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MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Babylon University
College of Materials Engineering
Subject: Engineering Mechanics Date:
Time:2 Hrs stage: first
Department :ceramic & polymer . Number of Lecture: 2 ,3
(Force Vectors) 2.1 Scalar and vectors A scalar is any positive or negative physical quantity that can be
completely specified by its magnitude.
A vector is any physical quantity that requires both a magnitude and
direction for its complete description. A vector is shown graphically by
an arrow. The length of the arrow represents the magnitude of the
vector, and a fixed axis defines the direction of its line of action .The
head of the arrow indicates the sense of direction of the vector (Fig 2-1)
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
System of Forces:
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Fig(a)
2.2 Vector operations:-
Multiplication and division of vector by a scalar:
If a vector is multiplied by a positive scalar, its magnitude is increased
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
by that amount. When multiplied by a negative scalar it will also
change the directional sense of the vector (Fig 2-2).
Amrani
addition Vector
All vector quantities obey the parallelogram law of addition. Fig 2-3 and
Fig 2-4 and Fig 2-5 illustrates addition of vectors 𝐀 and 𝐁 to obtain a
resultant 𝐑 .
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
The process of finding out the resultant force of a number of given forces is
called composition of forces or compounding of forces.
Graphical
Methods:
Analytical
Graphical (Vector) Method:-
Two forces P & Q are represented by the line segments AB & AD.
Resultant Force:-
R2 = P2 + Q2 - 2PQ cos θ
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Ex(1):
Forces of 6N, 3N, and 4N act as shown in the diagram. Find graphically the
magnitude and the direction of the resultant of these forces.
Ans.
By using scale: 1 N=0.5 cm
By measurement, AD=2.35 cm and angle DAB=49º
The resultant is 4.7 N and makes an angle of 49º with the x-axis.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Referring to Fig. 1.8 (b), we can get the resultant AD by constructing triangle
ABD. Line AB is drawn to represent F1 and BD to represent F2. Then AD
should represent the resultant of F1 and F2. Then we have derived triangle
law of forces from fundamental law parallelogram law of forces. The
Triangle Law of Forces may be stated as If two forces acting on a body are
represented one after another by the sides of a triangle, their resultant is
represented by the closing side of the triangle taken from first point to the
last point.
If more than two concurrent forces are acting on a body, two forces at a
time can be combined by triangle law of forces and finally resultant of all
the forces acting on the body may be obtained. A system of 4 concurrent
forces acting on a body are shown in Fig. 1.6. AB represents F1 and BC
represents F2. Hence according to triangle law of forces AC represents the
resultant of F1 and F2, say, R1.
2. DERIVED LAWS:
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Fig (1.6)
If CD is drawn to represent F3, then from triangle law of forces AD
represents, the resultant of R1 and F3. In other words AD represents the
resultant of F1, F2 and F3. Let it be called as R2.On the same line logic can
be extended to say that AE represents the resultant of F1, F2, F3 and F4 if
DE represents F4. Thus resultant R is represented by the closing line of the
polygon ABCDE in the direction AE. Thus we have derived polygon of law of
forces and it may bestated as ‘If a number of concurrent forces acting
simultaneously on a body are represented in magnitude and direction by
the sides of a polygon, taken in a order, then the resultant is represented in
magnitude and direction by the closing side of the polygon, taken from first
point to last point.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Examples(2):
Read Example • 2.1 in book.
• Exercise 2.1
P = 75 N Q = 125 N
Determine the resultant using
(1) Parallelogram Law,
(2) Triangle Rule,
(3) Trigonometry.
.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Ex (4):
Two forces act on a body so that the resultant is a force of 50
pounds.The measures of the angles between the resultant and the
forces are 25o and 38o. Find to the nearest pound the magnitude of
the larger applied force?
X
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
2.3 vector addition of forces: Experimental evidence has shown that a force is a vector quantity since
it has a specified magnitude, direction, and sense and it adds according
to the parallelogram law.
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Finding a resultant force: The two component forces 𝐅 𝟏 and 𝐅𝟐 acting on the pin in Fig 2-7 can be
added together to form the resultant force
𝐅𝐑 = 𝐅𝟏 + 𝐅𝟐
MSC Baraa Hassun Mechanic Engineering2014
Original text book: Engineering Mechanics -Statics, Twelfth Edition, R. C. Hibbeler, 2009.
Finding the components of a force: Sometimes it is necessary to resolve a force into two components in order to
study its pulling and pushing effect in two specific directions.
For example, in Fig 2.8, F is to be resolved into two components along
two members, defined by u and v (Fig 2.8)