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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.