this lesson is an important one since it will deal with ... mechanics 102.pdf · common point...
TRANSCRIPT
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This lesson is an important one since it will
deal with forces acting in conjunction with one
another, against one another, and the resultant
of a number of forces acting through a
common point (known as coplanar forces) or
acting other than through a common point
(known as non-coplanar forces). As we are
using the Systeme International, commonly
known as SI, mass will be given in
Kilogrammes whereas force will be given in
Newtons.
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Here you see a man with a force of 20 N
pulling on a rope which is fixed to a wall. In a
problem such as this we find it is more
convenient to represent a force by a straight
line, drawn to a convenient scale. For this
purpose the straight line must show: first, the
direction of the line of action; second, the
position at which the force is applied; and third,
the magnitude of the force, usually stated in
Newtons or kiloNewtons.
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Let us go back to our problem of the man pulling on the rope. Line XY
represents the wall drawn to a scale of 1 cm = 0.5 m. The wall
is 5 m long, therefore xy to scale is 10 cm. The man is pulling on the
rope at an angle of 45° to the wall at the middle position A. The line AB
is drawn at an angle of 45° from point A towards the right as shown.
The length AB is again drawn to scale but this time 1 cm equals 4 N so
it is 5 cm in length.
The line AB is called a Vector Quantity because it represents the pull of
the man on the rope in both magnitude and direction in relation to
position A, the point on the wall. A Vector Line is a line drawn to scale
to represent the magnitude of a force, its point of application
(sometimes called sense) and its direction in relation to a fixed position.
Line AB is a vector line.
In structural problems it is sometimes necessary to determine the
resultant of non-parallel forces acting at a point or on a body. This is
best explained by means of a simple experiment.
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Here is a simple arrangement of two pulleys supporting three weights. Begin by attaching three pieces of string to a small metal ring of about 3cm in diameter, two of the strings being, placed over two small pulleys fixed to a vertical board. The ends of these strings support weights W1 and W2. A third weight W is fastened to the third string so that all the weights are at rest-- that is motionless. If you did this experiment using different weights you would find that, for example, the system would be at rest if W1 = 5 N, W2 = 3 N and W = 5 N when there is an angle of
approximately 120O between the inclined pulley strings.
Since weight W is equal to the combined effects of weights W1 and
W2 it must be equal and opposite to their resultant and its line of action must be that of the third string. Now draw in pencil or in chalk on the surface of the board to which the pulleys are attached straight lines corresponding to the strings radiating from the metal ring. Put arrows on these lines showing which way the strings are being pulled away from the ring. Put an 0 by the metal ring, A at the end of the left hand string, B at the end of the right-hand string and C at the end of the middle string. Having done this remove the string, the metal ring and the weights.
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This slide should be like the pencil or chalk
lines which you have drawn on the pulley
board.
OA represents the string from the ring at 0 to
the left-hand pulley. OB represents the string
from the ring at 0 to the right-hand pulley. DC
represents the string acting vertically
downwards.
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Now choose a suitable scale 1 cm = 1 N force and
draw vectors as follows:
First, along OA mark off OD to represent the force
acting in the string, that is force W1equal to 5 N
force, that is 5 cm.
Second, along OB mark off OE to represent the
force acting in the string, that is force W2 equal to 4
N force, that is 4 cm.
Third, along DC mark off OF to represent the force
acting in the string, that is force W equal to 5 N
force, that is 5 cm.
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Draw DX parallel to OE and EX parallel to OD. Therefore ODXE is a
parallelogram. If FO is extended upwards it will pass through X,
hence OX is a diagonal of the parallelogram. Measure OX. You will find that it
is the same length as OF. To the same scale in which OD represents W1 and
OE represents W2, OX represents W in magnitude. Note that W is the force
which is in equilibrium with W1and W2 just as the force W, represented by the
vector OF acting downwards, maintains equilibrium and the vector line OX
represents in magnitude and direction the Resultant of the forces W1 and W2
represented by the vectors OD and DE.
Repeat this experiment yourself several times using different
values for W1 and W2 and record the different values for W. Bear
in mind that you must make allowances for slight errors due to the frictional
resistance of the pulleys. Nevertheless, you should come to the same
conclusion, which is as follows:
If two forces, represented by the vectors OD and DE, act at point 0, and
include between them the angle DOE, then if the parallelogram ODXE be
completed, the diagonal OX of the parallelogram which passes through O,
represents in magnitude and direction the resultant of the forces.
This is the theorem known as the Parallelogram of Forces and is defined on
the next slide.
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This theorem is important and
you should write it in your
notebook
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Let us now attempt a simple example. Here is a stake in the
ground
to which two ropes are attached pulled by two men. The angle
between the ropes is 45°. Man P pulls with the force of 6 N on
one rope and man Q pulls with a force of 10 N on the second
rope. If we wish to find the resultant of these two forces we
proceed as shown in the space and force diagrams, Figs 2 and 3.
From a convenient point 0, draw OA to a suitable scale, to
represent
force P. As force P is 6 Nand 1 cm = 2 N then a vector line of
3 cm is drawn. Draw OB to the same scale, that is 5 cm long,
since force Q is 10 N and at 45° to OA. Complete the
parallelogram OACB and draw in the diagonal DC. DC will
represent (to the force scale already adopted) the resultant of the
forces P and Q. If it is drawn correctly it will be found to be a force
of 14.8 N.
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Let us take a second example. A sailing boat is to
be pulled ashore by two men hauling on two ropes
which are at the same height above
the shore. The angle between the ropes is 30°. One
man pulls with a force of 20 N on one rope and the
other man pulls with a force of 30 N on the second
rope. Determine the resultant force on the boat. As
you will see from the force diagram, Fig 3, the
resultant DC equals 48.2 N.
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In the first of the two examples showing two men
pulling ropes attached to a stake, the resultant force
was obtained by constructing a parallelogram of forces.
The resultant of two forces can be obtained by another
method which also uses vectors but without setting up
a complete parallelogram. The new method is
constructed in the following way. Look at Fig 2. Draw
OA representing force P (10 cm long if 1 cm = 1 N). We
now draw a vector AC representing force Q (6 cm long
since force Q = 6 N). Now join the final point C to the
starting point 0 and reverse the arrow to point towards
c.
OC is the resultant R force and equals 48.2 N. Study
Fig 2 carefully.
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Obviously a force E, equal in magnitude to the
resultant force R and acting in the opposite
direction, would balance force Rand therefore would
also balance the forces P and Q as is shown here.
OC is a vector representing force R and OE is a
vector of the same magnitude representing the
balancing force or the Equilibrant as it is generally
called. Forces P, Q and E acting together would
constitute a system of three concurrent forces in
equilibrium. We have now come to something which
is important. It should now be clear that if vector
quantities are drawn in succession to represent the
forces P, Q and E a triangle will be formed.
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This triangle is known as the Triangle of Forces
for the three given forces. The theorem
appertaining to it should be written in your
notebook. Any triangle whose sides are
parallel to the lines
of action of the three given forces will have its
sides respectively proportional to the
magnitude of the forces. If, therefore, we
represent one of the forces to a particular
scale, the remaining sides of the triangle will
represent the other two respective forces to the
same scale.
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Here in Fig 1 we see three forces, one of which, force A, can be represented by a vector quantity
because we know its magnitude, the direction of its line of action and the position at which the
force is applied. Forces Band C cannot be represented as vector quantities because we only
know their points of application (ie sense) and line of action, but we can determine their
magnitudes and their directions by drawing lines parallel to each of them and so complet- ing the
triangle of forces, as in Fig 2.
Draw OD as a vector representing force A, put on the arrow indicating its direction. Through D
draw a line DX parallel to force B. Through O draw a line OY parallel to the force C. Where these
intersect call the point E.
Now look at Fig 3. From point O continue around the three vectors by following the arrow on
vector OD and place new arrows on vectors DE and EO and return to point O. By using the same
scale as was used for the vector OD we can measure the magnitude of the vectors DE and EO to
give the forces B and C. Working back from Fig 3, the triangle of forces and the known
magnitudes and directions of the forces, we can draw a complete force diagram as shown in Fig
4.
The above problem is typical of the use of the triangle of forces. As you see it has enabled us to
find the following: first, the magnitude of the two unknown forces; and second, the sense in which
the unknown forces act, ie whether they act respectively towards or away from O, the point of
concurrence. This acting towards or away from point O decides whether the force in the member
is pushing or thrusting into point O or pulling away from point O. This action determines whether
the member is in compression or in tension. A member in compression is called a Strut while a
member in tension is called a Tie.
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Here we see a simple frame which is pin-jointed being
subjected to forces pushing at A, B and C. The member
AB (or BC) is therefore in a state of compression, as
shown in Fig 1. To each' external force there must be an
equal and opposite reaction from the member to keep it
motionless. Hence the internal forces of the member must
push against each pin; therefore the internal forces of the
member on the pins must be as shown in Fig 2. Hence the
member is in a state of compression and is therefore a
strut. Similarly, the frame can be subjected to forces
pulling at A, Band C as shown in Fig 3 and the member AB
(or BC) must pull on the pins at A and B to keep it
motionless as shown in Fig 4. Hence the member is in a
state of tension and is therefore a tie.
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We can now tackle a simple example. Here a wall bracket is used to
carry a weight of 60 kN. Determine the force in each projecting arm
of the bracket and state the nature of the force. First, set up the
space diagram (Fig 1) to a convenient scale and show accurately the
direction of the three forces A, B and E acting at the end of the
bracket as shown in Fig 2.
Next, draw the force diagram Fig 3 by starting with the downward
load,of 60 KN to some convenient scale. This is represented by
AB. then draw BC parallel to the lower bracket member and AC
parallel to the upper bracket member. The force diagram is now
complete. Put in the arrows on the members by starting with ab
down and ending at position a. On the free body diagram (Fig 4) put
in the direction of the arrows. If these arrows are transferred to the
force diagram (Fig 5) and equal and opposite arrows given at the
other ends of the members, it will be clearly seen that bc is a strut
taking a compressive force of 30 kN and ac a tie, taking a tensile
force of 52.5 kN.
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At the beginning of this lesson you will remember
we talked about vectors--the geometrical
representation of a force in terms of its direction,
where it acts and its magnitude. Here you see two
forces pushing on point o. These two forces can be
represented as two pulls on point O instead of two
pushes so long as we keep the direction of the line
of action, the point of application and the values of
the two forces P and Q the same. Sometimes when
solving problems it is convenient to change the
thrusts into pulls and vice versa.
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This slide shows two forces acting on point O. Force
P is a thrust while force Q is a pull. To use the
parallelogram of forces to solve this problem, the
system of forces must be changed into two thrusts
or two pulls as indicated in Figs 3 and 5. Suppose
we now try an example.
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A rafter in a roof truss exerts a thrust of 100 N and a
horizontal tie pulls with a force of 86.5 N as shown
in Figs 1 and 2. Find the resultant force the truss
transmits to the supporting wall if the angle between
the tie and the rafter is 30°.
Study carefully the method shown on this slide for
solving the problem. Note in Fig 3 that the tie is
converted into a thrust at the top of the wall to set
up the parallelogram. The force diagram Fig 4
indicates that the wall will have to carry a vertical
force of 50 N.
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The second special case is when we have two
forces which are pulling on a body at two different
points and not at a common point. To determine the
resultant force on the body the two forces P and Q
must be projected backwards until a point of
intersection 0 is found as shown on Fig 2. The
parallelogram of forces can then be applied in the
usual way. The resultant R will pass through the
intersection point 0 of the lines of action of forces P
and Q. You should note that in all problems of this
kind concerning equilibrium, a force may be
assumed to be acting at any point in its own line of
action.
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This slide shows how the previous method of drawing successive
vectors can be applied to any number of concurrent forces. In Fig 1
you will see four forces P, Q, Sand T acting at a point O. Now look
at Fig 2. Ob represents the resultant of the forces P and Q. This
resultant R1 is then combined with force S giving Oc as the vector
line representing the resultant of forces P, Q and S.
Resultant R2 is then combined with force T giving Od as the
resultant of all the forces. How can we determine the final resultant
Od more quickly? This is achieved by constructing vectors
representing the forces P, Q, Sand T to some suitable scale.
Now look at Fig 3. Oa represents force P, ab represents force Q, bc
represents force S and cd represents force T. Finally, join d to 0
and reverse the direction of the arrow, that is, draw the arrow
pointing from 0 to d. You will notice that with this method there is no
need to draw in the intermediate resultant vectors Ob and Oc since
the resultant Od was found by simply constructing the polygonal
outline Oabcd.
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This brings us to another rule which is given
here. Write it in your notebook.
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Let us end this lesson with one more example. Here
in Fig 1 you see four forces acting about point 0
Determine the resultant of the forces about o. The
forces may be represented in the vector diagram
(Fig 2) in any order you wish providing the arrows
follow each other round the diagram. The resultant
is a force of 30.5 kN acting in the direction indicated.
Its point of application is the point of concurrence or
meeting point of the forces forming the system.
You should note that if the vector lines cross one
another as in this example, this has no effect on the
determination of the resultant force.
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We shall end this lesson by considering how the
theory which we have been discussing is applied in
real building. This slide shows a cross-section
through a Gothic cathedral and indicates the two
stone buttresses which support the roof and the
stone vaulting over the nave of the church. The
thrusts (T), outward forces, from the stone vaulting
is counteracted by the vertical dead weight (w) of
the stone in the pinnacles and the dead weight (w)
of the stone in the buttresses themselves. To obtain
the resultants of these dead weights (w) and the
thrusts (T), parallelograms of forces are set up at
positions A and B.