Civil Engineering Dept.
CE 201 Engineering
Statics
Dr. Rashid Allayla
Lecture 1 General Principles (1.11.6)
Fundamentals:
Scalar
versus Vector:
Scalar quantity is a quantity that has magnitude only and is independent of direction. Examples include: Time, Speed, Volume and Temperature. On the other hand, vector quantity has both magnitude and direction. Examples include: Force, Velocity and Acceleration.
Graphical
representation of a vector:
The symbol → above the letter q indicates that q is vector. The magnitude of q is designated as by the symbol ׀q׀.
Basic
definitions:
Length: Designated by the letter L (cm,
mm, m, km, inch, ft, mile)
Mass:
Designated by the letter M (kg, lb)
Force: Designated
by the letter F (N “
Particle: A particle is a mass of negligible size with no
particular geometry.
Rigid Body: It is a combination of large number of particles that occupy more than one point in space and located a fixed distance from each other both before and after applying a load.
Concentrated
Force: All loads are acting on a point
on a very small body.
First Law:
“A particle
in a state of uniform motion or at rest tends to remain in that state unless
subjected to an external force".
Example:
A 10 N object is moving at constant speed of 10 km / hr on
a friction free surface. Which one of the horizontal forces is necessary to
maintain this state of motion?
a) 0 N b)
1 N c)
2 N ?
Answer:
It does not take any force to maintain the motion as long as
the surface is friction free. Any additional force will accelerate or
decelerate the motion depending on the force applied.
Second Law:
“The
acceleration of a particle is proportional to the resultant force acting on it
and moves in the same direction of this force”
f = ma
Where “f” is the force, “m” is the mass and “a” is the acceleration. In this notes, instead of placing arrows above forces, they will be written in bold letters instead.
Third Law:
“For every action there is reaction. The mutual forces of action
and reaction are equal in magnitude and opposite in direction and collinear in
orientation".
F (Action)
F (Reaction)
Online Conversion Unit: Go to http://www.onlineconversion.com/
SI Units:
SI is known as the International System of Units where
Length is in meters (m), time is in seconds (s),
and mass is in kilograms (kg) and force is in
Newton (N) (1 Newton is the force required to give 1 kilogram of mass
an acceleration of 1 m/s^{2}).
US Customary System of Units (FPS); is the system of
units where length is in feet (ft), time is in seconds (s),
and force is in pounds (lb).
The unit mass is called a slug (1 pound is the force required to give
one slug of mass an acceleration of 1 ft/s^{2}).
Conversion of Units:
Force; 1
lb (FPS Unit) = 4.4482 N (SI Unit)
Mass; slug (FPS Unit) = 14.5938 kg (SI
Unit)
Length; ft (FPS Unit) = 0.304 m (SI Unit)
Prefixes:
Giga = G = 10^{9} = 1 000 000 000 Milli = m = 10^{3} =
0.001
Mega = M
= 10^{6} = 1 000 000 Micro
= μ = 10^{6}
= 0.000 001
Kilo = k = 10^{3} = 1 000 Nano = η = 10^{9}
= 0.000 000 001
Example:
If one lb of an object has a mass of 0.4536 kg, find
the weight in
Solution: Mass Acceleration Force
Weight in
A force represents an action of one body on another. A force is defined by the following components:
a) Point of application b) Magnitude c) Direction
Forces F_{1} and F_{2} acting on a particle may be replaced by a single (resultant) force R which will have the same effect on the particle. The resultant force R can be found by constructing a parallelogram. So it is evident that vector addition does not obey ordinary arithmetic addition, that is, two forces of 9 and 3 lb magnitudes do not add up to 12 lb. On the other hand, if the two vectors are collinear (i.e. acting on the same line), arithmetic addition (or scalar addition) will apply.
Vector Addition Using Triangular
Construction:
Required: Add the two vectors A and B
Resultant Resultant B A
Method: We can add the two vectors by connecting the tail of B to the head of A or connecting the head of B to the tail
Vector Subtraction Using Triangular
Construction:
Vector subtraction is a special case of vector addition. It is carried out by reversing the sign of the vector to be subtracted and performing the same rule of vector addition
Required: Subtract vector B from A
B Resultant
Resolution of a Vector:
Resolution of a vector into two vectors acting along any two given lines is carried out by constructing parallelogram as shown in the illustration below:
Vector Addition of Number of Forces:
Vector addition of n forces is accomplished by successive application of parallelogram
law as described above and as shown in the following illustration:
Law of Sine and Cosine:
The magnitude of the resultant force can be obtained using the law of cosines and the direction can be obtained using the law of sines.
Given: force A and Force B as shown below
Required: The resultant force and its direction using Sine & Cosine laws.
Cosine Law: R = SQRT (A^{2} + B^{2} – 2 AB Cos β)
Sine Law: A/Sin γ = B / Sin α = R/ Sin β
B B
Resolving
Resultant to Components Using Law of Sine:
A_{y}^{’} β α Ax^{’} Y^{’} A α
^{’}
A_{x} =  A Cos α = A Cos (180  α)
A_{y} = A Sin α = A Sin
(180 – α) Note that: A_{x}^{’}_{ }≠ A
EXAMPLE:
Determine the magnitude and
direction of force P such that the resultant of the two forces on the pulling
tug boat (P & T) is equal to 4.00 kN.
Solution:
Using Cosine Law: P = SQRT [ 4^{2} + (2.6)^{2}
 2 x 4 x 2.6 cos 20^{o}]
Gives: P = 1.8 kN
Using Sine Law: 2.6 / Sin θ = 1.8 / Sin 20^{o} Gives: θ = 30^{o}
P
θ
P 2.6 N 20^{o }
2.6 N
θ 4.0 KN
The
resultant is found using triangular law (see figure) R = 4.0 KN
EXAMPLE: (Beer & Johnston)
Two
forces A = 40N and B = 60N acting on bolt C. Determine the magnitude and the
direction of the resultant R using law of Cosine & Sine.
B = 60 N
25^{o}
A = 40 N
20^{o}
Solution:
Drawing the system using triangular rule and applying the
law of cosine: A = 40 N
25^{O}
R^{2} = A^{2} + B^{2} – 2 AB Cos [β)]
But: β = 18025=155
B=60 N R
= (40^{2})
+ (60^{2}) – 2 (Cos 155)
β
θ
α
= 97.7 N
Applying
the law of Sines:
A / Sin α = R / Sin 155 where
α is the angle opposite to vector A.
40
/ Sin α =
97.7 / sin 155^{o} then α =
Sin^{1} (40) Sin 155 / 97.7 = 0.173 = 10^{o}
Then θ = (25+20) – 10 =
35^{o}
EXAMPLE: (Beer & Johnston)
Two forces are applied as
shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14lb, determine (a) the required
angle α if the resultant R of the two forces applied to the
support to be horizontal, (b) the corresponding magnitude of R.
Solution: 20 lb 30^{o}
Force
Triangle: R α α
20 lb
P =14 lb
P = 14 β
R α 30^{o}^{ }
P = 14 lb
Using law of sines:
20 / Sin α =
P / Sin 30 = R / Sin β
Since P = 14 lb,
then: Sin α = (20 / 14) Sin 30 =
0.71428 → α = 45.6^{o}
The value of β: β + α + 30 = 180 → β = 104.4 then 14 /
Sin 30 = R / Sin 104.4 Gives
R = 27.1 lb
Lecture 3 Force Vectors Addition of System of Coplanar
Forces (2.4)
The successive application of parallelogram method to find the resultant of set of forces is often tedious. Instead, it would be easier to find the components of the forces along specified axis algebraically and then find the resultant.
It is often desirable to resolve a force into two components which are perpendicular to each other as shown below.
Unit Vector
In order to obtain the resultant of a set of coplanar forces, each force is resolved into x and y components and then added algebraically to obtain the resultant. In the figure below, F_{1, }F_{2} and F_{3} are a set of coplanar forces. In Cartesian vector notation, the forces are written as _{}
F_{1} F_{2}
F_{3y }F_{3x}
F_{3}
F_{1}
=  F_{1x} + F_{1y} , F_{2
}= F_{2x} + F_{2y} ,
F_{3} = F_{3x}  F_{3y}
The resultant is: F_{R} = F_{1} + F_{2} + F_{3 Angle resultant makes with + x axis}
_{ }
F_{R}
= (F_{1x} + F_{2x} + F_{3x}) i + (F_{1y}
+ F_{2y} – F_{3y}) j
& ІF_{R }І
= SQRT ( F_{Rx}^{2} + F_{Ry}^{2} ) θ = Tan^{1} (ІF_{Ry }І
/ ІF_{Rx }І)
Unit
Vector:
A unit vector is a vector directed along the positive x and y axis having dimensionless magnitude of unity. Any vector can be expressed in terms of the unit vector as, F = F_{x} i + F_{y }j
Where i and j are the unit vectors in x and y direction and F_{x} and F_{y} are the “scalar” magnitudes of F in x and y direction. The two magnitudes can be either positive or negative depending on the sense of F_{x} and F_{y}.
If θ is measured counterclockwise from the positive x axis, the magnitude of the force is measured as
F_{x} = F Cos θ and
F_{y} = F Sin θ
EXAMPLE: (Hibbeler)
Determine the x and y
components of F_{1} and F_{2} acting on the boom and express
each force as a Cartesian vector.
1) Scalar Notation:
F_{1x}
= 200 Sin 30 = 100 N
F_{1y}
= 200
F_{2x}
/ 260 = 12/13 Gives: F_{2x} =
240 N
F_{2y}
/ 260 = 5/13 Gives: F_{2y} = 100 N
2)
Cartesian Notation:
Having
determined the magnitudes of forces and their directions, then:
F_{1}
= 100 i + 173 j
F_{2}
= 240 i – 100 j
EXAMPLE: (Beer & Johnston)
Four
forces act on bolt A, determine the resultant of the forces on the bolt.
Force Magnitude N XComponent N YComponent F_{1} 150 +129.9 + 75.0 F_{2} 80 27.4 + 75.2 F_{3} 110 0.0  110.0 F_{4} 100 +96.6  25.9 R_{x}
= +199.9 R_{y} =
+14.3
F_{2 }F_{1}
F_{4}
F_{3} F_{1} Cos
30 i
F_{4} Cos 15 i
F_{2} Cos 20 i
θ = Tan^{1} ( 14.3/199.9) = 4.1^{o}^{}
EXAMPLE:
Determine
magnitude and direction cosine of resultant (R) of the following force vectors:
F1 = 5i + 15 j + 30 k (N)
F2 = 25i +
30 j  40 k (N)
F3 =  25j

50 k (N)
Solution:
R = ∑ F_{i} = F_{1}
+ F_{2} + F_{3}
R
= 30 i + 20 j  60 k R = SQRT
[(30)^{2} + (20)^{2} + (60)^{2}] = 70 N
Cos α = R_{x} / │R │ = 30 / 70 = 0.42857 α = 64.6^{o}
Cos β = R_{y} /
│R │ = 20 / 70 = 0. 28571 β =
73.4^{o}
Cos γ = Rz / │R │ = 60 / 70 = 0.8571 γ = 149.0^{o}
Check
the result Cos^{2} α + Cos^{2}
β + Cos^{2} α = 1
(0.42857)^{2 }+ (0.28571)^{2} + (0.8571)^{2} = 1 OK
Lecture 4 Force Vectors Cartesian Vectors (2.5, 2.6)
Cartesian Vectors
Cartesian vector is a set of unit vectors i, j and k that defines the direction of a given vector. It locates a point in space relative to a second point. Unit vector in the direction of a given vector (such as the one shown in the figure) is obtained by dividing the position vector r_{AB} by the magnitude of r_{AB}:
z
B (x_{B}, y_{B}, z_{B})
A
Y r_{AB }= (X_{B}
– X_{A}) i + (Y_{B} – Y_{A}) j + (Z_{B} – Z_{A})
k │r_{AB}│_{
}= SQRT [ (X_{B} – X_{A})^{2} + (Y_{B}
– Y_{A})^{2} + (Z_{B} – Z_{A})^{2 }]
Unit Vector u_{AB} =
r_{AB }/ │r_{AB}│ = (X_{B} – X_{A})
i + (Y_{B} – Y_{A}) j + (Z_{B} – Z_{A}) k / SQRT
[ (X_{B} – X_{A})^{2} + (Y_{B} – Y_{A})^{2}
+ (Z_{A} – Z_{B})^{2 }] z_{A} x_{A} _{ } y_{A} _{ } _{ }x_{}
C
Unit vector is useful to express a force in a vector form. When a unit vector acting in the same direction of the force is multiplied by the magnitude of the force, a vector representation of the force is accomplished.
F = │F│u_{AB }and, therefore, u_{x} = F_{x}
/ │F│ u_{y} = F_{x} / │F│ u_{x} = F_{z} / │F│
U_{F}
= (F_{x} / │F│) i + (F_{y} /
│F│) j + (F_{z} / │F│) k Then:
U = Cos α i + Cos β j +
Note that the sum of squares of
direction cosines is unity because │u_{F}│
= 1
Cos α^{2} i + Cos^{2}
β j + Cos^{2}
γ k = 1
EXAMPLE: (From umr)
Determine
the distance between point A and B located as shown using a position vector.
4 m
Solution:
The
position vector in the direction AB is
r_{AB}
= (X_{B} – X_{A} ) i + ((Y_{B} – Y_{A} )
j + (Z_{B} – Z_{A} ) k
= (4 – (2)) i +(12 – (6)) j
+ (2 – 3 ) k = 6 i + 18 j – 5 k m
The
distance from A to B is │r │ = SQRT[( 6^{2} +18^{2}
+ 5^{2} )]= 19.62 m
EXAMPLE: (From Hibbeler)
Determine the magnitude and the coordinate direction
angle of the resultant force acting on the ring.
The
resultant force F_{R} = (50+0) i + (100+40) j + (100+80)
k = 50 i – 40 j + 180 k
The
magnitude = SQRT [ (50)^{2} + (40)^{2} + (180)^{2} ] =
191.0 lb
U_{FR}
= (50 / 191.0 i – (40 / 191.0) i + ( 180 / ( 191.0) k =
0.2617 i – 0.2094 j + 0.9422 k
Then
and: Cos γ = 0.9422 γ= 19.6^{o}
Lecture 5 Force Vectors Position Vector, Force alona Line
(2.7, 2.8)
We have shown that the unit vector along a line AB is:
Called position vector r
u_{ = } (X_{B} – X_{A} ) i +
((Y_{B} – Y_{A} ) j + (Z_{B} – Z_{A} ) k
/ SQRT{(X_{B} – X_{A} )^{2}
+ ((Y_{B} – Y_{A} )^{2} + (Z_{B} – Z_{A}
)^{2}} or: u = r / │r│
If we have a force F with magnitude of │F│acting along the line AB, then the vector F is defined as:
F = u │F│
Where: u is the unit
vector acting along the line AB as defined above.
EXAMPLE: (From umr)
Write
a unit vector in the direction from B to A
Solution:
The
unit vector from B toward A U_{BA} = r_{BA }/ │ r_{BA}
│ and the position vector is:
r_{BA}
= (X_{A} – X_{B}) i + (Y_{A} – Y_{B}) j + (Z_{A}
– Z_{B}) k
= (6 – 3) i + (8 – (4)) j + (5 (2))
k
= 9 i + 12 j + 7 k m
The
magnitude of the unit vector u_{BA} is: U_{BA} = r_{BA }/
│ r_{BA} │
Where,
r_{BA} = SQRT [(9)^{2} + (12)^{2} + (7)^{2}] = 16.553 Then:
U_{BA}
= (9 i + 12 j +
7 k) / 16.553 = 0.5437 i + 0.7249 j + 0.4229 k
EXAMPLE : (Beer & Johnston)
A
towe guy wire is anchored by means of bolt at A. The reaction in the wire is
2500 N. Determine a) the components F_{x}, F_{y} and F_{z},
b) The angles α, β and γ
The
distance from A to B = SQRT [ (40)^{2} + (80)^{2} + (30)^{2}]
= 94.3 m Then:
Position
vector: r_{AB} =  40 i + 80 j_{ }+ 30 k ,
The unit vector u_{AB} =  (0.4242) i + (0.8484) j + (0.3181) k
The
vector Along AB = [ (2500) u_{AB }] =
 (1060.5 N) i + (2121 N) j +
(795.33 N) k
Direction of force: α = Cos^{1} [1060 / 2500] = 115.1^{o}, β = Cos^{1} [2120 / 2500] = 32.0^{o}, γ = Cos^{1} [795 / 2500] = 71.5^{o}
^{ }
Example: (Hibbeler)
A
roof is supported by cables as shown. If the cables exert forces F_{AB}
= 100 N and F_{AC} 120 N on the wall hook at A as shown. Determine the
resultant force at A and its magnitude.
The
position vector AB: r_{AB} = (4 m – 0) i + (0 – 0) j + (0 – 4 m) k = 4 i – 4 k
׀r_{AB}׀ = SQRT [ (4)^{2}
+ (4)^{2}] = 5.66
m
Then: F_{AB}
= (100 N) [ r_{AB} / ׀r_{AB}׀
] = (100 N) [ (4 / 5.66) i – (4 / 5.66) k ]
F_{AB}
= [70.7 i – 70.7 k] N
The
position vector AC r_{AB} = (4 m – 0) i + (2 m – 0) j + (0 – 4 m) k = 4 i + 2j – 4 k
׀r_{AC}׀ = SQRT [ (4)^{2}
+ (2)^{2} + (4)^{2}] = 6 m
Then: F_{AC}
= (120 N) [ r_{AC} / ׀r_{AC}׀
] = (120 N) [ (4 / 6) i + (2 / 6) – (4 / 6) k ]
F_{AC}
= [80 i + 40 j – 80 k] N
The
resultant force is:
F_{R}
= F_{AB} + F_{AC} = [70.7 i – 70.7 k] + [80 i + 40 j – 80 k]
= [150.7 i + 40 j – 150.7 k] N
׀F_{R}׀
= SQRT [ (150.7)^{2} + (40)^{2} + (150.7)^{2} ] = 217 N
Lecture 6 Force Vectors Dot Product (2.9)
Dot product of two vectors P and Q (otherwise known as scalar product) is defined as the product of the scalar magnitudes of the two vectors and the cosine of the angle formed by the vectors. Dot product is useful for:
a) Determining the angle between two vectors, and,
b) Determining
the projection of a vector along a specified line.
Let: P = Px_{ }i + P_{y}
j + P_{z} k
and: Q
= Q_{x }i + Q_{y} j + Q_{z} k
Then: P.Q = │P││Q│
cos θ Ξ Px Q_{x} + P_{y}Q_{y} + P_{z}
Q_{z}
_{ }
Rules:
1)
Dot product follows commutative law: Q . P = P.Q
2)
Dot Product follows distributive law: P. (Q_{1} + Q_{2}) = P. Q_{1} + P. Q_{2}
_{ }3) Multiplication by a scalar: a (P.Q) =
(a P) . (Q) = (P) . (a Q) = (P.Q) a
u_{P = P /│P│}
P.Q =
│P││Q│ cos θ or:
cos θ = [P.Q /
│P││Q│] or: θ
Ξ cos1 (u_{P} . u_{Q})
Since │P_{aa}│_{
}=_{ }│P│cos θ
Then: │P_{aa}│_{
}= P. u_{aa }and P_{aa} = [ P . u_{aa} ]_{
}u_{aa}
_{ }
Usefulness of Dot Product:
Vector form of
projection of F into x axis
 Angle between two intersecting vectors can be determined:
θ =
cos1 [P.Q / │P││Q│]
 The component of a vector parallel and perpendicular to a line can be determined if the unit vector along this line is known:
F ║ = F cos θ = F. u
Since F
= F ║+ F _{┴}_{ } Then:
F _{┴ } = F  F ║
Projecting a Force along a Line:
Given: A force F_{AB} = A i + B j + C k _{ }along line AB
Required: The projection of this force along line AC
Method
of solution:
1) Find the unit vector along the line AC
U_{AC } = [(x_{C} – x_{A}) i + (y_{C} – y_{A}) j + (z_{C} – z_{A})] / SQRT [[(x_{C}–x_{A})^{2}+[(y_{C} – y_{A})^{2}+ [(z_{C} – z_{A})^{2}_{]}_{}
2) Use the dot product to find the projection of the force along AC:
׀F_{AC}׀ = U_{AC} . F_{AB}
_{ }
׀F_{AC}׀ = { [(x_{C}
– x_{A}) i + (y_{C} – y_{A}) j + (z_{C} – z_{A})]
/ SQRT [[(x_{C}–x_{A})^{2}+[(y_{C} – y_{A})^{2}+
[(z_{C} – z_{A})^{2}]}
. { A i + B j + C k }
This is scalar value which is the projection of force F into line AC
The Cartesian vector of the projection of F into AC is:
F_{AC} = U_{AC} ׀F_{AC}׀ _{}
_{ }
_{}_{}
_{ }
_{ }
_{ }
_{ }
_{ }
_{ }
_{ }
_{ }
EXAMPLE 1: (from umr)
The force F = 50i + 75 j + 100 k acts on a pole AB as shown. Determine the projected component of F along AB and component of the force perpendicular to AB.
The
unit vector along AB = r_{AB} /│ r │ = {(43) i +
[4(2)] j + (60) k} / SQRT[1^{2} + 6^{2} + 6^{2}
]
Then:
u_{AB} = (0.117 i + 0.702 j + 0.702 k) (This is the
unit vector along AB)
F_{AB}
(the projection of F on AB) = F . u_{AB} = ( 50i + 75 j +
100 k) . (0.117 i + 0.702 j + 0.702 k)
= 23.41 lb
The
Cartesian vector from the parallel component is F_{AB} . u_{AB}
= 23.41 (0.117 i + 0.702 j +
0.702 k)
F_{║} = 2.74 i + 16.44 j + 16.44 k lb
The
component of the force perpendicular to AB is = F – F_{AB} = (50i
+ 75 j +100 k) – (2.74 i + 16.44 j + 16.44 k) lb
F_{┴}
= 47.3 I –
91.4 j + 85.6 k lb
EXAMPLE 2:
Find
the a) angle between cable BD and the boom AB and b) the projection on AB of
cable BD at point B.
Unit
vector in AB direction = ( 6 i + 4.5 j ) / SQRT ( 6^{2} +
4.5^{2} ) = 0.8 I + 0.6 j
The angle between BD and AB
is = cos^{1} [ u_{AB}
. u_{BD} ] = cos^{1} [(0.8i +0.6j) . (0.67i
+0.33j 0.67 k) ]
= cos^{1}_{ }[ 0.536 + 0.19] = 110.24^{o}
Force BD = (180) (u_{BD})
= (180) (0.67i +0.33j 0.67 k) =  120.6
i + 59.4 j + 120.6 k
The projection of BD on AB =
u_{AB} . F_{BD }= (0.8i +0.6j ).( 120.6 I + 59.4 j + 120.6 k)
=  96.48 + 35.6
= 60.88^{‘ }
^{ }
Lecture 7 Equilibrium of Particle Free Body Diagram (3.1, 3.2)
Conditions for Equilibrium of a Particle:
A particle is said to be at equilibrium if the resultant of all forces acting on it is zero. Another case of equilibrium is illustrated in the figure below. If the four forces acting on a particle at point O are at equilibrium, then starting from point O with F_{1} and arranging the forces in tip to tail fashion, the tip of F_{4} will coincide with the tail of force F_{1} and the resultant of the four forces will be zero. The graphical representation is expressed mathematically as:
∑ F = 0
FreeBody
Diagram:
What?  It
is a drawing that shows all external forces acting on the particle.
Why?  It helps
you write the equations of equilibrium used to solve for the unknowns (usually forces or angles) (Hibbeler)
Therefore:
 Free
body diagram is a method of isolating all external forces acting on a body from its surroundings.
Procedure for Drawing Free Body Diagram (FBD):
1) Isolate the particle from its
surroundings.
2) Sketch all forces that act on the
particle while observing
3)
Known
forces are labeled with their magnitudes
and directions. Assign letters to the unknown forces with assumed
directions. The body’s weight must be included if applicable.
EXAMPLE:
Draw the free body
diagram of the two structures shown
Springs:
The magnitude of the force exerted on the linear elastic spring is:
F = Ks
Where K is the stiffness of the spring (measured in N/m), s is the deformation (which is a measure of the difference between the deformed length L and the undeformed length L_{0}). Note that if s is negative, F must push on the spring and if s is positive, F must pull on the spring to bring it to the desired length. K is also defined as the force required to deform the spring a unit distance.
EXAMPLE:
A spring has undeformed length of 0.4 meters and
stiffness k = 500 N/m. What is the force needed to stretch the spring to a length
of 0.6 m? and what force is required to compress the spring to a length = 0.2
m?
SOLUTION:
F = K s
F = (500 N/m) (0.6 m – 0.4 m) = 100 N (s is positive, force is pulling spring)
F = (500 N/m) (0.2 m – 0.4 m) = 100 N (s is negative, force is pushing spring)
Cables and Pulleys:
When a cable is passing over a frictionless pulley, the force along the cable is always in tension and constant in magnitude. This is necessary condition to keep the cable in equilibrium.
T T
EXAMPLE:
The springs
have stiffness of 500 N/m each and length of 3 m each. Determine the horizontal
force F applied to the cord so that the displacement of the pulley from the
wall is d = 1.5 m.
SOLUTION:
T
AC = 3.3541 m
F 2T
Cos θ
Σ F_{x} = 0 If the tension at each spring is T Then: T_{x} = (1.5/3.3541) T T_{x}
Then, 2(1.5 / 3.3541) (T) – F = 0 since: s = 3.35413=0.3541 m T
T = K S = (500)(
3.3541  3) =
177.05 N Then: F = 158 N
Lecture 8 Equilibrium of Particle Coplanar Force System (3.3)
Procedure for Solution of Problems in
Equilibrium:
EXAMPLE : (From Higdon &
Stiles)
A 500 N
shaft A and 300 N shaft B are supported as shown. Neglecting friction at all
contact points find the reactions at points R and S at shaft A.
The first FBD ( Shaft A ) has three unknowns Q, R and S and only two independent equations of equilibrium. The next step is to draw FBD of shaft B. The force on shaft B exerted by shaft A is Q directed to the upper right. Writing the equation:
∑
F_{y} = 0 then: Q sin 40^{o} – 300 = 0 then Q = 467 N
on B
From the FBD of shaft A:
∑F_{y} = 0 then: S – 500 – Q sin 40^{o} = 0 then S = 800 N directed upward.
∑F_{x}
= 0 then: R – Q cos 40^{o} = 0 then: R = 467 cos 40^{o} = 358 N directed to the right
EXAMPLE:
(From Hibbeler)
Determine the required length of cord AC so that the 8
kg lamp is suspended in the position shown. The undeformed length of the spring
is L^{’}_{AB} = 0.4 m and its stiffness is 300 N/m
Y T_{AC} _{ 30}^{0}_{
}T_{AB} _{ } _{ }78.5 _{ } _{ } 78.5 N
Weight
of lamp W = 8 (9.81) = 78.5 N
∑Fx
= 0 T_{AB} – T_{AC}
cos 30^{o} = 0
∑F_{y} = 0
T_{AC} sin 30^{o} – 78.5 = 0
Then:
T_{AC} = 157.0 N and T_{AB} = 136.0 N which is the stretch of spring AB
T_{AB} = K s or:
136.0 = 300 s Then s = 0.453 m (this is the amount of stretch
on the spring)
The
stretch length is L_{AB} = L_{AB}^{’ } + S_{AB} = 0.4 + 0.453 = 0.853 m
The
horizontal distance CB requires that 2
= L_{AC} cos 30^{o} + 0.853
Then L_{AC} = 1.32.m
EXAMPLE: (from umr)
The pulley system is used to rise a 50 lb weight.
Determine the tension T necessary to hold the weight in equilibrium.
∑ F_{y} = 0 Then: from FBD (3): T + T – T_{B } = 0 Or: T_{B} = 2 T Since: ∑ F_{y} = 0 Then from the FBD (4) T_{B} + T – 50 = 0 or: T = 50 / 3 = 16.67 lb
Example:
Calculate
the tension T in the cable which supports the 1000 lb load with the system of
pulley shown. Ignore the weight of the pulley. Find F at Pulley C.
Solution:
Start
with the free body diagram (pulley A) because it has the only known force:
∑ F_{y}
= 0
Then: T_{1} + T_{2}
= 1000
Since
the cord around pulley A is continuous,
T_{1} = T_{2}
Then: T_{1} = T_{2}
= 500 lb ↑
Since
the cord around pulley B is
continuous, T_{3} = T_{4}
= T_{2} / 2 = 250 lb↑
Again,
since the cord around pulley C is continues T_{3} = T Then: T = T_{3} = 250 lb ↑
Evaluation of
reaction forces at pulley C:
∑ Fx = 0 250
cos 30^{o} – F_{x} = 0
Then: F_{x} = 217 lb ←
∑ F_{y}
= 0
 F_{y} + 250 sin 30^{o} 250 = 0 Then F_{y} =  125 lb ↓ or: +125 lb ↑
F
= SQRT [ (217)^{2} + (125)^{2} ] = 250 l
Example: (Hebbeler)
The
sack at A weighs 20 lb, determine the weight of the sack at B and the force in
each cord needed to hold the system in the equilibrium position shown.
There
are three forces acting on E: T_{EG, }T_{EC} and 20 lb and on
C: T_{CE}, T_{CD} and W_{B}
From
FBD No. 1
_{ }
∑
F_{X} = 0 Then: T_{EG}
sin 30^{O} – T_{EC} cos 45^{O} = 0
∑
F_{y} = 0 Then: T_{EG} cos 30^{O} – T_{EC}
sin 45^{O} – 20 lb = 0 Then: T_{EC}
= 38.6 lb and T_{EG}
= 54.6 lb
From
FBD No. 3: T_{EC} = T_{CE} = 38.6 lb
From
FBD No. 2
∑
F_{X} = 0 Then: 38.6 cos 45^{O}
– (4/5) T_{CD} = 0
∑
Fy = 0 Then: (3/5) T_{CD} + 38.6
sin 45^{O} – W_{B} = 0
Then: T_{CD} = 34.2 lb and: W_{B} = 47.8 lb
Lecture 9 Equilibrium of Particle Three Dimensional Force Systems
(3.4)
The
requirement for equilibrium of a particle is:
∑F
= 0 or: ∑F_{x} + ∑F_{y}
+ ∑F_{z} = 0 The method of solution is summarized as
follows:
Draw
Free body diagram and:
 Establish appropriate x,y and z axis.
 Label all
known and unknown forces and assume the direction of unknown forces.
 Apply the equations of
equilibrium ∑F_{x} = 0, ∑F_{y} = 0, ∑F_{z
}= 0.
 Reverse the directions of
unknown forces if solution yields negative result.
Example: (From
Hibbeler)
Determine
the magnitudes of F_{1}, F_{2} and F_{3} for
equilibrium of the particle.
∑ F_{z}
= 0 F_{1} sin 30^{O} – 2.8 = 0 Then:
F_{1} = 5.6 KN
∑ Fy = 0  F_{2
}(24/25) + (8.5) cos 15 = 0 Gives: F_{2} = 8.55 k
F_{3} – 5.6 cos 30^{O} – 8.55 (7/25) – 8.5 sin 15^{O}
= 0 Then:
F_{3} = 9.44 k
The 100 kg crate is
supported by three cords, one of which is connected to a spring. Determine the
tensions in AC and AD and the stretch of the spring
Note: F_{C } is directed in  x,  y, +z zone, F_{D}_{ }is directed_{ }in
–x, +y, +z zone and F_{B}_{ }is directed in +x
zone, then:
F_{B }_{= }F_{B} i
F_{C} = F_{C} cos 120 i + F_{C} cos 135 j + F_{C} cos 60 k
= 0.5 F_{C
} i
 0.707 F_{C} j + 0.5 F_{C} k
F_{D} = F_{D} ( 1 i + 2j +
2k) / SQRT (1^{2} +2^{2} + 2^{2}) = 0.333 F_{D} i + 0.667 F_{D}_{ }j + 0.667 F_{D} k
W =  981 k
Σ F = 0 Then:
Σ F = 0, then: F_{B }i _{ }_{ }0.5 F_{C}_{ }i
– 0.707 F_{C} j + 0.5 F_{D} i + 0.667 F_{D }j + 0.667 F_{D} k – 981 k = 0
Setting Σ F_{x} = 0 , Σ F_{y} = 0 , Σ F_{z} = 0 We
have three unknowns and three equations, then:
F_{C} = 813 N
F_{D} = 862 N
F_{B} = 693.7 N
The stretch of the spring is
F_{B} = k s
693.7 = 1500 s Then s = 0.462
m
EXAMPLE: (From
Hibbeler)
A 200 lb box is
supported by cables DA, DB and DC find the forces in these cables
Σ F_{x }=
0 Then:
(4.5–1.5) i / SQRT[( 4.51.5)^{2}+(1.5)^{2}+(3)^{2 }] F_{DA}_{ }+ (1.5) i / SQRT[ (1.5)^{2 }+ (2.51.5)^{2 }+(3)^{2 }] F_{DC}
= 3 / 4.5 F_{DA} – 1.5
/ 3.5 F_{DC}_{ } = 0.667 F_{DA}  0.429 F_{DC} = 0
Σ F_{y}=0 Then:
(–1.5) j / SQRT [( 4.51.5)^{2}+(1.5)^{2}+(3)^{2 }] F_{DA }+ (2.51.5) j / SQRT [(1.5)^{2}+(3)^{2}+(2.51.5)^{2}] F_{DC}
_{ }
_{ }_{+} (1.5) j / SQRT[( 1.5 )^{2 }+ 0 + 0) ] F_{DB}
= 1.5/ 4.5 F_{DA} + 1 /
3.5 F_{DC}_{ } F_{DB} =  0.33 F_{DA}_{ }+ 0.286 F_{DC}  F_{DB} = 0
Σ F_{z} =
0 then:
(3) k /SQRT[(4.51.5)^{2}+(1.5)^{2}+(3)^{2}]
F_{DA}+(3)
k /SQRT[(1.5)^{2}+(3)^{2}+(2.51.5)^{2}]
F_{DC}_{ }– W k
= 3 / 4.5 F_{DA} + 3 /
3.5 F_{DC}_{ }  200 = 0.667 F_{DA} + 0.857 F_{DC} – 200 = 0
The above are three
equations and three unknowns F_{DA}, F_{DB} and F_{DC}.
Solving yields:
F_{DA} = 100 lb F_{DB}
= 11.1 lb
, and: F_{DC} = 155.6 lb
Lecture 10 Force System Resultant Moment of a ForceScalar
Formulation (4.1)
A moment of a force around a point (say O) is a measure of the tendency of the force to rotate the body around the point. The magnitude of the moment is found by multiplying the magnitude of the force with the shortest distance from the force to that point. The sense the moment vector is determined by the righthand rule defined earlier.
If F is a force along AA line, the magnitude of the moment of the force around point O is:
M_{o}
= │F│ d = │F││r│sin θ Where θ is the angle between the
two vectors
Here: d is the “moment arm” which is the perpendicular distance from O to the line AA.
F is
parallel to xy axis F is
parallel to zy axis F
passes through 0
EXAMPLE
(From Hibbeler)
Determine
the magnitude and the directional sense of the resultant moment of the force
about point P.
Solution:
+
M_{P} = (260) (5/13) [3] +(260) (12/13) [2] – (400) (sin 30^{o}) [42] + (400) (cos 30^{o}) [3+5]
= 300 + 480 – 400 + 2771.28= 3151 N
EXAMPLE:
A 100 lb vertical force applied to the end of a lever attached to a shaft. Determine 1) The moment about O; b) The magnitude of horizontal force applied at A which creates the same moment about O; c) The smallest force applied at A which creates the same moment about O; d) How far from the shaft a 240 lb vertical force must act to create the same M_{O}
a)
d = 24 cos 60 = 12 in
M_{O} = (100 lb) (12 in) = 1200 lbin Perpendicular to
the paper and pointed into the paper.
b)
d = (24 in) sin 60 = 20.8 in Then: M_{O} = 1200 = (20.8) F or: F = 57.8 lb
c)
The smallest value of F occurs when d is maximum which is 24 inches, then
1200 = F (24) Then:
F = 50 lb
d)
1200 = (240) d Then: d = 5 in
EXAMPLE:
(From Hibbeler)
Determine
the resultant moment of the four forces acting on the rod shown below.
Assuming
that positive moment acts on the +k direction (counterclockwise), then:
+M _{R0
}=_{ }Σ F d
M _{R0 }= 
50 N (2 m) + 60 N (0) + 20 N (3 Sin 30^{o} m) – 40 N (4 m + 3
=  334 N.m
= 334 N.m
Example:
(From Hibbeler)
The
70 N force acts on the end of the pipe at B. Determine the angle θ (0^{o}
< θ > 180) of the force that will produce maximum
moment about point A. What is the magnitude of the moment? (a = 0.9 m, b = 0.3
m and c = 0.7 m)
M_{A}
= F sin (θ)
c + F cos (θ) a
dM_{A}/d θ = 0 = F cos (θ) c  F sin (θ) a
θ_{max} = atan (c/a) Gives:
θ = 37.9^{o}
M_{max} = (70 N) sin (37.9^{o})
(0.7 m) + (70 N) cos (37.9^{o}) (0.9 m)
M_{max} = 79.812 N.m
Lecture 11 Force System Resultant Cross Product (4.2)
Direction
of Moment:
The direction of moment is determined using the right–hand rule. Starting from the position r and curling fingers into the force F, the direction of the moment is parallel to the direction of the thumb (i.e.: perpendicular to the plane formed by the vectors r and F). We can also use the principle of screw where the direction of the movement of the screw as it is driven, is the same as the direction of the moment when the turning action of the screw is directed from r to F.
Direction of moment z Direction of fingers from r to F x y r Position Vector r F
Cross Product:
Cross product of vectors Q and P is a vector V = Q X P directed perpendicular to the plane containing the two vectors. The magnitude of the resulting vector is the product of the magnitudes of the two vectors multiplied by the sine of the angle formed by Q and P. The direction of the resultant vector is determined using the righthand rule. This suggests that the commutative rule does not apply in cross products as:
V = P X Q Q P V = Q X P
Q
P X Q ≠ Q X P
However, distributive law can be applied as:
A X (B + C) = (A X B) +
(A X C)
The cross product of two vectors P X Q in Cartesian coordinate system is:
P X Q = (P_{x} i + P_{y} j + P_{z} k ) X ( Q_{x} i + Q_{y} j + Q_{z} k)
= P_{x} Q_{x} (i X i) + P_{x} Q_{y}_{ }(i X j) + P_{x} Q_{z}_{ }(i X k) + P_{y} Q_{x} (j X i) + P_{y} Q_{y}_{ }(j X j) + P_{y} Q_{z}_{ }(j X k)
+ P_{z} Q_{x} (k X i) + P_{z} Q_{y} (k X j) + P_{z} Q_{z} (k X k)
Noting that: (i X i) = (j X j) = (k X k) = 0 and: (i X j) = (j X k) =
(k X i) = 1
(j
X i) = (k X j) = (i X k) = 1
i
j k
P X Q = (P_{y} Q_{z} P_{z} Q_{y}) i – (P_{x} Q_{z}_{ }– P_{z} Q_{x}) j + (P_{x} Q_{y} – P_{y} Q_{x}) ki _{ = }│ P_{x } P_{y}
P_{z}_{ }│
Where the symbol │ … │ is known as the determinant Q_{x} Q_{y} Q_{z}
EXAMPLE:
A
300 lb force acts on bracket as shown. Find the moment M_{A} about
point A
The
moment is the vector M_{A} = r X F where r is the vector pointing from
A to C
M_{x}
= 300 cos 30^{O}i(6) =1558.8
inlb M_{y} =300 sin 30^{O }(7) = 1050
Then
M_{A} = 15581050 = 508.8 inlb Pointing
into paper
i j
k
M_{A}
= r X F =  7
6 0  = (7x150 – 6x259.8) k
= – 508.8 k () Pointing into the paper
259.8 150 0
Example:
Determine the direction θ for 0^{O} ≤ θ ≥ 180^{O
}of the force F so that F produces a) the maximum moment about point A, b)
the minimum moment about point A.
The
maximum moment produced when F ┴ BA
M_{A }= 400 [√ ( 3^{2} + 2^{2}) ] = 1442 Nm Since: Ф = tan^{1} (2/3) = 33.69^{O} Then θ = 90^{O} – 33.69= 56.3^{O}
b) M_{A} = 0 Then: θ = 180^{O} – 33.69^{O} = 146^{O}
Lecture 12,13 Force System Resultant Moment of a ForceVector
Formulation (4.3,4.4)
Principle
of Transmissibility:
The force applied at point A (shown above) produces a moment M about point O at the direction shown. The moment produced is:
M = F x r_{1}
= F x r_{2} = ׀_{ }F ׀
׀_{ }r ׀
sin θ = F d
It is evident that the moment produced is the same as long as the position vector r is taken from point O to any point along the line of action of force F (This is called principle of transmissibility)
Resultant
Moment of a Set of Forces:
The total moment above is:
(M_{R})_{O} = ∑(r X F)
The moment M of F about an axis through O is a vector normal to a plane formed by F and r and is calculated as:
i
j k
M =
│ r_{x }r_{y
}r_{z }│= [r_{y} F_{z }– r_{z}F_{y}]
i  [r_{x} F_{z }– r_{z}F_{x}] j +
[r_{x} F_{y }– r_{y}F_{x}] k Ξ r X
F
F_{x} F_{y} F_{z}
_{ }
│M
│ = │r│ │F│ sin θ = │F│d
The moment about point O in any given space can be interpreted as the moment about an axis passing through point O and perpendicular to a plane containing point O and force F. The Cartesian representation of moment vector is:
i
j k
M_{O}
= r X F = │ r_{x }r_{y }r_{z }│= [r_{y}
F_{z }– r_{z}F_{y}] i  [r_{x} F_{z }–
r_{z}(Fx)] j + [r_{x} F_{y }– r_{y}(F_{x})]
k
F_{x} F_{y} F_{z}
Where: r is the position vector connecting point O to any point on the line of action of F.
F
z
(M_{O})_{z}
r_{}
_{}
r_{z}
y
r_{x} _{ } (M_{O})_{y}_{}
r_{y}
x (M_{O})_{x}
The
i component
of the moment is the result of three moments: one due to F_{x}
about the x axis (but
producing no moment since F_{x} is parallel to xaxis), the second is due to F_{z} about
the xaxis (Ξ +r_{y}
F_{z} acts in the positive xaxis), the third
is due to F_{y} about the xaxis (Ξ  r_{z}F_{y} acts in the negative xaxis).
Similarly:
The j component of the moment is the result of three moments: one due to F_{y} about the y axis (but producing no moment since F_{y} is parallel to yaxis), the second is due to F_{z} about the yaxis (Ξ  r_{x} F_{z} acts in the negative yaxis), the third is due to F_{x} about the yaxis (Ξ  r_{z}F_{x} acts in the negative yaxis). Also: The k component of the moment is the result of three moments: one due to F_{z} about the z axis (but producing no moment since F_{z} is parallel to zaxis), the second is due to F_{y} about the zaxis (Ξ + r_{x} F_{y} acts in the positive zaxis), the third is due to F_{x} about the zaxis (Ξ + r_{y}F_{x} acts in the positive yaxis). _{}
Example:
A
force P and Q of magnitude 50 N and 100 N act in the directions shown. Determine
the moment of P about point F.
2 i  (95) j + 6 k
P
= 50 N [ ] = 50 (0.267 i – 0.535 j + 0.802 k) = 13.35 i
– 26.75 j + 40.10 k
SQRT( 4+16+36)
Q
=  100 N j
r_{FB}
= (010) i +(90) j + (04) k =  10 i + 9
j – 4 k
r_{FD}
= (010) i +(00) j +(04) k =  10 i – 4
k
i j k i j k
M_{F}
= r_{FB }x P + r_{FD} x Q
=  10 +9
 4
 +  10
0  4 
13.35 26.75
40.10 0
100 0
= [9x40.1
–(4x(26.75)]i–[10x40.1–(4x13.5) ] j+[10x(26.75)9x13.35]
+[(4)x(100)] i
+0 j +[10x(100)] k
= (360.9  107) i +(400.1  54) j + (267.5 – 120.15) k  400 i + 1000 k
= 253.9 i +
346.1 j +_{ }147.35 k  400 i + 1000 k
=  146.1 i+346.1 j+1147.4 k
Alternative
way: Using r_{FB }as an arm_{ }for both forces:
_{ }
i j k i j k
M_{F}
= r_{FB }x P + r_{FD} x Q
=  10
+9 4  + 
10 +9 4

13.35 26.75 40.10 0
100 0
= (360.9  107) i +(400.1
 54) j + (267.5 – 120.15) k  400 i + 1000 k
=  146.1 i+346.1 j+1147.4 k