as F1, x1 then it
will balance.
If you know 3 of the quantities in the above diagram it is possible to work out the 4th.
Here are a few questions for you to try. You can try them out as many times as you like
(You will be given different values each time). F1 and F2 are numbers of newtons. x1 and
x2 are numbers of metres.
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MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 5
5.0 Force resolution
5.1 The Parallelogram of forces
The parallelogram of forces is a method for solving (or visualizing) the results of
applying several different forces to an object. It utilizes the principles of vectors to solve
this problem called vector addition. For example, see Figure 1. This construction has the
same result as moving F2 so its tail coincides with the head of F1, and taking the net force
as the vector joining the tail of F1 to the head of F2. This procedure can be repeated to add
F3 to the resultant F1 + F2, and so forth.
Figure 1: Parallelogram construction for adding vectors
Parallelogram Of Forces
Fig. 2
In physics and applied mathematics, a method of calculating the resultant (combined
effect) of two different forces acting together on an object. Because a force has both
magnitude and direction it is a vector quantity and can be represented by a straight line. A
second force acting at the same point in a different direction can be represented by
another line drawn at an angle to the first. By completing the parallelogram (of which the
two lines are sides) a diagonal may be drawn from the original angle to the opposite
corner to represent the resultant force vector.
83
a parallelogram the diagonal of which represents the resultant of two velocities, forces,
accelerations, moments, etc., both in quantity and direction, when the velocities, forces,
accelerations, moments, etc., are represented in quantity and direction by the two
adjacent sides of the parallelogram.
In mathematics, the simplest form of the parallelogram law belongs to elementary
geometry. It states that the sum of the squares of the lengths of the four sides of a
parallelogram equals the sum of the squares of the lengths of the two diagonals. In case
the parallelogram is a rectangle, the two diagonals are of equal lengths and the statement
reduces to the Pythagorean theorem. But in general, the square of the length of neither
diagonal is the sum of the squares of the lengths of two sides.
Fig. 4 A parallelogram.
In geometry, a parallelogram is a quadrilateral with two sets of parallel sides. The
opposite sides of a parallelogram are of equal length, and the opposite angles of a
parallelogram are congruent. The three-dimensional counterpart of a parallelogram is a
parallelepiped.
84
5.2 Forces in a plane
5.2.1 Force on a particle. Resultant of two forces
A force represents the action of one body on another and is generally characterized by its
point of application, its magnitude, and its direction. Forces acting on a given particle,
however, have the same point of application. Each force considered in this chapter will
thus be completely defined by its magnitude and direction. The magnitude of a force is
characterized by a certain number of units. As indicated in Chap. 1, the SI units used by
engineers to measure the magnitude of a force are the newton (N) and its multiple the
kilonewton (kN), equal to 1000 N, while the U.S. customary units used for the same
purpose are the pound (lb) and its multiple the kilopound (kip), equal to 1000 lb. The
direction of a force is defined by the line of action and the sense of the force. The line of
action is the infinite straight line along which the force acts; it is characterized by the
angle it forms with some fixed axis (Fig. 5).
Fig.5
The force itself is represented by a segment of that line; through the use of an appropriate scale, the
length of this segment may be chosen to represent the magnitude of the force. Finally, the sense of the
force should be indicated by an arrowhead. It is important in defining a force to indicate its sense. Two
forces having the same magnitude and the same line of action but different sense, such as the forces
shown in Fig. 5. a and b, will have directly opposite effects on a particle. Experimental evidence
shows that two forces P and Q acting on
a particle A (Fig. 6a) can be replaced by a single force R which has the same effect on the particle
(Fig. 6c). This force is called the resultant of the forces P and Q and can be obtained, as shown in
Fig. 6b, by constructing a parallelogram, using P and Q as two adjacent sides of the parallelogram. The
diagonal that passes through A represents the resultant. This method for finding the resultant is
known as the parallelogram law for the addition of two forces. This law is based on experimental
evidence; it cannot be proved or derived mathematically.
85
Fig. 6
86
5.3 Derivation of the area formula
Area of the parallelogram is in blue
The area formula,
,
can be derived as follows:
The area of the parallelogram to the right (the blue area) is the total area of the rectangle
less the area of the two orange triangles. The area of the rectangle is
and the area of a single orange triangle is
Therefore, the area of the parallelogram is
87
Example
88
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 6
6.1 Resultant force
Referring to the Fig (6.1), it is often an advantage to replace the forces F1 ands F2 by a
single force, known as the resultant force or simply the resultant, which must
pass through the concurrent point. The resultant will have the same effect as the two
forces.
Example1
Determine the resultant of the two coplanar forces shown in the figure below.
b
10N
8N
8N
120o
Resultant
a
(b) space diagram
scale: 10mm = 1N
10N
(b) Vector diagram
Fig.6.1
O
(4) Draw vector oa 100mm long parallel to the 10Nforce in Fig.(a).
Note that oa means the direction of the vector is from o towards a. Compare this
with the direction of the 10N force in the space diagram.
(5) Draw vector ab 80mm long parallel to the 8Nforce, the direction from a towards
b being the same as the direction of the 8N force in the space diagram.
(6) Join ab, the vector ab represents the resultant force and is 92mm long.
Therefore: the resultant is 9.2N
It is important to note that the direction of the resultant vector, ab, is always in
opposition to the direction of the other vectors.
Example 2
89
Determine the magnitude and direction of the resultant of the two coplanar forces shown
in the figure below.
Referring to the vector diagram, Fig, 6(b). Let 5mm represent 1N
(4) Draw vector oa 80mm long, parallel to and in the same direction as the 16Nforce
in Fig.(a).in the space diagram.
(5) Draw vector ab 50mm long, parallel to the and in the same direction as the
10Nforce in the space diagram.
(6) Join ab, scaling the vector diagram, Fig.(b), the resultant force ab is 86.5mm long.
Therefore: the resultant is 17.3N
To find the direction of the resultant, measure the angle θ and refer this back to the space
diagram as shown in Fig,( c)
b
16N
R
100O
10N
(a)Space diagram
10N
θ
a
o
16N
(b) Vector diagram
Scale: 5mm = 1N
145.4o
R
θ= 34.6o
16N
100O
10N
(c)
Fig.6.2
Exercises
90
Determine the magnitude and direction of the resultant force for each of the system of
concurrent coplanar forces shown in fig.( ). State the direction relative to the larger force
in each case.
10N
90o
10N
5N
30o
15N
1
10N
120o
2
15N
3
Fig.6.3
6.2 Vectors - Fundamentals and Operations
6.3 Resultants
The resultant is the vector sum of two or more vectors. It is the result of adding two or
more vectors together. If displacement vectors A, B, and C are added together, the
result will be vector R. As shown in the diagram, vector R can be determined by the
use of an accurately drawn, scaled, vector addition diagram.
Fig.6.4
To say that vector R is the resultant displacement of
displacement vectors A, B, and C is to say that a person
who walked with displacements A, then B, and then C
would be displaced by the same amount as a person who
walked with displacement R.
91
Fig.6.5
Displacement vector R gives the same result as displacement vectors A + B + C. That
is why it can be said that
A+B+C=R
The above discussion pertains to the result of adding displacement vectors. When
displacement vectors are added, the result is a resultant displacement. But any two
vectors can be added as long as they are the same vector quantity. If two or more
velocity vectors are added, then the result is a resultant velocity. If two or more force
vectors are added, then the result is a resultant force. If two or more momentum
vectors are added, then the result is ...
In all such cases, the resultant vector (whether a displacement vector, force vector,
velocity vector, etc.) is the result of adding the individual vectors. It is the same thing
as adding A + B + C + ... . "To do A + B + C is the same as to
do R." As an example, consider a football player who gets hit
simultaneously by three players on the opposing team (players
A, B, and C). The football player experiences three different
applied forces. Each applied force contributes to a total or
resulting force.
Fig.6.6
If the three forces are added together using methods of vector addition (discussed
earlier), then the resultant vector R can be determined. In this case, to experience
the three forces A, B and C is the same as experiencing force R. To be hit by players
A, B, and C would result in the same force as being hit by one player applying force
R. "To do A + B + C is the same as to do R." Vector R is the same result as vectors A
+ B + C!!
Fig.6.7
In summary, the resultant is the vector sum of all the individual vectors. The
resultant is the result of combining the individual vectors together. The resultant can
92
be determined by adding the individual forces together using vector addition
methods.
6.4 Vectors - Fundamentals and Operations
6.5 Relative Velocity and Riverboat Problems
On occasion objects move within a medium which is moving with respect to an observer.
For example, an airplane usually encounters a wind - air which is moving with respect to
an observer on the ground below. As another example, a motor boat in a river is moving
amidst a river current - water which is moving with respect to an observer on dry land. In
such instances as this, the magnitude of the velocity of the moving object (whether it be a
plane or a motor boat) with respect to the observer on land will not be the same as the
speedometer reading of the vehicle. That is to say, the speedometer on the motor boat
might read 20 mi/hr; yet the motor boat might be moving relative to the observer on shore
at a speed of 25 mi/hr. Motion is relative to the observer. The observer on land, often
named (or misnamed) the "stationary observer" would measure the speed to be different
than that of the person on the boat. The observed speed of the boat must always be
described relative to who the observer is.
To illustrate this principle, consider a plane flying amidst a tailwind. A tailwind is merely
a wind which approaches the plane from behind, thus increasing its resulting velocity. If
the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind
velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the
ground below? The resultant velocity of the plane (that is, the result of the wind velocity
contributing to the velocity due to the plane's motor) is the vector sum of the velocity of
the plane and the velocity of the wind. This resultant velocity is quite easily determined if
the wind approaches the plane directly from behind. As shown in the diagram below, the
plane travels with a resulting velocity of 125 km/hr relative to the ground.
Fig.6.8
If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr.
Since a headwind is a wind which approaches the plane from the front, such a wind would
93
decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100
km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. In this case,
the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an
observer on the ground. This is depicted in the diagram below.
Fig.6.10
Now consider a plane traveling with a velocity of 100 km/hr, South which encounters a
side wind of 25 km/hr, West. Now what would the resulting velocity of the plane be?
This question can be answered in the same manner as the previous questions. The
resulting velocity of the plane is the vector sum of the two individual velocities. To
determine the resultant velocity, the plane velocity (relative to the air) must be added to
the wind velocity. This is the same procedure which was used above for the headwind and
the tailwind situations; only now, the resultant is not as easily computed. Since the two
vectors to be added - the southward plane velocity and the westward wind velocity - are at
right angles to each other, the Pythagorean theorem can be used. This is illustrated in the
diagram below.
Fig.6.12
In this situation of a side wind, the southward vector can be added to the westward vector
using the usual methods of vector addition. The magnitude of the resultant velocity is
determined using Pythagorean theorem. The algebraic steps are as follows:
94
(100 km/hr)2 + (25 km/hr)2 = R2
10 000 km2/hr2 + 625 km2/hr2 = R2
10 625 km2/hr2 = R2
SQRT(10 625 km2/hr2) = R
103.1 km/hr = R
The direction of the resulting velocity can be determined using a trigonometric function.
Since the plane velocity and the wind velocity form a right triangle when added together
in head-to-tail fashion, the angle between the resultant vector and the southward vector
can be determined using either the sine, cosine, or tangent functions. The tangent function
can be
used; this is shown below:
Fig.6.13
tan (theta) = (opposite/adjacent)
tan (theta) = (25/100)
theta = invtan (25/100)
theta = 14.0 degrees
If the resultant velocity of the plane makes a 14.0 degree angle with the southward
direction (theta in the above diagram), then the direction of the resultant is 256 degrees.
Like any vector, the resultant's direction is measured as a counterclockwise angle of
rotation from due East.
6.6 Analysis of a Riverboat's Motion
The affect of the wind upon the plane is similar to the affect of the river current upon the
motor boat. If a motor boat were to head straight across a river (that is, if the boat were to
point its bow straight towards the other side), it would not reach the shore directly across
from its starting point. The river current influences the motion of the boat and carries it
downstream. The motor boat may be moving with a velocity of 4 m/s directly across the
river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the
downstream direction. While the speedometer of the boat may read 4 m/s, its speed with
respect to an observer on the shore will be greater than 4 m/s.
95
Fig.6.14
The resultant velocity of the motor boat can be determined in the same manner as was
done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity
and the river velocity. Since the boat heads straight across the river and since the current
is always directed straight downstream, the two vectors are at right angles to each other.
Thus, the Pythagorean theorem can be used to determine the resultant velocity. Suppose
that the river was moving with a velocity of 3 m/s, North and the motor boat was moving
with a velocity of 4 m/s, East. What would be the resultant velocity of the motor boat (i.e.,
the velocity relative to an observer on the shore)? The magnitude of the resultant can be
found as follows:
(4.0 m/s)2 + (3.0 m/s)2 = R2
16 m2/s2 + 9 m2/s2 = R2
25 m2/s2 = R2
SQRT (25 m2/s2) = R
5.0 m/s = R
The direction of the resultant is the counterclockwise angle of rotation which the resultant
vector makes with due East. This angle can be determined using a trigonometric function
as shown below.
tan (theta) = (opposite/adjacent)
tan (theta) = (3/4)
theta = invtan (3/4)
theta = 36.9 degrees
Fig.6.15
96
Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant
velocity of the boat will be 5 m/s at 36.9 degrees.
Motor boat problems such as these are typically accompanied by three separate
questions:
a.
What is the resultant velocity (both magnitude and direction) of the boat?
b. If the width of the river is X meters wide, then how much time does it take the boat to
travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
The first of these three questions was answered above; the resultant velocity of the boat
can be determined using the Pythagorean theorem (magnitude) and a trigonometric
function (direction). The second and third of these questions can be answered using the
average speed equation (and a lot of logic).
ave. speed = distance/time
Consider the following example.
Example 1
A motor boat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
a.
What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters wide, then how much &me does it take the boat to
travel shore to shore?
c.
What distance downstream does the boat reach the opposite shore?
The solution to the first question has already been shown in the above discussion. The
resultant velocity of the boat is 5 m/s at 36.9 degrees. We will start in on the second
question.
The river is 80-meters wide. That is, the distance from shore to shore as measured straight
across the river is 80 meters. The time to cross this 80-meter wide river can be determined
by rearranging and substituting into the average speed equation.
time = distance /(ave. speed)
The distance of 80 m can be substituted into the numerator. But what about the
denominator? What value should be used for average speed? Should 3 m/s (the current
velocity), 4 m/s (the boat velocity), or 5 m/s (the resultant velocity) be used as the average
speed value for covering the 80 meters? With what average speed is the boat traversing
the 80 meter wide river? Most students want to use the resultant velocity in the equation
97
since that is the actual velocity of the boat with respect to the shore. Yet the value of 5
m/s is the speed at which the boat covers the diagonal dimension of the river. And the
diagonal distance across the river is not known in this case. If one knew the distance C in
the diagram below, then the average speed C could be used to calculate the time to reach
the opposite shore. Similarly, if one knew the distance B in the diagram below, then the
average speed B could be used to calculate the time to reach the opposite shore. And
finally, if one knew the distance A in the diagram below, then the average speed A could
be used to calculate the time to reach the opposite shore.
Fig.6.16
In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s
(average speed in the direction straight across the river) should be substituted into the
equation to determine the time.
%me = (80 m)/(4 m/s) = 20 s
It requires 20 s for the boat to travel across the river. During this 20 s of crossing the
river, the boat also drifts downstream. Part c of the problem asks "What distance
downstream does the boat reach the opposite shore?" The same equation must be used to
calculate this downstream distance. And once more, the question arises, which one of the
three average speed values must be used in the equation to calculate the distance
downstream? The distance downstream corresponds to Distance B on the above diagram.
The speed at which the boat covers this distance corresponds to Average Speed B on the
diagram above (i.e., the speed at which the current moves - 3 m/s). And so the average
speed of 3 m/s (average speed in the downstream direction) should be substituted into the
equation to determine the distance.
distance = ave. speed * %me = (3 m/s) * (20 s)
distance = 60 m
The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.
The mathematics of the above problem is no more difficult than dividing or multiplying
two numerical quantities by each other. The mathematics is easy! The difficulty of the
problem is conceptual in nature; the difficulty lies in deciding which numbers to use in
the equations. That decision emerges from one's conceptual understanding (or
unfortunately, one's misunderstanding) of the complex motion which is occurring. The
98
motion of the river boat can be divided into two simultaneous parts - a motion in the
direction straight across the river and and a motion in the downstream direction. These
two parts (or components) of the motion occur simultaneously for the same time duration
(which was 20 seconds in the above problem). The decision as to which velocity value or
distance value to use in the equation must be consistent with the diagram above. The
boat's motor is what carries the boat across the river the Distance A; and so any
calculation involving the Distance A must involve the speed value labeled as Speed A
(the boat speed relative to the water). Similarly, it is the current of the river which carries
the boat downstream for the Distance B; and so any calculation involving the Distance B
must involve the speed value labeled as Speed B (the river speed). Together, these two
parts (or components) add up to give the resulting motion of the boat. That is, the acrossthe-river component of displacement adds to the downstream displacement to equal the
resulting displacement. And likewise, the boat velocity (across the river) adds to the river
velocity (down the river) to equal the resulting velocity. And so any calculation of the
Distance C or the Average Speed C ("Resultant Velocity") can be performed using the
Pythagorean theorem.
Now to illustrate an important point, let's try a second example problem which is similar
to the first example problem. Make an attempt to answer the three questions and then
click the button to check your answer.
Example 2
A motor boat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
a.
What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters wide, then how much &me does it take the boat to
travel shore to shore?
c.
What distance downstream does the boat reach the opposite shore?
An important concept emerges from the analysis of the two example problems above. In
Example 1, the time to cross the 80-meter wide river (when moving 4 m/s) was 20
seconds. This was in the presence of a 3 m/s current velocity. In Example 2, the current
velocity was much greater - 7 m/s - yet the time to cross the river remained unchanged. In
fact, the current velocity itself has no affect upon the time required for a boat to cross the
river. The river moves downstream parallel to the banks of the river. As such, there is no
way that the current is capable of assisting a boat in crossing a river. While the increased
current may affect the resultant velocity - making the boat travel with a greater speed with
respect to an observer on the ground - it does not increase the speed in the direction across
the river. The component of the resultant velocity which is increased is the component
which is in a direction pointing down the river. It is often said that "perpendicular
99
components of motion are independent of each other." As applied to river boat problems,
this would mean that an across-the-river variable would be independent of (i.e., not be
affected by) a downstream variable. The time to cross the river is dependent upon the
velocity at which the boat crosses the river. It is only the component of motion directed
across the river (i.e., the boat velocity) which affects the time to travel the distance
directly across the river (80 m in this case). The component of motion perpendicular to
this direction - the current velocity - only affects the distance which the boat travels down
the river. This concept of perpendicular components of motion will be investigated in
more detail in the next part of Lesson 1.
6.7 Check Your Understanding
1. A plane can travel with a speed of 80 mi/hr with respect to the air. Determine the
resultant velocity of the plane (magnitude only) if it encounters a
a. 10 mi/hr headwind.
b. 10 mi/hr tailwind.
c. 10 mi/hr crosswind.
d. 60 mi/hr crosswind.
2. A motor boat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North.
a. What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters wide, then how much time does it take the boat to
travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
100
3. A motor boat traveling 5 m/s, East encounters a current traveling 2.5 m/s, South.
a. What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters wide, then how much time does it take the boat to
travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
4. A motor boat traveling 6 m/s, East encounters a current traveling 3.8 m/s, South.
a. What is the resultant velocity of the motor boat?
b. If the width of the river is 120 meters wide, then how much time does it take the boat to
travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
5. If the current velocity in question #4 were increased to 5 m/s, then
a. how much time would be required to cross the same 120-m wide river?
b. what distance downstream would the boat travel during this time?
101
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 7
7.1 Polygon of Forces
7.2 Principle of polygon of Forces
When more than two coplanar forces act on a body with their lines of action passing
through a concurrent point, the resultant is again determine by the vectorial addition of
the forces. The ensuing vector diagram is known as the polygon of forces.
Example 1
Determine the resultant of the three concurrent coplanar forces shown in fig.(7.1 ).
8N
b
120o
10N
120o
a
o
12N
R
(a) space diagram
(b) vector diagram
Scale: 10mm = 1N
Fig.7.1
In determining the resultant of a multi-force, coplanar system, the force vectors may be
drawn in any order. However, it is convenient to consider the forces in a clockwise
direction as shown in the vector diagram, fig.(b).
Referring to the vector diagram, Fig,(b). let 10mm represent 1N.
(1) Draw vector oa 100mm long, parallel to and in the same direction as the 10Nforce
in Fig.(a).in the space diagram.
(2) Draw vector ab 80mm long, parallel to the and in the same direction as the
8Nforce in the space diagram.
(3) Draw vector bc 120mm long, parallel to the and in the same direction as the
12Nforce in the space diagram.
(4) Join oc, scaling the vector diagram, Fig.(b), the resultant force oc is 34mm long.
Therefore: the resultant is 3.4N
It is important to note that the direction of the resultant vector, oc, is always in
opposition to the direction of the other vectors.
102
c
Example 2
Determine the magnitude and direction of the single force (i.e. the resultant) which will
replace the concurrent coplanar forces shown in fig.(a).
o
θ
R
d
10N
15N
o
40o
60
c
20N
30N
a
(a) space diagram
(b) vector diagram
10N
o
b
15N
Scale: 5mm = 1N
40o
60
20N
30N
76o
R=3.6N
(c)
Fig.7.2
Scaling the vector diagram, fig.(b), in which 5mm = 1N, the resultant, od, is 18mm long.
Therefore: the resultant is 3.6N, and also the direction of the resultant, measure the angle
θ = 76o
i.e. the magnitude and direction of the single force to replace the given concurrent
coplanar forces are as shown in fig. (c).
103
Exercises
Determine the magnitude and direction of the resultant force for each of the system of
concurrent coplanar forces shown in fig.( ). State the direction relative to the larger force
in each case.
10N
90o
230O
5N
30o
130O
10N
20N
15N
1
5N
2
10N
120o
120o
15N
12N
3
Fig.7.3
104
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 8
8.1 Lami’s Theorem
If three forces acting at a point are in equilibrium each is proportional to the sine of the
angle included between the other two forces.
F2
β
F3
F2
γ
θ
C
(a)Space diagram
β
θ
(b)Force diagram
F1
F1
γ
F3
Now:
Fig.8.1
F
1
Sinβ
=
F
2
Sinθ
=
F
3
Sinγ
Exercise
A mass of 10kg is supported by two strings; OA, OB, knotted at O and attached at A and
B to a horizontal beam. The angle between the strings is 120o; that between OA and the
vertical string holding the weight is 135o, and that between OB and the vertical string is
105o. Find the tensions in the strings OA and OB.
8.2 Resolving a Force
Consider a force R, acting along and represented by OC; it can be resolved in to two
forces; F1 and F2 ,one of which makes an angle θ with R; and the other which makes an
angle of ( 90o – θ ) with R.
105
A
C
F1
R
θ
O
B
F2
Fig.8.2
Then:
OB
= Cosθ
OC
→ OB = OC Cosθ
→ F2 = RCosθ
BC
= Sinθ
OC
→ BC = RSinθ
→ F1 = RSinθ
Therefore:
Resolved part of R along OB = RCosθ
Resolved part of R along OA = RSinθ
Where:
RCosθ is the total effect of R in the direction OB
RSinθ is the total effect of R in the direction OA
8.3 Independence of Perpendicular Components of Motion
A force vector which is directed upward and rightward has two parts - an upward
part and a rightward part. That is to say, if you pull upon an object in an upward
and rightward direction, then you are exerting an influence upon the object in two
separate directions - an upward direction and a rightward direction. These two
parts of the two-dimensional vector are referred to as components. A component
describes the affect of a single vector in a given direction. Any force vector which is
exerted at an angle to the horizontal can be considered as having two parts or
106
components. The vector sum of these two components is always equal to the force at
the given angle. This is depicted in the diagram below.
Fig.8.3
Any vector - whether it be a force vector, displacement vector, velocity vector, etc. directed at an angle can be thought of as being composed of two perpendicular
components. These two components can be represented as legs of a right triangle
formed by projecting the vector onto the x- and y-axis.
Fig.8.4
The two perpendicular parts or components of a vector are independent of each
other. Consider the pull upon Fido as an example. If the horizontal pull upon Fido
increases, then Fido would be accelerated at a greater rate to the right; yet this
107
greater horizontal pull would not exert any vertical influence upon Fido. Pulling
horizontally with more force does not lift Fido vertically off the ground. A change in
the horizontal component does not affect the vertical component. This is what is
meant by the phrase "perpendicular components of vectors are independent of each
other." A change in one component does not affect the other component. Changing a
component will affect the motion in that specific direction. While the change in one
of the components will alter the magnitude of the resulting force, it does not alter the
magnitude of the other component.
Fig.8.5
The resulting motion of a plane flying in the presence of a crosswind is the
combination (or sum) of two simultaneous velocity vectors which are perpendicular
to each other. Suppose that a plane is attempting to fly northward from Chicago to
the Canada border by simply directing the plane due northward. If the plane
encounters a crosswind directed towards the west, then the resulting velocity of the
plane would be northwest. The northwest velocity vector consists of two components
- a north component resulting from the plane's motor (the plane velocity) and a
westward component resulting from the crosswind (the wind velocity). These two
components are independent of each other. An alteration in one of the components
will not affect the other component. For instance, if the wind velocity increased, then
the plane would still be covering ground in the northerly direction at the same rate.
It is true that the alteration of the wind velocity would cause the plane to travel more
westward; however, the plane still flies northward at the same speed. Perpendicular
components of motion do not affect each other.
108
Fig.8.6
Now consider an air balloon descending through the air
toward the ground in the presence of a wind which blows
eastward. Suppose that the downward velocity of the balloon
is 3 m/s and that the wind is blowing east with a velocity of 4
m/s. The resulting velocity of the air balloon would be the combination (i.e., the
vector sum) of these two simultaneous and independent velocity vectors. The air
balloon would be moving downward and eastward.
If the wind velocity increased, the air balloon would begin moving faster in the
eastward direction, but its downward velocity would not be altered. If the balloon
were located 60 meters above the ground and was moving downward at 3 m/s, then
it would take a time of 20 seconds to travel this vertical distance.
d=v•t
So t = d / v = (60 m) / (3 m/s) = 20 seconds
During the 20 seconds taken by the air balloon to fall the 60 meters to the ground,
the wind would be carrying the balloon in the eastward direction. With a wind speed
of 4 m/s, the distance traveled eastward in 20 seconds would be 80 meters. If the
wind speed increased from the value of 4 m/s to a value of 6 m/s, then it would still
take 20 seconds for the balloon to fall the 60 meters of downward distance. A motion
in the downward direction is affected only by downward components of motion. An
alteration in a horizontal component of motion (such as the eastward wind velocity)
will have no affect upon vertical motion. Perpendicular components of motion are
independent of each other. A variation of the eastward wind speed from a value of 4
m/s to a value of 6 m/s would only cause the balloon to be blown eastward a distance
of 120 meters instead of the original 80 meters.
In the most recent section of Lesson 1, the topic of relative velocity and riverboat
motion was discussed. A boat on a river often heads straight across the river,
perpendicular to its banks. Yet because of the flow of
water (i.e., the current) moving parallel to the river
banks, the boat does not land on the bank directly
across from the starting location.
Fig.8.7
109
The resulting motion of the boat is the combination (i.e., the vector sum) of these two
simultaneous and independent velocity vectors - the boat velocity plus the river
velocity. In the diagram at the right, the boat is depicted as moving eastward across
the river while the river flows southward. The boat starts at Point A and heads itself
towards Point B. But because of the flow of the river southward, the boat reaches the
opposite bank of the river at Point C. The time required for the boat to cross the
river from one side to the other side is dependent upon the boat velocity and the
width of the river. Only an eastward component of motion could affect the time to
move eastward across a river.
Suppose that the boat velocity is 4 m/s and the river velocity is 3 m/s. The magnitude
of the resultant velocity could be determined to be 5 m/s using the Pythagorean
Theorem. The time required for the boat to cross a 60-meter wide river would be
dependent upon the boat velocity of 4 m/s. It would require 15 seconds to cross the
60-meter wide river.
d=v•t
So t = d / v = (60 m) / (4 m/s) = 15 seconds
The southward river velocity will not affect the time required for the boat to travel
in the eastward direction. If the current increased such that the river velocity
became 5 m/s, then it would still require 15 seconds to cross the river. Perpendicular
components of motion are independent of each other. An increase in the river
velocity would simply cause the boat to travel further in the southward direction
during these 15 seconds of motion. An alteration in a southward component of
motion only affects the southward motion.
Fig.8.8
All vectors can be thought of as having perpendicular components. In fact, any
motion that is at an angle to the horizontal or the vertical can be thought of as
having two perpendicular motions occurring simultaneously. These perpendicular
components of motion occur independently of each other. Any component of motion
occurring strictly in the horizontal direction will have no affect upon the motion in
the vertical direction. Any alteration in one set of these components will have no
affect on the other set. In Lesson 2, we will apply this principle to the motion of
projectiles which typically encounter both horizontal and vertical motion.
110
8.4 Check Your Understanding
1. A plane flies northwest out of O'Hare Airport in Chicago at a
speed of 400 km/hr in a direction of 150 degrees (i.e., 30 degrees
north of west). The Canadian border is located a distance of
1500 km due north of Chicago. The plane will cross into Canada
after approximately ____ hours.
Fig.8.9
a. 0.13
b. 0.23
c. 0.27
d. 3.75
e. 4.33
f. 6.49
g. 7.50
h. None of these are even close.
2. Suppose that the plane in question 1 was flying with a velocity of 358 km/hr in a
direction of 146 degrees (i.e., 34 degrees north of west). If the Canadian border is
still located a distance of 1500 km north of Chicago, then how much time would it
take to cross the border?
3. TRUE or FALSE:
A boat heads straight across a river. The river flows north at a speed of 3 m/s. If the
river current was greater, then the time required for the boat to reach the opposite
shore would not change.
a. True
b. False
111
4. A boat begins at point A and heads straight across a 60meter wide river with a speed of 4 m/s (relative to the
water). The river water flows north at a speed of 3 m/s
(relative to the shore). The boat reaches the opposite
shore at point C. Which of the following would cause the
boat to reach the opposite shore at a location SOUTH of
C?
Fig.8.10
a. The boat heads across the river at 5 m/s.
b. The boat heads across the river at 3 m/s.
c. The river flows north at 4 m/s.
d. The river flows north at 2 m/s.
e. Nonsense! None of these affect the location where the boat lands.
112
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 9
9.1 Equilibrium
A body is in equilibrium when the forces acting on it compose a balanced system.
The body may be stationary or alternatively it may be moving with uniform velocity, in
which
case the applied force is just equal to the resistance as shown in the fig(9.1)
Since motion is uniform the force system is balanced. R = F and P = W
P (normal reaction)
Uniform velocity
R (resistance)
F (applied force)
W (weight)
Fig.9.1
113
The diagram below clearly shows a state of equilibrium. The cars on either side of the
seesaw
are exactly the same in weight and height, in fact they are the same model. As a result the
seesaw
stays level.
The centre of the seesaw is called the ‘fulcrum’, seen here as a triangle and this is where
the
beam that the cars are on tilts backwards and forwards. However, because of the state of
equilibrium they remain completely still.
The weight (the cars) is called the effort.
The cars are in a 'state of equilibrium' because the weight on either side is exactly
the same
Fig.9.2
If an extra car is added to the right hand
side
(see diagram to right) then the seesaw
will
turn in a clockwise direction - called a
clockwise moment.
114
Fig.9.3
Alternatively, if more cars are added to
the
left hand side the seesaw will turn in an
anticlockwise direction - called an
anticlockwise moment.
A clockwise moment as an extra car is
added to the right side
The cars are in a 'state of equilibrium' because the weight on either side is exactly
the same
The moment of a force P about a pivot is the product of the force
and the perpendicular distance x that the force is away from the
pivot.
Why not try out a calculation yourself? If P is a number of newtons
and x is a number of meters what how many newton meters (Nm)
will the moment be?
You can try it out as many times as you like.
Fig.9.4
115
Fig.9.5
9.2 Balance of Moments
The above problem is based on this principle: take any set of forces that don't act on the same point;
calculate the moments of all these forces about a pivot and for balance we must have:
clockwise moments (positive)= anticlockwise moments (negative)
Fig.9.6
116
9.3 Clockwise Moment.
It is the moment of a force, whose effect is to turn or rotates the body, in the same direction in which the
hands of a
clock moves, as shown in the figure below.
o
M
Fig.9.7
9.4 Anticlockwise Moment.
It is the moment of a force, whose effect is to turn or rotates the body, in the opposite direction in which the
hands
of a clock moves, as shown in the figure below
o
M
Fig.9.8
9.5 Varignon’s Principle of Moments (or Law of Moments)
It states, ‘’ If a number of coplanar forces are acting simultaneously on a particle, the
algebraic sum of the moments of all the forces about any point is equal to the sum of their
resultant force about the same point.’’
117
15N
15N
O
A
60
O
0.8m
B
0.8m
(a)
(b)
.If this force is applied at an angle of 60
o
to the edge of the same door, as
shown in fig. (b), find the moment of this force.
118
o
QUESTIONS
1. The diagram below shows a lever
where an effort of 200 N balances a
load of 600 N. The effort force is 6
metres from the fulcrum. The load
force is two metres from the fulcrum.
Fig.9.9
Clockwise moment = 600 x 2 Nm
Anti-clockwise moment = 200 x 6 Nm
In a state of equilibrium,
clockwise moments = anti-clockwise moments
600 X 2 Nm = 200 x 6 Nm
1200 = 1200
119
2. In the diagram below a crow-bar is
used to move a 400n load. What effort
is required to move the load?
Fig.9.10
Clockwise moments = 400 N x 0.6 m
Anticlockwise moments = effort x 1.5m
In equilibrium;
clockwise moments = anti-clockwise moments
400 x 0.6 = effort x 1.5
effort = 400 x 0.6
1.5
effort = 240
1.5
= 160 N
3. Another crow-bar is used to lever
a load of 120N. The load is 2m from
the fulcrum and the effort is 6m
from the fulcrum. What effort is
required to move the load?
120
Fig.9.11
Clockwise moments = 120 N x 2 m
Anticlockwise moments = effort x 6m
In equilibrium;
clockwise moments = anti-clockwise moments
120 x 2 = effort x 6
effort = 120 x 2
6
effort = 240
6
= 40 N
An effort of over 40 N is required to move the load.
4. A wheel-barrow is used to lift a
load of 150N. The wheel acts as the
fulcrum. Calculate the effort
required.
Fig.9.12
Clockwise moments = 150 N x .5 m
Anticlockwise moments = effort x 1.5m
In equilibrium;
clockwise moments = anti-clockwise moments
150 x .5 = effort x 1.5
effort = 150 x .5
1.5
121
effort = 75
1.5
= 50 N
An effort of over 50 N is required to lift the wheel-barrow.
EXERCISES
TRY THE FOLLOWING QUESTIONS
5. A wheel-barrow is used to lift a
load of 200N. The wheel acts as the
fulcrum. Calculate the effort
required.
Fig.9.13
6. A metal bar is used to lever a load
of 150N. The load is 1m from the
fulcrum and the effort is 6m from the
fulcrum. What effort is required to
move the load ?
Fig.9.14
7. Another metal bar is used to lever
a load of 200N. The load is 3m from
the fulcrum and the effort is 2m from
the fulcrum. What effort is required
to move the load ?
Fig.9.15
122
A metre scale is supported at the centre. It is balanced by two weights A and B as shown
in figure below, find the distance of B from the pivot.
Fig.9.16
Clockwise moment and anticlockwise moment about 50 cm divisions are equal.
20 x d = 40 x 20
Therefore,
123
Hence the 20 N force of B is acting from 90 cm mark.
The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a
wedge placed under the 40 cm mark and carrying a weight of 70 N hanging from the 10
cm mark. The ruler is balanced horizontally by a weight W hanging from the 100 cm
mark. Calculate the value of the weight W.
Fig.9.17
W x (100 - 40) + 30 (50 - 40) = 70 x (40 - 10)
60 x W + 30 x 10 = 70 x 30
60 W = 2100 - 300
Therefore,
124
The illustration in figure below represents a metre scale balancing on a knife edge at 20
cm mark when a weight of 60 N is suspended from 10 cm mark. Calculate the weight of
the ruler.
Fig.9.18
Weight of the ruler is acting from the centre of gravity of the ruler (i.e., the mid point 50
cm)
W x 30 = 60 x 10
Therefore,
9.6 Principle of Moments
125
If a body is in equilibrium under the action of a number of forces, then the algebraic sum
of the moments of the forces about any point is equal to zero.
In other words, the sum of the clockwise moments equals sum of the anticlockwise
moments when the body is in equilibrium.
Fig.9.19
Clockwise moments equal to anticlockwise moments
In the figure above,
Sum of the anticlockwise moments = Sum of the clockwise moments
i.e., (50 x 40) + (100 x 20) + (60 x 10) = (30 x 20) + (100 x 40)
126
A metre scale is supported at the centre. It is balanced by two weights A and B as shown
in figure below, find the distance of B from the pivot.
Fig.9.20
Clockwise moment and anticlockwise moment about 50 cm divisions are equal.
20 x d = 40 x 20
Therefore,
Hence the 20 N force of B is acting from 90 cm mark.
The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a
127
wedge placed under the 40 cm mark and carrying a weight of 70 N hanging from the 10
cm mark. The ruler is balanced horizontally by a weight W hanging from the 100 cm
mark. Calculate the value of the weight W.
Fig.9.21
W x (100 - 40) + 30 (50 - 40) = 70 x (40 - 10)
60 x W + 30 x 10 = 70 x 30
60 W = 2100 - 300
Therefore,
The illustration in figure below represents a metre scale balancing on a knife edge at 20
cm mark when a weight of 60 N is suspended from 10 cm mark. Calculate the weight of
128
the ruler.
Fig.9.22
Weight of the ruler is acting from the centre of gravity of the ruler (i.e., the mid point 50
cm)
W x 30 = 60 x 10
Therefore,
129
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 10
10.0 Friction
10.1 Fundamentals and application of friction
10.2 Friction
Frictional resistance to the relative motion of two solid objects is usually proportional to the force
which presses the surfaces together as well as the roughness of the surfaces. Since it is the force
perpendicular or "normal" to the surfaces which affects the frictional resistance, this force is
typically called the "normal force" and designated by N. The frictional resistance force may then
be written:
ffriction = µ N
µ = coefficient of friction
Standard model
µk = coefficient of kinetic friction
of friction
µs = coefficient of static friction
The frictional force is also presumed to be proportional to the coefficient of friction. However, the
amount of force required to move an object starting from rest is usually greater than the force
required to keep it moving at constant velocity once it is started. Therefore two coefficients of
friction are sometimes quoted for a given pair of surfaces - a coefficient of static friction and a
coefficent of kinetic friction. The force expression above can be called the standard model of
surface friction and is dependent upon several assumptions about friction.
While this general description of friction (which I will refer to as the standard model) has practical
utility, it is by no means a precise description of friction. Friction is in fact a very complex
phenomenon which cannot be represented by a simple model. Almost every simple statement you
make about friction can be countered with specific examples to the contrary. Saying that rougher
surfaces experience more friction sounds safe enough - two pieces of coarse sandpaper will
obviously be harder to move relative to each other than two pieces of fine sandpaper. But if two
pieces of flat metal are made progressively smoother, you will reach a point where the resistance to
relative movement increases. If you make them very flat and smooth, and remove all surface
contaminants in a vacuum, the smooth flat surfaces will actually adhere to each other, making what
is called a "cold weld".
130
10.3 Friction and Surface Roughness
Fig.10.1
In general, the coefficients of friction for static and kinetic friction are different.
Like all simple statements about friction, this picture of friction is too simplistic. Saying that
rougher surfaces experience more friction sounds safe enough - two pieces of coarse sandpaper
will obviously be harder to move relative to each other than two pieces of fine sandpaper. But
if two pieces of flat metal are made progressively smoother, you will reach a point where the
resistance to relative movement increases. If you make them very flat and smooth, and remove
all surface contaminants in a vacuum, the smooth flat surfaces will actually adhere to each
other, making what is called a "cold weld". Once you reach a certain degree of mechanical
smoothness, the frictional resistance is found to depend on the nature of the molecular forces in
the area of contact, so that substances of comparable "smoothness" can have significantly
different coefficients of friction.
An easily observed counterexample to the idea that rougher surfaces exhibit more friction is
that of ground glass versus smooth glass. Smooth glass plates in contact exhibit much more
frictional resistance to relative motion than the rougher ground glass.
131
10.4 Static Friction
Static frictional forces from the interlocking of the irregularities of two surfaces will increase to
prevent any relative motion up until some limit where motion occurs. It is that threshold of
motion which is characterized by the coefficient of static friction. The coefficient of static friction
is typically larger than the coefficient of kinetic friction.
Fig.10.2
In making a distinction between static and kinetic coefficients of friction, we are dealing with an
aspect of "real world" common experience with a phenomenon which cannot be simply
characterized. The difference between static and kinetic coefficients obtained in simple
experiments like wooden blocks sliding on wooden inclines roughly follows the model depicted
in the friction plot from which the illustration above is taken. This difference may arise from
irregularities, surface contaminants, etc. which defy precise description. When such experiments
are carried out with smooth metal blocks which are carefully cleaned, the difference between
static and kinetic coefficients tends to disappear. When coefficients of friction are quoted for
specific surface combinations are quoted, it is the kinetic coefficient which is generally quoted
since it is the more reliable number.
Another important example of static friction is the force that prevents a car wheel from slipping
as it rolls on the ground. Even though the wheel is in motion, the patch of the tire in contact with
the ground is stationary relative to the ground, so it is static rather than kinetic friction.
The maximum value of static friction, when motion is impending, is sometimes referred to as
limiting friction, although this term is not used universally. The value is given by the product of
the normal force and coefficient of static friction.
132
10.5 Kinetic Friction
Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and
rub together (like a sled on the ground).
When two surfaces are moving with respect to one another, the frictional resistance is almost
constant over a wide range of low speeds, and in the standard model of friction the frictional
force is described by the relationship below. The coefficient is typically less than the
coefficient of static friction, reflecting the common experience that it is easier to keep
something in motion across a horizontal surface than to start it in motion from rest.
Fig.10.3
The coefficient of kinetic friction is typically denoted as µk, and is usually less than the
coefficient of static friction.
133
10.6 Friction Plot
static friction resistance will match the applied force up until the threshold of motion. Then the
kinetic frictional resistance stays about constant. This plot illustrates the standard model of
friction.
Fig.10.4
The above plot, though representing a simplistic view of friction, agrees fairly well with the
results of simple experiments with wooden blocks on wooden inclines. The experimental
procedure described below equates the vector component of the weight down the incline to the
coefficient of friction times the normal force produced by the weight on the incline.
Fig.10.5
Having taken a large number of students through this experiment, I can report that the coefficient
134
of static friction obtained is almost always greater than the coefficient of kinetic friction. Typical
results for the woods I have used are 0.4 for the static coefficient and 0.3 for the kinetic
coefficient.
When carefully standardized surfaces are used to measure the friction coefficients, the difference
between static and kinetic coefficients tends to disappear, indicating that the difference may have
to do with irregular surfaces, impurities, or other factors which can be frustratingly nonreproducible. To quote a view counter to the above model of friction:
"Many people believe that the friction to be overcome to get something started (static friction)
exceeds the force required to keep it sliding (sliding friction), but with dry metals it is very hard to
show any difference. The opinion probably arises from experiences where small bits of oil or
lubricant are present, or where blocks, for example, are supported by springs or other flexible
supports so that they appear to bind." R. P. Feynman, R. P. Leighton, and M. Sands, The Feynman
Lectures on Physics, Vol. I, p. 12-5, Addison-Wesley, 1964.
10.7 Rolling Friction
Rolling resistance is the force that resists the rolling of a wheel or other circular objects
along a surface caused by deformations in the object and/or surface. Generally the force
of rolling resistance is less than that associated with kinetic friction. Typical values for
the coefficient of rolling resistance are 0.001. One of the most common examples of
rolling resistance is the movement of motor vehicle tires on a road, a process which
generates heat and sound as by-products.
A rolling wheel requires a certain amount of friction so that the point of contact of the wheel
with the surface will not slip. The amount of traction which can be obtained for an auto tire is
determined by the coefficient of static friction between the tire and the road. If the wheel is
locked and sliding, the force of friction is determined by the coefficient of kinetic friction and
is usually significantly less.
Assuming that a wheel is rolling without slipping, the surface friction does no work against the
motion of the wheel and no energy is lost at that point. However, there is some loss of energy
and some deceleration from friction for any real wheel, and this is sometimes referred to as
rolling friction. It is partly friction at the axle and can be partly due to flexing of the wheel
which will dissipate some energy. Figures of 0.02 to 0.06 have been reported as effective
coefficients of rolling friction for automobile tires, compared to about 0.8 for the maximum
static friction coefficient between the tire and the road.
135
10.8 Coulomb friction
Coulomb friction, named after Charles-Augutin de Coulomb, is a model to describe
friction forces. It is described by the equation:
Ff = µ Fn
Where
•
•
•
Ff is either the force exerted by friction, or, in the case of equality, the maximum
possible magnitude of this force.
µ is the coefficient of friction which is an empirical property of the contacting
materials,
Fn is the normal force exerted between the surfaces.
For surfaces at rest relative to each other µ = µs, where µs is the coefficient of static
friction. This is usually larger than its kinetic counterpart. The Coulomb friction may take
any value from zero up to Ff, and the direction of the frictional force against a surface is
opposite to the motion that surface would experience in the absence of friction. Thus, in
the static case, the frictional force is exactly what it must be in order to prevent motion
between the surfaces; it balances the net force tending to cause such motion. In this case,
rather than providing an estimate of the actual frictional force, the Coulomb
approximation provides a threshold value for this force, above which motion would
commence.
For surfaces in relative motion µ = µk, where µk is the coefficient of kinetic friction. The
Coulomb friction is equal to Ff, and the frictional force on each surface is exerted in the
direction opposite to its motion relative to the other surface.
This approximation mathematically follows from the assumptions that surfaces are in
atomically close contact only over a small fraction of their overall area, that this contact
area is proportional to the normal force (until saturation, which takes place when all area
is in atomic contact), and that frictional force is proportional to the applied normal force,
independently of the contact area (you can see the experiments on friction from Leonardo
Da Vinci). Such reasoning aside, however, the approximation is fundamentally an
empirical construction. It is a rule of thumb describing the approximate outcome of an
extremely complicated physical interaction. The strength of the approximation is its
simplicity and versatility – though in general the relationship between normal force and
frictional force is not exactly linear (and so the frictional force is not entirely independent
of the contact area of the surfaces), the Coulomb approximation is an adequate
representation of friction for the analysis of many physical systems.
136
When the surfaces are conjoined, Coulomb friction becomes a very poor approximation
(for example, Scotch tape resists sliding even when there is no normal force, or a
negative normal force). In this case, the frictional force may depend strongly on the area
of contact. Some drag racing tires are adhesive in this way.
The force of friction is always exerted in a direction that opposes movement (for kinetic
friction) or potential movement (for static friction) between the two surfaces. For
example, a curling stone sliding along the ice experiences a kinetic force slowing it down.
For an example of potential movement, the drive wheels of an accelerating car experience
a frictional force pointing forward; if they did not, the wheels would spin, and the rubber
would slide backwards along the pavement. Note that it is not the direction of movement
of the vehicle they oppose, it is the direction of (potential) sliding between tire and road.
The coefficient of friction is an empirical measurement – it has to be measured
experimentally, and cannot be found through calculations. Rougher surfaces tend to have
higher effective values. Most dry materials in combination have friction coefficient values
between 0.3 and 0.6. Values outside this range are rarer, but Teplon, for example, can
have a coefficient as low as 0.04. A value of zero would mean no friction at all. Rubber in
contact with other surfaces can yield friction coefficients from 1.0 to 2.
Examples of kinetic friction:
•
Sliding friction (also called dry friction) is when two objects are rubbing against
each other. Putting a book flat on a desk and moving it around is an example of
sliding friction.
•
Fluid friction is the interaction between a solid object and a fluid (liquid or gas),
as the object moves through the fluid. The drag of air on an airplane or of water on
a swimmer are two examples of fluid friction. This kind of friction is not only due
to rubbing, which generates a force tangent to the surface of the object (such as
sliding friction). It is also due to forces that are orthogonal to the surface of the
object. These orthogonal forces significantly (and mainly, if relative velocity is
high enough) contribute to fluid friction. Fluid friction is the classic name of this
force. This name is no longer used in modern fluid dynamics. Since rubbing is not
its only cause, in modern fluid dynamics the same force is typically referred to as
drag or fluid resistance, while the force component due to rubbing is called skin
friction. Notice that a fluid can in some cases exert, together with drag, a force
orthogonal to the direction of the relative motion of the object (lift). The net force
exerted by a fluid (drag + lift) is called fluid dynamics force (aerodynamic if the
fluid is a gas, or idrodynamic is the fluid is a liquid).
137
10.9 Other types of friction
Rubbing dissimilar materials against one another can cause a build-up of
electrostatics charge, , which can be hazardous if flammable gases or vapours are
present. When the static build-up discharges, explosions can be caused by ignition of
the flammable mixture.
10.10 Reducing friction
10.11 Devices Related to Friction
Devices such as tires, ball bearings, air cushion or roller bearing can change sliding
friction into a much smaller type of rolling friction. Many thermoplastic materials such as
nylon, HDPE and PTFE are commonly used for low friction bearings. They are especially
useful because the coefficient of friction falls with increasing imposed load.
10.12 Lubricants
A common way to reduce friction is by using a lubricant, such as oil, water, or grease,
which is placed between the two surfaces, often dramatically lessening the coefficient of
friction. The science of friction and lubrication is called tribology. Lubricant technology
is when lubricants are mixed with the application of science, especially to industrial or
commercial objectives.
Super lubricity, a recently-discovered effect, has been observed in graphite: it is the
substantial decrease of friction between two sliding objects, approaching zero levels. A
very small amount of frictional energy would still be dissipated.
Lubricants to overcome friction need not always be thin, turbulent fluids or powdery
solids such as graphite and talc; acoustic lubrication actually uses sound as a lubricant.
Another way to reduce friction between two parts is to superimpose micro-scale vibration
to one of the parts. This can be sinusoidal vibration as used in ultrasound-assisted cutting
or vibration noise, known as dither.
10.13 Energy of friction
According to the law of conservation of energy, no energy is destroyed due to friction,
though it may be lost to the system of concern. Energy is transformed from other forms
into heat. A sliding hockey puck comes to rest because friction converts its kinetic energy
into heat.
138
Since heat quickly dissipates, many early philosophers, including Aristotle, wrongly
concluded that moving objects lose energy without a driving force.
When an object is pushed along a surface, the energy converted to heat is given by:
where
Fn is the normal force,
µk is the coefficient of kinetic friction,
x is the coordinate along which the object transverses.
Physical wear is associated with friction. While this can be beneficial, as in polishing, it is
often a problem. As materials are worn away, fit and finish of a object can degrade until it
no longer functions properly.
In the reference frame of the interface between two surfaces, static friction always
does no work, because there is never any displacement. In the same reference frame,
kinetic friction is always in the direction opposite the motion and so does negative
work. However, friction can do positive work in certain inertial frames of reference.
One can see this by placing a heavy box on a rug, then pulling on the rug quickly. In
this case, the box slides backwards relative to the rug, but moves forward relative to
the floor, an inertial frame of reference. Thus, the kinetic friction between the box
and rug accelerates the box in the same direction that the box moves, doing positive
work.
The work done by friction can translate into deformation, wear, and heat that can affect
the contact surface's material properties (and even the coefficient of friction itself). The
work done by friction can also be used to mix materials such as in the technique of
friction welding. Erosion of mating surfaces occurs when frictional forces rise to
unacceptable levels through loss of lubricant for example. Bearing seizure can also occur.
Corrosion can occur when water enters a dry bearing, and exacerbates wear by fretting of
hard corrosion particles caught between the mating surfaces.
Example1
139
Two blocks A and B of weights 1kN and 2Kn respectively are in equilibrium position as
shown in Fig.6.
30o
A
B
F
Fig.10.5
If the coefficient of friction between the two blocks as well as the block B and the floor is
0.3, find the force (P) required to move the block B.
140
Solution:
Given: WA = 1kN; WB = 2kN;
µ = 0.3
R1
T
30o
F1
1kN
(a) Block A
T
R2
30o
F1
P
F2
2kN+ R1
Fig.10.5
(b) Block B
The forces acting on the two blocks A and B are as shown in Fig.10.5 (a) and (b)
respectively. First of all, consider the block A.
Resolving the forces vertically,
R1 + Tsin30o = 1kN
Or
…………… (i)
Tsin30o = 1- R1
And now resolving the forces horizontally,
141
Tcos30o = F1 = µR1 = 0.3R1 ……………………… (ii)
Dividing equation (i) and (ii),
o
tan30 =
1 − R1
Since :
0.3 R1
o
 sin
30  = tan 30o

o
 cos
30 

Therefore:
1.173 R1 = 1 implies
And
R1 = 0.85kN
F1 = µR1 = 0.3 x 0.85 = 0.255 Kn
…………….. (iii)
Now , consider the block B. A little consideration will show that the downward force
of the block A ( equal to R1 ) will also act along the weight of the block. Resolving the
forces vertically,
R2 = 2 + R1 = 2 + 0.85 = 2.85kN
F2 = µR2 = 0.3 x 0.85 = 0.885 kN
…………….. (iv)
And now resolving the forces horizontally,
P = F1 + F2 = 0.255 + 0.885 = 1.11KN
Answer.
Example2
A body of weight 500N is lying on a rough plane inclined at an angle of 25o with the
horizontal. It is supported by an effort (P) parallel to the plane. Determine the minimum
and maximum value of P, for which the equilibrium can exist, the angle of friction is 20o.
Answer: Minimum value of P = 46.4N; Maximum value of P = 376.2N
Solution:
An object of weight 100N is kept in position on a plane inclined 30o to the horizontal by a
horizontally applied force (F). The coefficient of friction of the surface of the inclined
plane is 0.25. Determine the minimum magnitude of the force (F).
Solution:
Given: W = 100N; α = 30o ; µ = 0.25 = tanφ or φ = 14.o 2’
We know that the minimum magnitude of the force to keep the object in position ( when
it is at the point of sliding downwards),
F = W tan ( α-φ) = 100 tan (30o - 14.o 2’) N
142
= 100 tan 15.o 58’ = 100 x 0.2852
= 28.52N Answer.
Exercise
Find the horizontal force, required to drag a body of weight 100N along the horizontal
plane. If the plane, when gradually raised up to 15o , the body will begin to slide.
Answer: 26.79N
Exercise
A body of weight 50N is hauled along a rough horizontal plane by a pull of 18N acting at
an angle of 14o with the horizontal. Find the coefficient of friction.
Answer: 0.383
Exercise
What do you understand by the term friction? Explain clearly why it comes in to play?
143
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK11
11.1 Fundamentals and application of friction (continues)
11.2Coefficient of static friction
The coefficient of static friction is defined as the ratio of the maximum static friction
force (F) between the surfaces in contact to the normal (N) force. The coefficient of
kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces
in contact to the normal force:
Both static and kinetic coefficients of friction depend on the pair of surfaces in contact;
their values are usually approximately determined experimentally. For a given pair of
surfaces, the coefficient of static friction is usually larger than that of kinetic friction; in
some sets the two coefficients are equal, such as teflon-on-teflon.
The friction force is directed in the opposite direction of the resultant force acting on a
body. In the case of kinetic friction, the direction of the friction force may or may not
match the direction of motion: a block sliding atop a table with rectilinear motion is
subject to friction directed along the line of motion; an automobile making a turn is
subject to friction acting perpendicular to the line of motion (in which case it is said to be
'normal' to it). A motionless body is subject to static friction. The direction of the static
friction force can be visualized as directly opposed to the force that would otherwise
cause motion, were it not for the static friction preventing motion. In this case, the friction
force exactly cancels the applied force, so the net force given by the Vector sum, equals
zero. It is important to note that in all cases, Newton's first law of motion holds.
While it is often stated that the coefficient of friction (COF) is a "material property," it is
better categorized as a "system property." Unlike true material properties (such as
conductivity, dielectric constant, yield strength), the COF for any two materials depends
on system variables like temperature, velocity, atmosphere and also what are now
popularly described as aging and deaging times; as well as on geometric properties of the
interface between the materials. For example, a copper pin sliding against a thick copper
plate can have a COF that varies from 0.6 at low speeds (metal sliding against metal) to
below 0.2 at high speeds when the copper surface begins to melt due to frictional heating.
The latter speed, of course, does not determine the COF uniquely; if the pin diameter is
increased so that the frictional heating is removed rapidly, the temperature drops, the pin
remains solid and the COF rises to that of a 'low speed' test.
144
Fig.11.
1
Block on a ramp (top) and corresponding free body diagram of just the block (bottom).
11.3 The normal force
The normal force is defined as the net force compressing two parallel surfaces together;
and its direction is perpendicular to the surfaces. In the simple case of a mass resting on a
horizontal surface, the only component of the normal force is the force due to gravity,
where N = mg. In this case, the magnitude of the friction force is the product of the mass
of the object, the acceleration due to gravity, and the coefficient of friction. However, the
coefficient of friction is not a function of mass or volume; it depends only on the material.
For instance, a large aluminum block has the same coefficient of friction as a small
aluminum block. However, the magnitude of the friction force itself depends on the
normal force, and hence the mass of the block.
If an object is on a level surface and the force tending to cause it to slide is horizontal, the
normal force N between the object and the surface is just its weight, which is equal to its
mass multiplied by the acceleration due to earth's gravity, g. If the object is on a tilted
surface such as an inclined plane, the normal force is less, because less of the force of
gravity is perpendicular to the face of the plane. Therefore, the normal force, and
ultimately the frictional force, is determined using vector analysis, usually via a free body
diagram. Depending on the situation, the calculation of the normal force may include
forces other than gravity.
145
11.4 Frequent mistake
Occasionally it is maintained that µ is always < 1, but this is not true. While in most
relevant applications µ < 1, a value above 1 merely implies that the force required to slide
an object along the surface is greater than the normal force of the surface on the object.
For example, silicone rubber or acrylic rubber-coated surfaces have a coefficient of
friction that can be substantially larger than 1.
11.5 Engineering coefficients of friction
Material1 Material2 static friction lubricated
Steel
Steel
0.6
0.16
Teflon
Teflon
0.04
0.04
Table 11.2
11.6 Coefficients of Friction
Friction is typically characterized by a coefficient of friction which is the ratio of
the frictional resistance force to the normal force which presses the surfaces
together. In this case the normal force is the weight of the block. Typically there is a
significant difference between the coefficients of static friction and kinetic friction.
Fig.11.3
Note that the static friction coefficient does not characterize static friction in
general, but represents the conditions at the threshold of motion only.
146
11.7 Normal Force
Frictional resistance forces are typically proportional to the force which presses
the surfaces together. This force which will affect frictional resistance is the
component of applied force which acts perpendicular or "normal" to the
surfaces which are in contact and is typically referred to as the normal force. In
many common situations, the normal force is just the weight of the object
which is sitting on some surface, but if an object is on an incline or has
components of applied force perpendicular to the surface, then it is not equal to
the weight.
Fig.11.4
The above cases are the common ly encountered situations for objects at rest or
in straight line motion. For curved motion, there are cases like a car on a
banked curve where the normal force is determined by the dynamics of the
situation. In that case, the normal force depends upon the speed of the car as
well as the angle of the bank.
11.8 The microscopic origin of friction forces
Friction is weird. In particular, we need to explain
(i)
why friction forces are independent of the contact area
(ii)
why friction forces are proportional to the normal force.
Coulomb grappled with these problems and came up with an incorrect
explanation. A truly satisfactory explanation for these observations was only
found 20 years or so ago.
To understand friction, we must take a close look at the nature of surfaces.
Coulomb/Amonton friction laws are due to two properties of surfaces:
147
(1) All surfaces are rough;
(2) All surfaces are covered with a thin film of oxide, an adsorbed
layer of water, or an organic film.
Surface roughness can be controlled to some extent – a cast surface is usually very
rough; if the surface is machined the roughness is somewhat less; roughness can
be reduced further by grinding, lapping or polishing the surfaces. But you can’t
get rid of it altogether. Many surfaces can be thought of as having a fractal
geometry. This means that the roughness is statistically self-similar with length
scale – as you zoom in on the surface, it always looks the same.
Fig.11.5
Of course no surface can be truly fractal: roughness can’t be smaller than the size
of an atom and can’t be larger than the component; but most surfaces look fractal
over quite a large range of lengths. Various statistical measures are used to
quantify surface roughness, but a discussion of these parameters is beyond the
scope of this course.
Now, visualize what the contact between two rough surfaces looks like. The
surfaces will only touch at high spots (these are known in the trade as `asperities’)
on the two surfaces. Experiments suggest that there are huge numbers of these
contacts (nobody has really been able to determine with certainty how many there
actually are). The asperity tips are squashed flat where they contact, so that there
is a finite total area of contact between the two surfaces. However, the true
contact area (at asperity tips) is much smaller than the nominal contact area.
148
N
Nominal contact area Anom
True contact
area Atrue
Fig.11.6
The true contact area can be estimated by measuring the surface roughness, and
then calculating how the surfaces deform when brought into contact. At present
there is some uncertainty as to how this should be done – this is arguably the most
important unsolved problem in the field. The best estimates we have today all
agree that:
The true area of contact between two rough surfaces is proportional to the normal
force pressing them together.
At present, there is no way to measure or calculate the contact C accurately.
This is true for all materials (except for rubbers, which are so compliant that the
true contact area is close to the nominal contact area), and is just a consequence of
the statistical properties of surface roughness. The reason that the true contact
area increases in proportion to the load is that as the surfaces are pushed into
contact, the number of asperity contacts increases, but the average size of the
contacts remains the same, because of the fractal self-similarity of the two
surfaces.
Finally, to understand the cause of the Coulomb/Amonton friction law, we need to
visualize what happens when two rough surfaces slide against each other.
149
Surface film,
shear strengthτ0
Fig.11.7
Each asperity tip is covered with a thin layer of oxide, adsorbed water, or grease.
It’s possible to remove this film in a lab experiment – in which case friction
behavior changes dramatically and no longer follows Coulomb/Amonton law –
but for real engineering surfaces it’s always present.
The film usually has a low mechanical strength. It will start to deform, and so
allow the two asperities to slide past each other, when the tangential force per unit
area acting on the film reaches the shear strength of the film
.
The tangential friction force due to shearing the film on the surface of all the
contacting asperities is therefore
Combining this with the earlier result for the true contact area gives
Thus, the friction force is proportional to the normal force. This simple argument
also explains why friction force is independent of contact area; why it is so
sensitive to surface films, and why it can be influenced (albeit only slightly) by
surface roughness.
Exercise
150
A man is walking over a dome of 10 m radius. How far he can descend from the top of
the dome without slipping? .take the coefficient of friction between the surface of the
dome and shoes of the man as 0.6.
Answer:
1.413
Exercise
A force of 250 N pulls a body of weight 500 N up an inclined plane, the force being
applied parallel to the plane. If the inclination of the plane to the horizontal is 15o , find
the coefficient of friction.
Answer: 0.25
151
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK12
12.1Fundamentals and application of friction (continues)
12.2 Angle of repose
Angle of repose (sometimes incorrectly confused with the 'Angle of Internal Friction') is
an engineering property of granular materials. The angle of repose is the maximum angle
of a stable slope determined by friction, cohesion and the shapes of the particles.
When bulk granular materials are poured onto a horizontal surface, a conical pile will
form. The internal angle between the surface of the pile and the horizontal surface is
known as the angle of repose and is related to the density, surface area, and coefficient of
friction of the material. Material with a low angle of repose forms flatter piles than
material with a high angle of repose. In other words, the angle of repose is the angle a pile
forms with the ground.
Fig.12.0
Angle of repose
The particular angle of inclination of the plane is known as the angle of repose. If the
plane is inclined at an angle greater than the angle of repose the body will slide down; if
the inclination is less than the angle of repose the body will remain at rest.
12.3 Applications of theory
The angle of repose is sometimes used in the design of equipment for the processing of
particulate solids. For example, it may be used to design an appropriate hopper or silo to
store the material. It can also be used to size a conveyor belt for transporting the material.
It can also be used in determining whether or not a slope (of a stockpile, or uncompacted
gravel bank, for example) will likely collapse; the talus slope is derived from angle of
repose and represents the steepest slope a pile of granular material will take. This angle of
repose is also crucial in determining the correct calculus of stability in vessels.
152
12.4 Measurement
There are numerous methods for measuring angle of repose and each produces slightly
different results. Results are also sensitive to the exact methodology of the experimenter.
As a result, data from different labs is not always comparable.
An alternative measurement, useful for many of the same purposes, is testing with a
specialized instrument called a shear cell.
12.5 Exploitation by antlion larvae
Fig. 12.1
Sand pit trap of the antlion
The larva of the antlion traps small insects such as ants by digging a conical pit in loose
sand, such that the slope of the walls is very close to the angle of repose for the sand.
Thus, when a small insect blunders into the pit, its weight causes the sand to collapse
below it, drawing the ant toward the center where the antlion larva lies in wait. The
antlion larva assists this process by vigorously flicking sand out from the center of the pit
when it detects a disturbance, undermining the pit walls and causing them to collapse
toward the center, bringing the prey with them.
The angle of repose plays a part in several topics of technology and science, including:
Example 1.
The normal reaction between a sliding crate and the floor is 900 N. determine the
frictional resistance to sliding if the coefficient of friction is 0.33.
Solution.
F=µ N
where µ = 0.33 (no units) and N = 900 N
F = 0.33 x 900 = 300 N
I.e the frictional resistance to sliding is 300 N.
Example 2.
153
A milling-machine table and component have a mass of 160 kg. if the coefficient of
friction between the table and saddle is 0.02, determine the frictional resistance to motion.
Solution.
F=µ N
Where N , the normal reaction, is equal to the downward force being exerted by the table
and component.
N = mg = 160 kg x 9.81 m/s2 = 1569.6 N
µ = 0.02 (no units)
F = 0.02 x 1569.6 N = 31.39 N
I.e. the frictional resistance to motion is 31.4 N.
Example 3.
In an experiment to determine the coefficient of friction, it was found that a force of 25 N
was required to just move the sliding object. If the normal reaction between the slider and
the horizontal friction plane was 100 N, determine the coefficient of friction.
Solution.
F=µ N
implies that
µ = F/N
Where F = 25 N and N = 100 N
µ = 25N/100N = 0.25
I.e. the coefficient of friction is 0.25
Exercise
The brakes on a motor car are applied suddenly, causing the wheels to lock and the car to
skid. If the mass of the car is one tone (1000 kg) and the coefficient of friction between
the tyres and the load surface is 0.4, determine the retarding force on the car.
154
Answer: the retarding force is 3924 N
Exercise
The sliding member on a machine tool has a mass of 200 kg. if the coefficient of friction
is 0.05, determine the total force required to give the sliding member a uniform
acceleration of 2 m/s2..
Answer: the total force required is 498.1 N
155
MECHANICAL ENGINEERING. SCIENCE (STATICS) – MEC111
WEEK13
13.0 Frames, Struts and Ties
13.0 Frame and structures
13.1 Simple frames
Any structure, be it a building, a machine tool, a lifting device, or even a human being is
build around a frame. The frame can be simple or complex. As shown below.
(a) Simple roof frame
(b) complex roof
Fig.13-1
When designing a framed structure, it is necessary to analyses the forces being exerted on
each of the members which make up the frame. This analysis is done to avoid:
(a) Making the frame too strong, thus wasting materials.
(b) Making the frame too weak, thus causing it to collapse.
There are three types of members in any frame:
(i) struts – these carry external compressive forces, as shown in fig.(a )
(ii) ties -- these carry external tensile forces, as shown in fig.(b )
(iii) redundant members – these do not, in theory, carry any tensile or compressive
redundant members are used to prevent the frame from twisting, and need not
be present in every frame.
INTERNAL FORCE
EXTERNAL FORCE
TIE
Fig.(a)
TIE
156
EXTERNAL FORCE
Fig.13.2
INTERNAL FORCE
EXTERNAL FORCE
EXTERNAL FORCE
STRUT
STRUT
Fig.(b)
Fig.13.3
13.2 Struts and Ties
You must understand the meaning of ‘struts, and ‘ties’ as these are always
mentioned in examinations. All structures have forces acting on them. You should
have an understanding of tensile, compressive and shear forces (see previous sheet).
The part of the structure that has a tensile force acting on it is called a TIE and the
part that has a compressive force acting on it is called a STRUT.
WALL
The beam is held in
position by a steel rod. The
weight of the beam is
stretching the rod (tensile
force).
ROOF
The roof beams are under
pressure from the weight
of the tiles on the roof
(compressive force).
The floor beam is being
stretched (tensile force).
(a)
(b)
Fig.13.4
157
FLAGPOLE
The wires on either side of
the flagpole are being
stretched (tensile force).
Why is the pole under a
compressive force ?
(c)
In the diagram opposite, forces act across
the entire length of the beam (it bends
because of the 'ton' weight). When a
structure bends like this it is in tension as
it is being stretched.
1. Draw an example of a structure and
identify which part(s) are in tension and
which are in compression.
2. Explain the difference between tension
and compression.
Fig.13.5
13.3 Example: Frame (non-truss, rigid body) Problems
In this example, we will examine a problem in which it will initially seem that there are
too many unknowns to allow us to determine the external support forces acting on the
structure, however, by a slight variation of our approach we will find that we can
determine all the unknowns. The problem then is this, for the structure shown in Diagram
1 below, determine the external support forces acting on the structure, and additionally,
determine the force in member CF. Our usual procedure to find the external support
reactions consists of three basic steps.
•
I. Draw a Free Body Diagram of the entire structure showing and labeling all external
load forces and support forces, include any needed dimensions and angles.
II. Resolve (break) all forces into their x and y-components.
•
•
III. Apply the Equilibrium Equations (
and solve for the unknown forces.
•
)
We note that the structure is composed of members ABC, DEF, DE, and CF. These
members are pinned together at several points as shown in Diagram 1. A load of 6000 lb.
is acting on member DEF at point E, and a load of 3000 lb. is applied at point F. These
forces are already shown by the downward arrows. We next look at the forces exerted on
the structure by the supports. Since each support is a pinned joint, the worst case we could
have is an unknown x and y-force acting on the structure at each support point. We also
must choose directions for the x and y support forces. In some problems the directions of
158
the support forces are clear from the nature of the problem. In other problems the
directions the support forces act is not clear at all. However, this is not really a problem.
We simple make our best guess for the directions of the support reactions. If our guess is
wrong, when we solve for the value of the support forces, that value will be negative.
Fig.13.6
This is important. A negative value when solving for a force does not mean the force
necessarily acts in the negative direction, rather it means that the force acts in the
direction OPPOSITE to the one we initially chose.
Step 1: Free Body Diagram (FBD).
In the FBD (Diagram 2), we have shown unknown x and y support forces acting on the
structure at pinned support points A and D, (Ax, Ay, Dx, Dy). If we think ahead somewhat,
we realize that there could be a problem. We have four unknown forces supporting the
structure, but there are only three equations in our Equilibrium Conditions,
.
159
Fig.13.7
Normally, one can not solve for more unknowns then there are independent
equations. Our first reaction should be to see if we can draw a better FBD. Perhaps
we can replace the two unknowns at either point A or D by one unknown acting at a
known angle (which is possible if we have a single axial member acting at the
support point). However in this case both member ABC and member DEF are nonaxial members and the forces in them (and on them from the wall) do not act along
the axis of the member. Thus, we already have the best FBD possible - we can not
reduce the number of external unknowns acting on the structure.
At this point we will simply continue with our normal analysis procedure and see what
results.
Step II: Resolve any forces not in the x or y-direction into x and y-components. (All
forces are already in either the x or y-direction.
Step III. Apply the Equilibrium conditions.
(Here we sum the x-forces, keeping track of their
direction signs, forces to right, +, to left, -)
(Sum of y-forces, including load forces. Again keeping track of direction signs.)
Sum of Torque about a point. We choose point D. Forces Ay, Dx and Dy do not
produce torque since their lines of action pass through point A. Thus, the torque
equation has only one unknown, Ax. We solve for Ax, and then use it in the sum of
forces in the x-direction equation to find the unknown, Dx . And if we do so, we find:
Ax = +18000 lb. Dx = +18000 lb. (The positive signs indicate we initially chose the
correct direction for the forces.) However, please notice that while we found Ax and
160
Dx, we can not find Ay and Dy. There are still two unknowns in the y-equation and
not enough information to determine then at this point. Thus, analysis of the
structure as a whole has enabled us to determine several of the external support
forces, but not all of them. What now?
First, an overview. There are problems for which the static equilibrium conditions are not
enough to enable one to solve the problem. They are called Statically Indeterminate
Problems, and we will be considering these a bit later. Then there are problems which, on
first glance, appear to be statically indeterminate, but are not. That is the case here. To
find the remaining unknown support forces (and at the same time, determining the force
in member CF), we will now take out a member of the structure and apply our statics
analysis procedure to the selected member of the structure (rather than the entire
structure). We will select member ABC to analyze. (See Diagram 3)
Step 1: Free Body Diagram (FBD)
In Diagram 3, we have drawn a FBD of member ABC, showing and labeling all forces
external to member ABC which act on it. That is at point A we have the forces of the wall
acting on ABC, at point B we have an axial force on ABC due to member BE, and at
point C we have a axial force on ABC due to the member CF. Both member BE and CF
are axial members and so we know the directions their forces act - along their axis. We do
have to guess if they push or pull on member ABC, and we have chosen those directions
as shown, (if the direction chosen is wrong, the force's value will be negative when we
solve). We also are happy with the FBD as it has only three unknowns acting on the
member, which indicates that we should be able to solve completely for the unknowns.
Fig.13.8
Step 2: Resolve forces into x/y components.
Here we have resolved force CF into its horizontal and vertical components as shown in
Diagram 4. All other forces are already in x or y-direction.
Fig.13.9
Step 3. Apply the Equilibrium conditions.
161
We now solve the first equation for CF, then use that value in the last equation to find BE,
and use both values in the middle equation to find Ay, giving us: CF = 30,000 lb., BE =
36,000 lb., Ay = -12, 000 lb. Please note that the value of Ay is negative, which indicates
the direction we chose was incorrect. Ay acts downward rather than in the positive ydirection we initial chose. Additionally, we now can return to the equations for the entire
structure (see below), and knowing the value for Ay, we can use it in the y-forces equation
to solve for the value of Dy.
Equilibrium Conditions for entire structure (from first part of problem)
•
•
•
From the y-forces equation: (-12,000 lb.) - Dy -6000 lb. -3000 lb = 0; Solving Dy =
-21,000 lb.
Fig.13.10
Once again, the negative sign indicates we selected the wrong direction for Dy, rather than
acting downward it actually acts upward. See Diagram 5 for final force values and
directions.
Ax = 18000 lb.
Ay = 12,000 lb. Dx = 18000 lb. Dy = 21,000 lb. CF = 30,000 lb. BE = 36,000 lb.
162
QUESTIONS ON FRAME STRUCTURES
Question
(a) State three assumptions made to simplify the analysis of statically determinate structures.
(b) Briefly explain the meaning of the following terms:
(i)
a tie – member
(ii)
a compression – member
(iii)
a cable
(iv)
a beam
( c) State the equilibrium conditions of a free – body under static equilibrium.
(d) For the framework shown in the figure below, find analytically the magnitude and nature
of the
Forces in members 1; 2; and 3. All members are of the same length, determine also
the
Supports reactions.
50kN
30kN
1
2
3
60o
Fig.13.11
163
60o
QUESTION ON FRAME STRUCTURES
Question
(a) Distinguish between an efficient structure and a deficient one; using one example to
illustrate each type.
(b) Briefly explain the followings:
(i) Statically determinate structures.
(ii) Statically indeterminate structures.
(iii)Built- in support.
(iv) Roller bearing support.
( c ) The framework shown in the figure below carries masses of 2 and 3 tonnes as
shown.
Find analytically the magnitude and nature of the forces in all members of the
frame and
State the reactions at the supports.
A
A
45o
A
45o
G
E
F
D
B
o
45
C
2t
3t
Fig.13.12
164
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 14
14.1 Structural Forces
Study the diagram of a computer
desk. Each member of the
structure is under some type of
force.
PART A: Is in tension because
the weight of the computer is
stretching it.
Fig.14.1
PART B: Is under compression
because the weight of the
computer unit and the members
above that make up the desk, are
pushing downwards and
compressing it.
PART C and B: This is the same
member but on the inside
compression is taking place and
on the outside it is being stretched
(under tension).
165
Fig.14.2
MONITOR WEIGHS
DOWN AND
STRETCHES SUPPORT
WEIGHT OF COMPUTER
UNIT COMPRESSES THE
SUPPORT BENEATH
THE SHELF
VERTICAL SUPPORT
BENDS INWARDS
BECAUSE OF
WEIGHT OF
COMPUTER. SIDE 'D'
IS STRETCHED
WHILST SIDE 'C' IS
COMPRESSED
1. Draw a piece of furniture and label the struts and ties. With the aid of
more detailed diagrams explain the forces acting on each of the important
members.
14.2 More forces in action
The bridge below is a common type called a Box Girder Bridge. These are usually made
of steel with all the members (parts) welded, bolted or held together with rivets. Usually
they are manufactured in a factory and transported to the site and dropped into position
by a large crane. The triangular shapes give the bridge immense strength and for short
spans this type of bridge is ideal.
Fig.14.3
166
Triangulation distributes the weight of any vehicle or pedestrian crossing the bridge. The
weight is distributed through all the members and so the bridge can cope with large
weights. This type of bridge is favoured by the British Army. The Royal Engineers have
transportable bridges like the one above that can be dismantled and transported anywhere
in the world and reassembled. They are bolted together and are semi-permanent
structures.
Fig.14.4
When a vehicle crosses the bridge each member experiences some type of force. The
diagram shows that the member that the car rests on is under tensile force (in tension) as
it stretches under the weight of the car. As the bridge bends, the top member is compress
(under a compressive force).
1. Draw a diagram that shows the forces the supporting pieces (triangles) are under
? HINT: make a model from art straws and sellotape. This may help you see the
forces in action as you add weights to the top.
167
The bracket holding up the hanging basket
is made of steel. It has been made by
heating up the steel until ‘red’ hot and then
bending it at 90 degrees.
1. What is the force exerting on the bracket
at point ‘A’?
2. What type of force is the chain under?
3. What type of force are the wires holding
the plant pot under?
4. What force is acting on the wall at point
‘B'?
5. As more weight is added to the plant pot
the bracket begins to bend too much. How
could the bracket be strengthened?
6. How could the bracket be fixed to the
wall? Use diagrams to illustrate your
answer.
Fig.14.5
ANSWERS TO QUESTIONS 1 TO 4
ANSWERS TO QUESTIONS 5 AND 6
Fig.14.6
A steel bracket is added. Two holes
are drilled, one at the top and
bottom. Countersunk screws and
168
wall plugs are used to fix the
strengthened bracket to the wall.
Two screws would be used.
QUESTION
a.
b.
(i)
(ii)
c.
Define a frame work and illustrate by means of diagram the three types of
support in a frame work.
Define the following terms.
Statistically determinate structure.
Statistically indeterminate structure.
Fig.14.7
For a frame works shown in fig. (4c), determine analytically the magnitude and nature
of forces in the members. (1), (2) (3) and (4).
Examples
169
14.3 Frames (non-truss, rigid body structures)
We are now ready to begin looking at somewhat more involved problems in statics. As
we do so there will be a number of concepts we need to keep in mind (and apply) as we
approach these problems. We will begin by looking at problems involving rigid bodies.
This simply means that at this point we will not be concerned with the fact that a body (or
member of a structure) may actually bend or deform (change length) length under the
applied loads. Bending and deformation effects will be considered in later materials. At
this point in the course we will also ignore the weight of the members, which are often
small compared to the loads applied to the structures.
As we begin to analyze structures there are two important considerations to keep in mind,
especially as we draw our free body diagrams. The first concern will be Structure
Supports. Different types of supports will result in different types of support forces
(reactions) acting on the structure. For example, a roller or bearing can only be placed in
compression, thus the force it will exert on a structure will be a normal or perpendicular
force only, while a hinged or pinned support point may exert both a horizontal and
vertical force. [Or to be more specific, a hinged or pinned support (or joint) will exert a
support force acting at a particular angle, this force may then always be broke into an x
and y-component. The net effect is that a hinged or pinned support may be replaced by x
and y support forces.] See Diagram 1 below.
Fig.14.8
In Diagram 1 we have shown a horizontal beam supported at point A by a roller and at
point C by a pinned support. Diagram 1a is the Free Body Diagram of the beam with the
roller and the pinned joint now replaced by the support forces which they apply on the
beam. The roller applies the vertical force Ay and the pinned support applies the forces Cx
and Cy. If we knew the value of the load, we could apply statics principles to find the
actual value of the support forces. We shall do this process in great detail later for a
somewhat more complex example.
170
Diagram 2 below shows examples of supports and the types of forces and/or torque which
they may exert on a structure.
Fig.14.9
The second important concept to keep in mind as we begin looking at our structure is the
type of members the structure is composed of - Axial or Non-Axial Members. (The
importance of this will be seen in more detail when we look at our first extended
example.)
In equilibrium or statics problems, an Axial Member is a member which is only in simple
tension or compression. The internal force in the member is constant and acts only along
the axis of the member. A simple way to tell if a member is an axial member is by the
number of Points at which forces act on a member. If forces (no matter how many) act at
only Two Points on the member - it is an axial member. That is, the resultant of the
forces must be two single equal forces acting in opposite directions along axis of the
member. See Diagram 3.
171
Fig.14.10
A Non-Axial Member is a member which is not simply in tension or compression. It may
have shear forces acting perpendicular to the member and/or there may be different values
of tension and compression forces in different parts of the member. A member with forces
acting at More Than Two Points (locations) on the member is a non-axial member. See
Diagram 4.
Fig.14.11
EXAMPLES: To show the application of the concepts discussed above and of our
general statics problem-solving technique, we will now look in careful detail at several
statics problems.
In Example 1 we will concentrate on finding the values of the external support forces
acting on the structure. See Example 1
In Example 2, we will examine a relatively straight forward problem which points out
several features concerning torques and beam loading. See Example 2
In Example 3, we will see how both the external support reactions and also the
internal forces in a member of the structure may be found. See Example 3
14.4 Additional Examples:
172
The following additional examples all demonstrate different aspects and features of a
variety of non-truss, or frame problems.
Example 1: Frames (non-truss, rigid body) problems
In this first example, we will proceed very carefully and methodically. It is important to
get the method and concepts we need to keep in mind firmly established.
In this problem we wish to determine all the external support forces (reactions) acting on
the structure shown in Diagram 1 below. Once again our procedure consists basically of
three steps.
1. Draw a Free Body Diagram of the entire structure showing and labeling all
external load forces and support forces, include any needed dimensions and
angles.
2. Resolve (break) all forces into their x and y-components.
3. Apply the Equilibrium Equations (
4. and solve for the unknown forces.
)
Step 1: Free Body Diagram (FBD). Making the FBD is probably the most important part
of the problem. A correct FBD usually leads to a quick solution, while an inaccurate FBD
can leave a student investing frustrating unsuccessful hours on a problem. With this in
mind we will discuss in near excruciating detail the process of making a good FBD.
We note that the structure is composed of members ABC, and CD. These two members
are pinned together at point C, and are pinned to the wall at points A and D. Loads of
4000 lb. and 2000 lb. are applied to member ABC as shown in Diagram1
Fig.14.12
173
In our example, the load forces are already shown by the downward arrows. We next look
at the forces exerted on the structure by the supports. Since each support is a pinned joint,
the worst case we could have is an unknown x and y-force acting on the structure at each
support point. We also must choose directions for the x and y support forces. In some
problems the directions of the support forces are clear from the nature of the problem. In
other problems the directions the support forces act is not clear at all. However, this is not
really a problem. We simple make our best guess for the directions of the support
reactions. If our guess is wrong, when we solve for the value of the support forces, that
value will be negative. This is important. A negative value when solving for a force
does not mean the force necessarily acts in the negative direction, rather it means
that the force acts in the direction OPPOSITE to the one we initially chose.
Thus, in our first FBD on the right (Diagram 2), we have shown unknown x and y support
forces acting on the structure at each support point.
Fig.14.13
This is an accurate FBD, but it is not the best. The difficulty is that for our problem, we
have three equilibrium conditions (
), but we have
four unknowns (Ax, Ay, Dx, Dy) in this FBD. And as we are well aware, we can not solve
for more unknowns than we have independent equations.
We can draw a better FBD by reflecting on the concept of axial and non-axial members.
Notice in our structure that member ABC is a non-axial member (since forces act on it at
more than two points), while member CD is an axial member (since if we drew a FBD of
member CD we would see forces act on it at only two points, D and C). This is important.
Since CD is an axial member the force acting on it from the wall (and in it) must act along
the direction of the member. This means that at point D, rather than having two unknown
forces, we can draw one unknown force acting at a known angle (force D acting at angle
of 37o, as shown in Diagram 3).
This means we have only three unknowns, Ax, Ay, and D. In Diagram 3, we have also
completed Step II, breaking any forces not in the x or y-direction into x and ycomponents. Thus, in Diagram 3, we have shown the two components of D (which act at
174
37o), D cos 37o being the x-component, and D sin 37o being the y-component. [Please
notice that there are not three forces at point D, there is either D acting at 37o or its two equivalent
components, D cos 37o and D sin 37o. In Diagram 3 at this point we really should cross out the D force,
which has been replaced by its components.]
Fig.14.14
Now before we proceed with the final step and determine the values of the support
reactions, we should deal with several conceptual questions which often arise at this
point. First, why can't we do at point A what we did at point D, that is put in one
force acting at a known angle. Member ABC is a horizontal member, doesn't the
wall just push horizontally on member ABC, can't we just drop the Ay force? The
answer is NO, because member ABC is not an axial member, it is not simply in
compression or tension, and the wall does not just push horizontally on member
ABC (as we will see in our solution). Thus the best we can do at point A is unknown
forces Ax and Ay.
A second question is often, what about the wall, aren't there forces acting on the wall
that we should consider? Well, yes and no. YES, there are forces acting on the wall
(as a matter of fact they are exactly equal and opposite to the forces acting on the
members, in compliance with Newton's Third Law). But NO we should not consider
them, because we are making a FBD of the STRUCTURE, not of the wall, so we
want to consider forces which act on the structure due to the wall, not forces on the
wall due to the structure.
Now Step III. Apply the Equilibrium conditions.
175
Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left,
Sum of y-forces, including load forces. Again keeping track of direction signs.
Sum of Torque about a point. We choose point A. Point D is also a good point to sum
torque about since unknowns act through both points A and D, and if a force acts through
a point, it does not produce a torque with respect to that point. Thus our torque equation
will have less unknowns in it, and will be easier to solve. Notice that with respect to point
A, forces Ax, Ay, and D sin 37o do not produce torque since their lines of action pass
through point A. Thus in this problem the torque equation has only one unknown, D. We
can solve for force D, and then use it in the two force equations to find the other
unknowns, Ax and Ay. (Completing the calculations, we arrive at the following answers.)
D = +7500 lb. Ax = +6000 lb. Ay = +1500 lb.
Note that all the support forces we solved for are positive, which means the directions we
choose for them initially are the actual directions they act. We have now solved our
problem. The support force at point D is 7500 lb. acting at 37o. The support forces at A
can be left as the two components, Ax = 6000 lb. and Ay = 1500 lb., or may be added (as
vectors) obtaining one force at a known angle, as shown in Diagram 4.
Fig.14.15
Thus the force at point A is 6185 lb. acting at 14o, as shown. This information would help
us purchase the correct size hinge (able to support 6185 lb. at A, and able to support 7500
lb. at D), or estimate if the wall is strong enough to support the structure. All very useful
and interesting information.
Example 2: Frame (non-truss, rigid body) Problems
176
Again in this second example, we will continue to proceed very carefully and
methodically. This second example is reasonably straight forward, but points out
some aspects of axial/non-axial forces, and torque. In this second example
(Diagram 1, below) we will want to determine the value of the external support
reactions. The general procedure to find the external support reactions consists of
three basic steps.
•
•
•
I. Draw a Free Body Diagram of the entire structure showing and labeling all
external load forces and support forces, include any needed dimensions and
angles.
II. Resolve (break) all forces into their x and y-components.
III. Apply the Equilibrium Equations (
and solve for the unknown forces.
)
We observe, as shown in Diagram 1, that the structure is composed of members AB and
BCD. These members are pinned together at point B, and are pinned to the floor at points
A and D. Additionally, point B supports a pulley with which a person is hoisting a 200 lb.
load. Member BCD has a weight of 160 lb., which may be considered to act at the center
of member BCD.
Fig.14.16
Step 1: Free Body Diagram (FBD).
We now proceed normally, that is we first draw our FBD (Diagram 2), showing and
labeling all loads and support reactions acting on the structure. As we do so we will note
several items: that member AB is an axial member (only in tension or compression), and
that the person / load / pulley combination at point B produces a net 400 lb. downward
load at point B. (This results since both sides of the rope have 200 lb. force in them one side due to the load, and the other side due to the pull of the person. Point B
must support both the load and the pull of the person which results in a total force of
400 lb. acting on point B)
In the FBD (Diagram 2), at point A we have shown one unknown support force 'A' acting
at a known angle (37o). We can do this at point A since we know member AB is an axial
member. In an axial member the force is along the direction of the member, thus the floor
177
must exert a force on the member also along the direction of the member (due to equal
and opposite forces principle). However, at point D, since member D is a non-axial
member, the best we can do is to show an unknown Dx and Dy support forces acting on
the structure at point D
[We simply make our best guess for the directions of the support reactions. If our
guess is wrong, when we solve for the value of the support forces, that value will be
negative - indicating our original direction was incorrect].
In Diagram 2, we have also included Step II: Resolve (break) any forces not in the x or
y-direction into x and y-components. Thus, we have shown the two components of A
(which act at 37o) - A cos 37o being the x-component, and A sin 37o being the ycomponent. .
Fig.14.17
Step III. Apply the Equilibrium conditions.
(Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left,
-)
(Sum of y-forces, including load forces. Again keeping track of direction
signs.)
Sum of Torque about a point. We chose Point D to calculate torque. Since two unknown
forces (Dx, Dy) are acting at Point D, and if a force acts through a point, it does not
produce a torque with respect to that point; thus our torque equation will have fewer
unknowns in it, and will be easier to solve. We now proceed through the structure,
looking a each force and calculating the torque due to that force with respect to the chosen
Point D, and entering it in our torque equation (above) with the correct sign (+ for
counterclockwise acting torque, - for clockwise acting torque). In this example, we must
be careful to use the correct distance in the torque relationship - Torque = Force x
Perpendicular Distance. (See Torque Review if needed.)
178
Fig.14.18
Finally, solving for our unknowns we obtain: A = +343 lb. Dx = +274 lb. Dy = +354 lb.
We observe that all the support forces we solved for are positive, which means the
directions we chose for them initially are the actual directions they act. (Notice that
means that A acts at 37o as shown, Dx act in the negative x-direction, and Dy acts in the
positive y - direction.)
We have now solved our problem - finding the support reactions (forces) acting on the
structure. We could add (as vectors) Dx & Dy to find one resultant force acting a some
angle on point D, as follows:
D = Square Root [Dx2 + Dy2] = Square Root [(-274 lb)2 + (354 lb)2] = 447.7 lb.
Tangent (angle) = Dy/Dx = 354lb/-274 lb = -1.29, so Angle = ArcTangent (-1.29) =
127.8o (from x -axis)
so support force at Point D could also be expressed as: D = 447.7 lb. @ 127.8o
(Please note that the force at D does not act along the direction of member BCD, which it
would do if BCD were an axial member.
Example 3: Frame (non-truss, rigid body) Problems
This third example is somewhat similar to example one, but we will extend the problem
by not only determining the external support reactions acting on the structure, but we will
also determine the internal force in member CD of the structure shown in Diagram 1.
We first calculate the support forces. The procedure to find the external support reactions
consists of our basic statics procedure.
179
I. Draw a Free Body Diagram of the entire structure showing and labeling all
external load forces and support forces, include any needed dimensions and angles.
II. Resolve (break) all forces into their x and y-components.
III. Apply the Equilibrium Equations (
)
and solve for the unknown forces.
We note that the structure is composed of members ABC, CD, AD, and cable DE. These
members are pinned together at several points as shown in Diagram 1. A load of 12,000
lb. is acting on member ABC at point B, and a load of 8000 lb. is applied at point C.
These forces are already shown by the downward arrows. We next look at the forces
exerted on the structure by the supports. Since each support is a pinned joint, the worst
case we could have is an unknown x and y-force acting on the structure at each support
point. We also must choose directions for the x and y support forces. In some problems
the directions of the support forces are clear from the nature of the problem. In other
problems the directions the support forces act is not clear at all. However, this is not really
a problem. We simply make our best guess for the directions of the support reactions. If
our guess is wrong, when we solve for the value of the support forces, that value will be
negative. This is important. A negative value when solving for a force does not mean
the force necessarily acts in the negative direction, rather it means that the force acts
in the direction OPPOSITE to the one we initially chose.
Fig.14.19
Step 1: Draw a Free Body Diagram of the entire structure. In the FBD (Diagram 2),
we have shown unknown x and y support forces acting on the structure at point A,
however, at point E we have shown one unknown force 'E' acting at a known angle (37o).
180
Fig.14.20
We can do this at point E since we know that ED is a cable, and a cable is an axial
member which can only be in tension. Since the cable pulls axially on the wall, the wall
pulls equally and in the opposite direction on the structure., as shown in Diagram 2.
In Diagram 2, we have also included Step II, Resolve any forces not in the x or ydirection into x and y-components. Thus, we have shown the two components of E
(which act at 37o) - E cos 37o being the x-component, and E sin 37o being the ycomponent. We have also used given angles and dimensions to calculate some distance,
as shown, which may be needed when we apply the equilibrium equations.
We also note that at point A we have two members pinned together to the wall, axial
member AD, and non-axial ABC. Because of these two members (as opposed to a single
axial member, such as at point E), the best we can do at point A is to replace the hinged
joint by an unknown Ax and Ay support forces acting on the structure as shown in
Diagram 2. However, we have a good FBD since we have only three unknown forces, and
we have three independent equations from our equilibrium conditions.
Step III. Apply the Equilibrium conditions.
(Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left,
181
-)
(Sum of y-forces, including load forces. Again keeping track of direction signs.)
Sum of Torque about a point. We choose point A. Point E is also a good point to sum
torque about since unknowns act through both points A and D, and if a force acts through
a point, it does not produce a torque with respect to that point - thus, our torque equation
will have less unknowns in it, and will be easier to solve. Notice that with respect to point
A, forces Ax, Ay, and E sin 37o do not produce torque since their lines of action pass
through point A. Thus, in this problem the torque equation has only one unknown - E. We
can solve for E, and then use it in the two force equations to find the other unknowns, Ax
and Ay. Doing the mathematics we arrive at the following answers.) E = +10,800 lb. Ax =
+8620 lb. Ay = +13500 lb.
We see that all the support forces we solved for are positive, which means the directions
we chose for them initially are the actual directions they act. We have now solved part
one of our problem. The support force at point E is 10,800 lb. acting at 37o. The support
force(s) at A can be left as the two components, Ax = 8,620 lb. and Ay = 13,500 lb., or
may be added (as vectors) obtaining one force at a known angle.
The second part of the problem is to determine the force in axial member CD. (We know
member CD is axial as there are only two points at which forces acts on CD, point C and
point D.) To determine the force in an internal member of a structure we use a procedure
similar to that used to find the external support reactions. That is, we draw a FBD, not of
the entire structure, but of a member of the structure, (choosing not the member we
wish to find the force in, but a member it acts on). Thus, if we wish to find the force in
member CD, we draw a FBD - not of member CD, but a member CD acts on, such as
member ABC, or member AD. In this example we will use member ABC to find the force
in member CD.
To find the force in a member of the structure we will use the following steps:
First, determine the external support reactions acting on the structure (as we did in the
first part of this example). Then continue with steps below
•
•
•
I. Draw a Free Body Diagram of a member of the structure showing and labeling
all external load forces and support forces acting on that member, include any
needed dimensions and angles. (The member selected should be one acted on by
the member in which we wish to find the force.)
II. Resolve (break) all forces into their x and y-components.
III. Apply the Equilibrium Equations (
solve for the unknown forces.
) and
Step 1: FBD of Member ABC. There are actually two good FBD for member ABC. In
Diagram 3 we have shown the first of these. Notice at the left end we have shown both
the wall support reactions at A, and also the force from axial member AD which acts on
182
member ABC. At point C we have shown the force from axial member CD which acts on
member ABC. That is, we have isolated member ABD and indicated the forces on it due
to the other members (and the wall) attached to it.
Fig.14.21
The second good FBD of member ABC is shown in Diagram 4. What we have done in
this diagram is to look more closely at the left end of member ABC and observe that the
effect of the wall forces and the effect of member AD, is to give some net x and y-force
acting on member ABC. Thus, rather than show both the wall forces and the force due to
AD on ABC, we simple show an ACx and an ACy force which is the net horizontal and
vertical force acting on ABC at the left end. This is fine to do, as we are looking for force
CD, and that is still present in our FBD. This second FBD is slightly easier than the first
in that it will result in one less force (AD) in the equilibrium equations. We will use the
second FBD in the rest of the problem.
Fig.14.22
Step 2: Resolve forces into their x and y-components (This is done in Diagram 5.) Notice
we have chosen directions earlier for the forces. These may not be the correct directions,
but our solution will tell us if we have the right or wrong directions for the unknown
forces.
183
Fig.14.23
Step 3: Apply the equilibrium conditions and solve for unknown forces.
Sum Fx: ACx + CD cos 53.8o = 0
Sum Fy: ACy -12,000 lb. + CD sin 53.8o = 0
Sum TA: -12,000 lb. (4 ft) + CD sin 53.8o (8 ft ) =0
Solving: CD = 7,440 lb., ACx = - 4390 lb. (- sign shows force acts opposite direction
chosen), ACy = 6000 lb.
We have determined the force in CD to be 7,440 lb. Since the force is positive, this
indicates that we have chosen the correct direction for the force CD (which indicates it is
in tension). This solves our problem. (For a little further analysis of forces at point A,
select MORE.)
Point A: As an aside, notice that the forces ACx and ACy (The horizontal and vertical
forces acting on member ABC at end A.) are not the same as the forces Ax and Ay acting
on the entire structure at joint A. This results since the forces of the wall at point A are not
just acting on member ABC, but are distributed to both members ABC and AD, as shown
in Diagram 6. Note in the diagram that if the forces ACx and ADx are summed (13010 lb.
- 4390 lb. = 8620 lb.), and if forces ACy and ADy are summed (6000 lb. + 7500 lb. =
13500 lb.), that their vector sums equal the external forces (Ax and Ay) acting on point A,
as we expect they should.
184
Fig.14.24
14.5 Problems Assignment - Frames 1
Draw a complete free body diagram as a part of the solution for each problem.
1. In the structure shown member AB is pinned at the floor, and member CB is a cable
pinned at the wall. Determine the values of the external support reactions, and the force in
member CB. (Ax = -78.3 lb., Ay = +178 lb., C = 455 lb., CB = 455 lb. (T))
Fig.14.25
2. The structure shown is composed of members ACD and BC pinned together at point
C.. The structure is pinned to ceiling at points A and B. Determine the values of the
external support reactions, and the force in member BC. (Ax = 0 lb., Ay = -1600 lb., By =
5600 lb., BC = 5600 lb.T)
Fig.14.26
185
3. The structure shown is composed of solid rigid members ABC, CD, and BDE pinned
together at points B, C, and D. The structure is supported by a roller at point E, and
pinned to the wall at point A.For the structure shown, determine the values of the external
support reactions, and the force in member DC. (Ax = 8,670 lb., Ay = 10,000 lb., Ex =8,670 lb., DC = 15,600 (T))
Fig.14.27
4. The structure shown is composed of solid rigid members ABC, CD, BD, and DE
pinned together at points B, C, and D. The structure is pinned to the floor at points A and
E. For the structure shown, determine the values of the external support reactions, and the
force in member CD. (Ax = 6375 lb., Ay = 10500 lb., E = 10,625 lb. CD = 9000 lb. C)
186
Fig.14.28
5. The structure shown is composed of solid rigid members AB and BC pinned together at
point B. The structure is pinned to the floor at points A and C. Determine the values of the
external support reactions (Ax = +419 lb., Ay = 4170 lb. Cx = - 5419 lb., Cy = 15,830 lb.)
Fig.14.29
14.6 Problems Assignment - Frames 2
Draw a complete free body diagram as a part of the solution for each problem.
1. Determine the reaction at supports A and B of the beam in the diagram shown.
Neglect the weight of the beam. (Ax = 938 lb., Ay = 728 lb., B = 814 lb.)
Fig.14.30
187
2. A 700 lb. weight is carried by a boom-and-cable arrangement, as shown in the
diagram. Determine the force in the cable and the reactions at point A. (Ax = 404 lb., Ay
= 296 lb., C = 572 lb.)
Fig.14.31
3. A brace is hinged at one end to a vertical wall and at the other end to a beam 14ft long.
The beam weighs 250 lb. and is also hinged to a vertical wall as shown. The beam
carries load of 500 lb. at the free end. What will be the compressive force in the brace,
and what will be the values of the vertical and horizontal components of the reaction at
hinge A? (Ax = 596 lb., Ay = 45 lb., B = 994 lb.)
188
Fig.14.32
4. A gate has a weight of 200 lb., which may be considered as uniformly distributed, see
the diagram shown. A small boy weighing 95 lb. climbs up on the gate at the point B.
What will be the reactions on the hinges? (Upper hinge horizontal force only = 318 lb.,
lower hinge horizontal force = 318 lb., vertical force = 295 lb.)
Fig.14.33
189
5. In an irrigation project, it was found necessary to cross low ground or else swing the
canal to the left by cutting into the solid rock. It was decided to run the canal as a flume
and support it on a number of frames as shown in the diagram. The two members rest in
sockets in solid rock at points A and B. These sockets may be considered as hinges.
What will be the vertical and horizontal components of the reactions at A and B? The
weight of the water in the flume supported by each frame is estimated as 18,200 lb. (This
is a somewhat more complicated problem then the others in this problem set, and it may
be skipped. Or contact your instructor for hints.) (Ax = 6389 lb., Ay = 5093 lb., Bx =
6389 lb., By = 13, 107 lb.)
Fig.14.34
6. An Ocean liner has an arrangement for supporting lifeboats and for lowering them over
the side as shown in the diagram. There is a socket at A and a smooth hole through the
deck rail at B. If the boat and its load weigh 2,000 lb., what are the reactions at A and B?
Two identical davits support each lifeboat. (Ay = 1000 lb., Ax = 1250 lb., Bx = 1250 lb.)
190
Fig.14.35
14.7 Problems Assignment - Frames 3
Draw a complete free body diagram as a part of the solution for each problem.
1. A pin-connected A-frame supports a load as shown. Compute the pin reactions at all
of the pins. Neglect the weight of the members. (Ay=1500 lb., Ey=1000 lb., Bx=1250 lb.,
By=1750 lb., Cx=1250 lb., Cy=250 lb., Dx = 1250 lb., Dy=750 lb., directions not
indicated)
Fig.14.36
191
2. A simple frame is pin connected at points A, B, and C and is subjected to loads as
shown. Compute the pin reactions at A, B, and C. Neglect the weight of the members.
(Ax=10,100 lb., Ay =2500 lb,. Bx =10,100 lb., By =2500 lb., Cx=1443 lb., Cy =7500 lb.,
directions not indicated)
Fig.14.37
3. A pin-connected crane framework is loaded and supported as shown. The member
weights are: post, 600 lbs.; boom, 700 lbs.; and brace, 800lbs. These weights may be
considered to be acting at the midpoint of the respective members. Calculate the pin
reactions at pins A, B, C, D, and E. (Ax=3824 lb., Ay=8100 lb., Bx=3824 lb., By=0,
Cx=7768 lb., Cy=3279 lb., ED = 12, 650 lb @ 52.1o , directions not indicated)
192
Fig.14.38
4. Sketch the structure and draw a free body diagram for the relevant member or the entire
structure. Write the appropriate moment or force equations and solve them for the
unknown forces. (Ax =500 lb., Ay =250 lb., Bx =500 lb., By =750 lb., Dx =100 lb., Dy
=750 lb., directions not indicated)
Fig.14.39
5. Calculate the pin reactions at each point of the pins in the frame shown below.
(Ax=300 lb., Ay =150 lb., Bx =300 lb., By =150 lb., Cx =300 lb., Cy = 0, Dx = 0, Dy
=150 lb., Ey =150 lb., directions not indicated)
193
Fig.14.40
6. The tongs shown are used to grip an object. For an input force of 15 lb. on each
handle, determine the forces exerted on the object and the forces exerted on the pin at A.
(F = 48 lb. on object from each jaw., Ax = 0, Ay = 63 lb.)
Fig.14.41
194
MECHANICAL ENGINEERING SCIENCE (STATICS) – MEC111
WEEK 15
15.0 Forces - More Questions
15.1 Statically Determinate - Example 2
In the structure shown in Diagram 1, members ABC and BDE are assumed to be solid
rigid members. Member BDE is supported by a roller at point E, and is pinned to member
ABC at point B. Member ABC is pinned to the wall at point A. Member ABC is an
aluminum rod with a diameter of 1 inch. (Young's Modulus for Aluminum is 10 x 106
lb/in2)
For this structure we would like to determine the axial stress in member ABC both in
section AB and section BC, and to determine the movement of point C due to the applied
loads.
Fig.15.1
Part I. To solve the problem we first need to determine the external support forces acting
on the structure. We proceed using our static equilibrium procedure (from Topic 1 Statics)
1: Draw a free body diagram showing and labeling all load forces and support (reaction)
forces, as well as any needed angles and dimensions.
195
(See Diagram2)
Fig.15.2
2: Resolve forces into x and y components.
3: Apply the equilibrium conditions.
Sum Fx = Ax = 0
Sum Fy = Ay + Ey -16,000 lbs - 12,000 lbs = 0
Sum TE = (16,000 lbs)(12 ft) + (12,000 lbs)(8 ft) - Ay(12 ft) = 0
Solving for the unknowns: Ay = 24,000 lbs; Ey = 4000 lbs
Part II. An interesting aspect to this problem is that member ABC is not an axial
member, and so it is not in simple uniform tension or compression. However, we are
fortunate in that it is not a complex non-axial member. It is not in shear, but rather simply
is in different amounts of tension above and below point B. Therefore to determine the
amount of stress in each part of ABC, we first make a free body diagram of member ABC
and apply static equilibrium principles.
1: FBD of member ABC. (Diagram 3)
Fig.15.3
196
2. Resolve all forces into x and y components.
3. Apply equilibrium conditions.
Sum Fx = Bx = 0
Sum Fy = 24,000 lb. - By - 16,000 lbs = 0
Solving for the unknowns:By = 8,000 lb.
Now to find the force in section AB of member ABC. Cut the member between points A
and B,and analyze the top section. We can do this since if a member is in static
equilibrium, then any portion of the member is also in static equilibrium.
Looking at Diagram 4 (which is the free body diagram of the upper section of member
ABC), we see that for the section of AB shown to be in equilibrium, the internal force
(which becomes external when we cut the member) must be equal and opposite to the
24,000 lb force of the wall on the member at point A.
Fig.15.4
Once we know the tension in section AB of member ABD, we find the stress from our
relationship Stress = F/A = 24,000 lb/(3.14 x .52) = 30,600 psi.
We then use the same approach with section BC of member ABC. We cut member ABC
between point B and point C, and apply static equilibrium principles to the top section.
Diagram 5 is the free body diagram of that section, and by simply summing forces in the
y-direction, we see that the internal force BC (which becomes an external force when we
cut the member) must be 16,000 lb. for equilibrium.
197
Fig.15.5
The stress in section BC is then given by Stress = F/A = 16,000 lb/(3.14 x .52)
Stress (BC) = 20,400 psi.
Part III. To determine the movement of point C is a relatively simple problem when we
realize that the movement of point C will be equal to the deformation (elongation) of
section AB plus the deformation (elongation) of section BC.
That is, the Movement of C = DefAB + DefBC
Movement of. C = [ (FL / EA)AB + (FL / EA)BC ]
Movement of C = [ (24,000 lbs)(72 in) / (10*106 psi)(3.14*(.5 in)2)]AB + [ (16,000 lbs)(48
in) /
(10*106 psi)(3.14*(.5 in)2)]BC
Movement of C = (.220 in) + (.0978 in) = .318 in
15.2 Statically Determinate - Example 3
In the structure shown in Diagram 1, member BCDFG is assumed to be a solid rigid
member. It is supported by a two cables, AB & DE. Cable AB is steel and cable DE is
aluminum. Both cables have a cross sectional area of .5 in2. For this structure we would
like to determine the axial stress in cable AB and DE. We would also like to determine
the movement of point F due to the applied loads. [Young's Modulus for steel: Est = 30 x
106 psi, Young's Modulus for Aluminum: Eal = 10 x 106 psi.]
198
Fig.15.6
Part I. To solve the problem we first need to determine the external support forces acting
on the structure. We proceed using our static equilibrium procedure (from Topic 1 Statics)
1: Draw a free body diagram showing and labeling all load forces and support (reaction)
forces, as well as any needed angles and dimensions.
Fig.15.7
2: Resolve forces into x and y components. All the forces are acting in the y-direction,
as is shown in the FBD.
3: Apply the equilibrium conditions.
Sum Fy = Ay + Ey - 30,000 lb. - 10,000 lb. = 0
Sum TB = (-30,000 lb.)(4 ft) + Ey(10 ft) - (10,000 lb.)(12ft) = 0
Solving for the unknowns: Ey = 24,000 lb.; Ay = 16,000 lb.
Part II. Since both the steel member AB and the aluminum member DE are single axial
members connected to the supporting ceiling, the external forces exerted by the ceiling on
the members is also equal to the internal forces in the members. Thus FAB = 16,000 lb.
(tension), FDE = 24,000 lb. (tension). To find the stress in each cable is now straight
forward. We apply the stress equation
(from the Stress / Strain / Hooke's Law relationships shown to the right).
So, Stress AB = F/A = 16,000 lbs/.5 in2 = 32,000 psi., Stress DE = F/A = 24,000 lbs/.5
in2 = 48,000 psi.
199
Part III To find the movement of point F requires us to use a bit of geometry. Point F
moves since both member AB and ED deform and member BCDFG moves downward
according to these deformations.
1: Calculate deformation of members AB and ED.
DefAB = (FL / AE)AB = (16,000 lbs)(120 in) / (30*106 lbs/in2)(.5 in2) = .128 in
DefED = (FL / AE)ED = (24,000 lbs)(120 in) / (10*106 lbs/in2)(.5 in2) = .576 in
2: Movement of point F. In Diagram 3 we have shown the initial and final position
(exaggerated) of member BCDFG.
Fig.15.8
Point B moves down .128 inches (the deformation of member AB), point D moves down
.576 inches (the deformation of member DE). Point F moves down an intermediate
amount. To determine this we have drawn a horizontal line from the final position of
point B across to the right side of the beam as shown. From this we see that the distance
point F moves down is .128 inches + "x" (where x is the distance below the horizontal
line as shown in the diagram). We can determine the value of x from proportionality
(since similar triangles are involved), and write: (x / 12 ft) = (.448 in / 10 ft). Solving for
x we find: x = .5376 inches.
So the Movement of F = .128 inches + .5376 inches = .666 inches
15.3 Trusses - Example 1
In the Statics – Truss page, we ended with a sample truss example in which we
determined the external forces acting on the truss and the internal forces in several
members by Method of Joints. For example 2, we would like to use the same truss and
solve for several internal forces by Method of Sections. And since we previously solved
for the external support reactions, we will not repeat that portion, but begin with the
external support forces given and move to determine the internal forces in the selected
members.
Example 1: In Diagram 1 we have a truss supported by a pinned joint at Point A and
supported by a roller at point D. A vertical load of 500 lb. acts at point F, and a horizontal
load of 800 lb. acts at Point C. The support reactions acting on the structure at points A
and D are shown. For this structure we wish to determine the values of the internal force
(tension/compression) in members BC, EC, and EF.
200
Fig.15.9
In Method of Sections, we will 'cut' the truss into two sections by drawing a line through
the truss. This line may be vertical, horizontal, at some angle, or even curved depending
on the problem. The criteria for this line is that we would like to cut through the unknown
members (whose internal force value we wish to determine), but not to cut through more
than three unknowns (since we will have three equilibrium conditions equations, we can
only solve for three unknowns). In this example, cutting the truss once will enable us to
find our selected unknowns, however, in some trusses, or for finding more internal forces,
one may have to repeat Method of Sections several times to determine all the unknowns.
In Diagram 2, we have cut through the original truss with a vertical line just to the right of
member BE. This vertical line cuts through members BC, EC, and EF (the selected
members whose internal forces we wish to determine). We have shown the section of the
truss to the left of the cut. We now treat this section of the truss as if it were a completely
new structure. The internal forces in members BC, EC, and EF now become external
forces with respect to this section. We have represented these forces with the arrows
shown. The forces must act along the direction of the cut member (since all members in a
truss are axial members), and we have selected an initial direction either into or away
from the section for each of the forces. If we have selected an incorrect initial direction
for a force, when we solve for the value of the force, the value will be negative indicating
the force acts in the opposite direction of the one chosen initially. We may now proceed
with the analysis of this structure using standard Static's techniques.
Fig.15.10
I. Draw a Free Body Diagram of the structure (section), showing and labeling all
external forces, and indicating needed dimensions and angles. (Diagram 3)
201
Fig.15.11
II. Resolve (break) all forces into their x and y-components.
III. Apply the Equilibrium Equations :
and solve for the unknown forces.
(Here we sum the x-forces)
(Sum of y-forces, including load forces.)
(Sum of Torque with respect to point E.)
Solving we obtain: BC = 44 lb. (T), EC = -50 lb. (T), EF = 712 lb. (T) (The negative
sign for force EC means that we initially chose it in the incorrect direction. EC acts out of
the section and so is in tension, not into the section as shown in the FBD). Thus, we have
solved for the internal forces in the members BC, EC, and EF by method of sections.
15.4 Trusses - Example 2
The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and
supported by a roller at point D. For this structure we wish to determine the value of all
the support forces acting on the structure, and to determine the force in member FC by
method of joints.
202
Fig.15.12
For the first part of the problem we proceed using our normal static's procedure.
STEP 1: Draw a free body diagram showing and labeling all load forces and support
(reaction) forces as well as any needed angles and dimensions. (Note in Diagram 2, we
have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced the
roller support by the vertical unknown force Dy.
Fig.15.13
STEP 2: Break any forces not already in x and y direction into their x and y
components.
STEP 3: Apply the equilibrium conditions
203
: Ax = 0
: Ay + Dy - 12,000 lbs - 20,000 lbs = 0
: (-12,000 lbs)(4 ft) - (20,000 lbs)(12 ft) + Dy(24 ft) = 0
Solving for the unknowns: Dy = 12,000 lbs; Ay = 20,000 lbs. These are the external
support reactions acting on the structure.
PART 2: Determine the internal force in member FC by method of joints. We begin at a
joint with only two unknowns acting, joint D.
JOINT D:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external
forces and load, and including any needed angles.(Diagram 3) We select an initial
direction for the unknowns, if their solution value is negative they act in a direction
opposite to the direction initially selected.
Fig.15.14
STEP 2: Resolve all forces into x and y components. (Diagram 3).
STEP 3: Apply equilibrium conditions:
: -CD + ED cos (66.4o) = 0
o
: 12,000 lbs - ED sin (66.4 )= 0
Solving for the unknowns: ED = 13,100 lbs (C); CD = 5,240 lbs (T)
Now that we have calculated the values for ED and CD we can move to joint E. We could
not solve joint E initially as it had too many unknowns forces acting on it.
JOINT E:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external
forces and loads, and including any needed angles. (Diagram 4) We select an initial
direction for the unknowns, if their solution value is negative they act in a direction
opposite to the direction initially selected.
STEP 2: Resolve all forces into x and y components. (Diagram 4).
204
Fig.15.15
STEP 3: Apply equilibrium conditions:
FE -(13,100 lbs) cos (66.4o) - CE cos (66.4o) = 0
o
o
:(13,100 lbs) sin (66.4 ) - CE sin (66.4 ) = 0
Solving for the unknowns: FE = 10,500 lbs (C); CE = 13,100 lbs (T) (Since force values
were positive, the initial direction chosen for the forces was correct.)
Now that we have calculated the values for FE and CE we move to joint C. We could not
solve joint C initially as it had too many unknowns forces acting on it.
JOINT C:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external
forces and loads, and including any need angles. (Diagram 5) We select an initial
direction for the unknowns, if their solution value is negative they act in a direction
opposite to the direction initially selected.
STEP 2: Resolve all forces into x and y components. (Diagram 5).
Fig.15.16
STEP 3: Apply equilibrium conditions:
: 5,450 + (13,100 lbs) cos (66.4o) + FC cos (66.4o) - BC = 0
o
o
: 13,100 lbs sin (66.4 ) - FC sin (66.4 ) = 0
Solving for the unknowns: FC = 13,100 lbs (c); BC = 15,950 lbs (t) Thus, member FC is
in compression with a force of 13, 100 lbs.
15.5 Trusses -Example 3
The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and
supported by a roller at point H. For this structure we wish to determine the value of all
the support forces acting on the structure, and to determine the force in member DG by
method of sections.
205
Fig.15.17
We begin by determining the external support reactions acting on the structure.
STEP 1: Draw a free body diagram showing and labeling all load forces and support
(reaction) forces, as well as any needed angles and dimensions. (Note in Diagram 1,
we have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced
the roller
support by the vertical unknown force Hy
STEP 2: Break any forces not already in x and y direction into their x and y
components. (All forces in x/y directions.)
STEP 3: Apply the equilibrium conditions.
: Ax = 0
: Ay + Hy -12,000 lbs - 20,000 lbs - 10,000 lbs = 0
: (-12,000 lbs)(20 ft) - (20,000 lbs)(40 ft) - (10,000 lbs)(60 ft) + Hy(80 ft) = 0
Solving for the unknowns: Ay = 21,500 lbs; Hy = 20,500 lbs. These are the external
support reactions acting on the structure.
Part 2: Now we will find internal force in member DG by method of sections. Cut the
truss vertically with a line passing through members DF, DG, and EG. We have shown
the section of the truss to the right of the cut. We now treat this section of the truss as if it
were a completely new structure. The internal forces in members DF, DG, and EG now
become external forces with respect to this section. We have represented these forces with
the arrows shown. The forces must act along the direction of the cut member (since all
members in a truss are axial members), and we have selected an initial direction either
into or away from the section for each of the forces. If we have selected an incorrect
initial direction for a force, when we solve for the value of the force, the value will be
negative indicating the force acts in the opposite direction of the one chosen initially. We
may now proceed with the analysis of this structure using standard Static's techniques.
206
Fig.15.18
I. Draw a Free Body Diagram of the structure (section), showing and labeling all
external forces, and indicating needed dimensions and angles. (Diagram 2)
II. Resolve (break) all forces into their x and y-components. (Diagram 2)
III. Apply the Equilibrium Equations (
)
:EG + DG cos (51.3o) + DF cos (22.6o) = 0
: -10,000 lbs + 20,500 lbs - DG sin (51.3o) - DF sin (26.6o) = 0
o
: -DF cos (26.6 )(15 ft) + (20,500 lbs)(20 ft) = 0
Solving for the unknowns: DF = 30,600 lbs (C); DG = -4,090 (opposite direction)=
4,090 lbs (T); EG = -24,800(opposite direction)= 24,800 lbs (T)
15.6 Problem Assignment - Trusses 1
1. The structure shown is a truss composed of axial members pinned together at the joints.
The stucture is pinned to the floor at points A and F. Determine the external force support
reactions, and the force in member BE by method of joints. (Ay = -9,000 lb, Fx = - 10,000 lb,
Fy = 9,000 lb; BE = 10,000 lb (T))
207
Fig.15.19
2. The structure shown is a truss composed of axial members pinned together at the joints.
The stucture is pinned to the floor at point A and supported by a roller at point D.
Determine the external force support reactions, and the force in member CD by method of
joints. (Ax = -4000 lb, Ay = 200 lb, Dy = 4800 lb; CD = 970 lb (C))
Fig.15.20
3. The structure shown is a truss composed of axial members pinned together at the joints.
The stucture is supported by a roller at point A and pinned to the floor at point G.
Determine the external force support reactions, and the force in members DE, DH and IH
by method of sections (Ay = 35,000 lb, Gy = 35,000 lb, DE = 25,000 lb (C), DH = 14,140 (C), IH =
35,000 lb (T))
Fig.15.21
4. The structure shown is a truss composed of axial members pinned together at the joints.
The stucture is pinned to the floor at point E and supported by a roller at point F.
Determine the external force support reactions and the force in members HI, HC and DC
208
by method of sections. (Fx = -28,000 lb, Fy = - 22,500 lb,Ey = 22,500 lb; HI = 12,650 lb
(T), HC = 3000 lb (T); CD = 10,500 lb (C))
Fig.15.22
5. The structure shown is a truss composed of axial members pinned together at the joints.
The stucture is pinned to the floor at point A and supported by a roller at point
L.Determine the external force support reactions and the force in members DF, DG and
EG by any method.
(Ay = 14,000 lb, Ly = 18,000 lb, DF = 30,600 (C), DG = 10, 420 (T), EG = 23,330 (T))
Fig.15.23
15.7 Problem Assignment - Trusses 2
Note the typical designation of pin and roller supports in the diagrams shown.
209
1. Calculate the forces in all members of the truss shown in the following diagram using
the method of joints. [AB =10,600 lb (C), CB = 10,600 lb (C)]
Fig.15.24
2. Calculate the forces in all member of the truss shown in the following diagram using
the method of joints. [Ay = 6,540 lb., Ax = -10,000 lb, Cy = 8460 lb., AB = 8,375 lb, AC
= 15,230 lb, BC=17,420 lb)
210
Fig.15.25
3. Calculate the forces in all members of the truss shown in the following diagram using
the method of joints. (Ay = 225 lb, Ax = -500 lb., Cy = 475 lb., AB = 450 lb., AD = 890
lb., BC = 950 lb.)
Fig.15.26
4. Calculate the forces in all members of the truss shown in the following diagram using
the method of joints. [Ay=Ey=40,000 lb., AB =56,580 lb. (C), AF = FG = 40,000 lb. (T),
BF=0, BC = CD =65,000 lb (T), BG = 35,360 (T), CG = 50,000 lb, right side same by
symmetry]
Fig.15.27
211
5. Calculate the forces in all members of the trusses shown in the following
diagram using the method of joints. [Ey =90,000 lb., Fy=-20,000 lb., Fx=40,000 lb., AB =
0, AC = 35,000 lb. (C), BC = 21,220 lb. (C), BD=20,000 lb. (C), CD = 25,000 lb. (C), DF
= 20,000 lb. (C), CE = 90,000 lb. (C), EF = 0, CF = 56,580 lb.(T)]
212
Fig.15.28
6. Calculate the forces in members BC, BG, and FG, by method of joints, for the
cantilever truss shown below. [Ax=-26,250 lb., Ay=15,000 lb., Ex=26,250 lb., BC =
8,750 lb. (T), BG=17,366 lb. (T), FG=17,500 lb (C)]
Fig.15.29
15.8 Problem Assignment - Trusses 3
1. Solve the following truss by the method of sections. Determine the truss reactions
and the force in members BD, BC, and AC.
Fig.15.30
213
2. Solve the following truss by the method of sections.
and the forces in members CD, ID, and IJ.
Determine the truss reactions
Fig.15.31
3. Solve the following truss by the method of sections. Determine the truss reactions and
the forces in members EF, FK, and KL.
Fig.15.32
4. Solve the following truss by the method of sections.
and the forces in members DE, JE, and MN.
214
Determine the truss reactions
Fig.15.33
5. Solve the following truss by the method of sections.
and the force in members DE, JE, AND JI.
Determine the truss reactions
Fig.15.34
6. Solve the following truss by the method of sections.
and the force in members CD, DH, and HI.
215
Determine the truss reactions
Fig.15.35
7. For the Howe roof truss shown below, determine the support reactions and the forces in
members BC, CI, and IJ by the method of sections.
Fig.15.36
216