Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

 Lecture 1 General Principles (1.1-1.6)

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 “Newton”, lbf “pound force”)

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.

Newton

Newton’s Laws of Motion:                          http://scienceworld.wolfram.com/biography/photo-credits.html - Newton

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 ?

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/s2).

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/s2).

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 = 109 = 1 000 000 000                Milli = m = 10-3 = 0.001

Mega = M = 106 = 1 000 000                      Micro = μ = 10-6 = 0.000 001

Kilo = k = 103 = 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 Newton's.

Solution:                                          Mass       Acceleration           Force

Weight in Newton's: (0.4536 kg) (9.807 m / s2) = 4.448 N

 Lecture 2 Force Vectors Vectors, Vector Operations and Vector Addition of Forces (2.1, 2.2 & 2.3)

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 F1 and F2 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.

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 (A2 + B2 – 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:

 Ay’ β α

 Ax’

 Y’

 A

 α

Ax = - A Cos α = A Cos (180 - α)

Ay = A Sin α = A Sin (180 – α)         Note that:     Ax ≠ A Cos α

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 [ 42 + (2.6)2 - 2 x 4 x 2.6 cos 20o]  Gives:  P = 1.8 kN

Using Sine Law:               2.6 / Sin θ = 1.8 / Sin 20o   Gives: θ = 30o

 P                                               θ                                         P                      2.6 N                                                     20o                                                                                               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

25o

A = 40 N

20o

Solution:

Drawing the system using triangular rule and applying the law of cosine:            A = 40 N

25O

R2 = A2 + B2 – 2 AB Cos [β)]    But: β = 180-25=155

B=60 N                        R

= (402) + (602) – 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 155o  then α = Sin-1 (40) Sin 155 / 97.7 = 0.173 = 10o

Then θ = (25+20) – 10 = 35o

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                               30o

Force Triangle:                                                                           R                       α      α

20 lb                                                                     P =14 lb

P  = 14                         β

R                                                                                      α                   30o

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 = 45.6o

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, F1, F2 and F3 are a set of coplanar forces. In Cartesian vector notation, the forces are written as

F1                              F2

F3y                                                                          F3x

F3

F1 = - F1x + F1y ,  F2 = F2x + F2y ,  F3 = F3x - F3y

The resultant is:                             FR = F1 + F2 + F3             Angle resultant makes with + x axis

FR  = (-F1x + F2x + F3x) i + (F1y + F2y – F3y) j    &  ІFR І = SQRT ( FRx2 + FRy2 )    θ = Tan-1 (ІFRy І / ІFRx 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 = Fx i + Fy j

Where i and j are the unit vectors in x and y direction and Fx and Fy 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 Fx and Fy.

If θ is measured counterclockwise from the positive x axis, the magnitude of the force is measured as

Fx = F Cos θ    and    Fy = F Sin θ

EXAMPLE: (Hibbeler)

Determine the x and y components of F1 and F2 acting on the boom and express each force as a Cartesian vector.

1) Scalar Notation:

F1x = -200 Sin 30 = -100 N

F1y = 200 Cos  30 =  173 N

F2x / 260 = 12/13  Gives: F2x = 240 N

F2y / 260 = 5/13   Gives:  F2y = 100 N

2) Cartesian Notation:

Having determined the magnitudes of forces and their directions, then:

F1 = --100 i + 173 j

F2 =   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         X-Component N     Y-Component       F1                150                             +129.9                      + 75.0   F2                  80                             -27.4                         + 75.2   F3                 110                             0.0                           - 110.0   F4                 100                            +96.6                        - 25.9                                                                                Rx = +199.9         Ry = +14.3

F2                                                                                F1

F4

F3           F1 Cos 30 i

F4 Cos 15 i

-F2 Cos 20 i

θ = Tan-1 ( 14.3/199.9) = 4.1o

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 = ∑ Fi = F1 + F2 + F3

R = 30 i + 20 j - 60 k        R = SQRT [(30)2 + (20)2 + (60)2] = 70 N

Cos α = Rx / │R │ = 30 / 70 = 0.42857    α = 64.6o

Cos β = Ry / │R │ = 20 / 70 = 0. 28571    β = 73.4o

Cos γ = Rz / │R │ = -60 / 70 = -0.8571    γ = 149.0o

Check the result     Cos2 α + Cos2 β + Cos2 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 rAB by the magnitude of rAB:

 z                                                B (xB, yB, zB)                              A                                                                      Y         rAB = (XB – XA) i + (YB – YA) j + (ZB – ZA) k │rAB│ = SQRT [ (XB – XA)2 + (YB – YA)2 + (ZB – ZA)2 ]                                                                     Unit Vector  uAB = rAB / │rAB│ = (XB – XA) i + (YB – YA) j + (ZB – ZA) k /                                                                                                                          SQRT [ (XB – XA)2 + (YB – YA)2 + (ZA – ZB)2 ]                                           zA           xA                                  yA    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│uAB   and, therefore,  ux = Fx / F      uy = Fx / F      ux = Fz / F

UF = (Fx / │F│) i + (Fy / │F│) j + (Fz / │F│) k     Then: U = Cos α i + Cos β j + Cos γ k

Note that the sum of squares of direction cosines is unity because │uF│ = 1

Cos α2 i + Cos2 β j + Cos2 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

rAB = (XB – XA ) i + ((YB – YA ) j + (ZB – ZA ) 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[( 62 +182 + 52 )]= 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 FR = (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

UFR = (50 / 191.0 i – (40 / 191.0) i + ( 180 / ( 191.0) k   =   0.2617 i – 0.2094 j + 0.9422 k

Then   Cos α = 0.2617    α = 74.8o    and:   Cos β = - 0.2094    β = 102o

and:   Cos γ = 0.9422     γ= 19.6o

 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 =  (XB – XA ) i + ((YB – YA ) j + (ZB – ZA ) k  / SQRT{(XB – XA )2 + ((YB – YA )2 + (ZB – ZA )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 UBA = rBA / │ rBA │ and the position vector is:

rBA = (XA – XB) i + (YA – YB) j + (ZA – ZB) k

= (-6 – 3) i + (8 – (-4)) j + (5- (-2)) k

= -9 i + 12 j + 7 k  m

The magnitude of the unit vector uBA is: UBA = rBA / │ rBA

Where, rBA = SQRT [(9)2 + (12)2 + (7)2] = 16.553 Then:

UBA = (-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 Fx, Fy and Fz, 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: rAB = - 40 i + 80 j + 30 k , The unit vector uAB = - (0.4242) i + (0.8484) j + (0.3181) k

The vector Along AB = [ (2500) uAB ] = - (1060.5 N) i + (2121 N) j + (795.33 N) k

Direction of force:  α =  Cos-1 [-1060 / 2500] = 115.1o, β = Cos-1 [2120 / 2500] = 32.0o, γ = Cos-1 [795 / 2500] = 71.5o

Example: (Hibbeler)

A roof is supported by cables as shown. If the cables exert forces FAB = 100 N and FAC 120 N on the wall hook at A as shown. Determine the resultant force at A and its magnitude.

The position vector AB: rAB = (4 m – 0) i + (0 – 0) j + (0 – 4 m) k = 4 i – 4 k

׀rAB׀ = SQRT [ (4)2 + (-4)2] = 5.66 m

Then:  FAB = (100 N)  [ rAB / ׀rAB׀ ] = (100 N) [ (4 / 5.66) i – (4 / 5.66) k ]

FAB = [70.7 i – 70.7 k] N

The position vector AC rAB = (4 m – 0) i + (2 m – 0) j + (0 – 4 m) k = 4 i + 2j – 4 k

׀rAC׀ = SQRT [ (4)2 + (2)2 + (-4)2] = 6 m

Then:  FAC = (120 N)  [ rAC / ׀rAC׀ ] = (120 N) [ (4 / 6) i + (2 / 6) – (4 / 6) k ]

FAC = [80 i + 40 j – 80 k] N

The resultant force is:

FR = FAB + FAC = [70.7 i – 70.7 k] + [80 i + 40 j – 80 k]

= [150.7 i + 40 j – 150.7 k] N

׀FR׀ = 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 + Py j + Pz k      and:   Q = Qx i + Qy j + Qz k

Then:   P.Q = │P││Q│ cos θ Ξ Px Qx + PyQy + Pz Qz

Rules:

1) Dot product follows commutative law:         Q . P = P.Q

2) Dot Product follows distributive law:            P. (Q1 + Q2) = P. Q1 + P. Q2

3) Multiplication by a scalar:                            a (P.Q) = (a P) . (Q) = (P) . (a Q) = (P.Q) a

uP = P /│P│

P.Q = │P││Q│ cos θ      or:   cos θ = [P.Q / │P││Q│]   or:     θ   Ξ   cos-1 (uP . uQ)

Since │Pa-a = │P│cos θ   Then: │Pa-a = P. ua-a    and      Pa-a = [ P . ua-a ] uaa

Usefulness of Dot Product:            Vector form of projection of F into x axis

- Angle between two intersecting vectors can be determined:

θ   =   cos-1 [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 FAB = 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

UAC  = [(xC – xA) i + (yC – yA) j + (zC – zA)] / SQRT [[(xC–xA)2+[(yC – yA)2+ [(zC – zA)2]

2) Use the dot product to find the projection of the force along AC:

׀FAC׀ = UAC . FAB

׀FAC׀ = { [(xC – xA) i + (yC – yA) j + (zC – zA)] / SQRT [[(xC–xA)2+[(yC – yA)2+ [(zC – zA)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:

FAC = UAC ׀FAC׀

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 = rAB /│ r │ = {(4-3) i + [4-(-2)] j + (6-0) k} / SQRT[12 + 62 + 62 ]

Then: uAB = (0.117 i + 0.702 j + 0.702 k)    (This is the unit vector along AB)

FAB (the projection of F on AB) = F . uAB = ( 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 FAB . uAB = 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 – FAB = (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 ( 62 + 4.52 ) = 0.8 I + 0.6 j

The angle between BD and AB is    = cos-1 [ uAB . uBD ] = cos-1 [(0.8i +0.6j) . (-0.67i +0.33j -0.67 k) ]

= cos-1 [- 0.536 + 0.19]  = 110.24o

Force BD = (180) (uBD) = (180) (-0.67i +0.33j -0.67 k) = - 120.6 i + 59.4 j + 120.6 k

The projection of BD on AB = uAB . FBD = (-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 F1 and arranging the forces in tip to tail fashion, the tip of F4 will coincide with the tail of force F1 and the resultant of the four forces will be zero. The graphical representation is expressed mathematically as:

∑ F = 0

Free-Body 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 Newton’s third law which                  notes the existence of equal and opposite reaction to every action.

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 L0). 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 Fx = 0      If the tension at each spring is T   Then:     Tx = (1.5/3.3541) T                                 Tx

Then, 2(1.5 / 3.3541) (T) – F = 0       since: s = 3.3541-3=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:

• Establish x-y Coordinate system. & Draw free body diagram.
• Label all known forces and unknown forces and assume  the direction of unknown forces.
• Apply the equations of equilibrium ∑ Fx = 0 and ∑ Fx = 0.
• Compare the number of unknowns on the free body diagram with the number of independent equations of equilibrium available.
• If there are more unknowns to be evaluated than the number of equations, draw a free body diagram of another body and repeat the steps described above.

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:

∑ Fy = 0   then:  Q sin 40o – 300 = 0    then Q = 467 N on B    From the FBD of shaft A:

∑Fy = 0  then:  S – 500 – Q sin 40o = 0  then  S = 800 N directed upward.

∑Fx = 0  then:   R – Q cos 40o = 0  then: R = 467 cos 40o = 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 LAB = 0.4 m and its stiffness is 300 N/m

 Y           TAC  300                                                                                                                               TAB            78.5                              78.5 N

Weight of lamp W = 8 (9.81) = 78.5 N

∑Fx = 0        TAB – TAC cos 30o = 0

∑Fy  = 0        TAC sin 30o – 78.5 = 0

Then:          TAC = 157.0 N   and  TAB = 136.0 N which is the stretch of spring AB

TAB = 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 LAB = LAB + SAB = 0.4 + 0.453 = 0.853 m

The horizontal distance CB requires that   2 = LAC cos 30o + 0.853   Then LAC = 1.32.

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.

∑ Fy = 0   Then:  from FBD (3):    T + T – TB  = 0  Or:        TB = 2 T . Since: ∑ Fy = 0   Then from the FBD (4) Then:   TB + 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:

Fy =  0      Then:   T1 + T2 = 1000

Since the cord around pulley A is continuous,  T1 = T2    Then:  T1 = T2 = 500 lb

Since the cord around  pulley B is continuous,  T3 = T4 = T2 / 2 = 250 lb

Again, since the cord around pulley C is continues T3 = T   Then: T = T3 = 250 lb

Evaluation of reaction forces at pulley C:

Fx =  0  250 cos 30o – Fx = 0   Then:  Fx = 217 lb

Fy =  0      - Fy + 250 sin 30o -250 = 0   Then Fy = - 125 lb   or:  +125 lb

F = SQRT [ (217)2 + (125)2 ] = 250 l

 Lecture 9 Equilibrium of Particle Three Dimensional Force Systems (3.4)

The requirement for equilibrium of a particle is:

∑F = 0   or:   ∑Fx + ∑Fy +∑Fz = 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 ∑Fx = 0, ∑Fy = 0, ∑Fz = 0.

- Reverse the directions of unknown forces if solution yields negative result.

Example: (From Hibbeler)

Determine the magnitudes of F1, F2 and F3 for equilibrium of the particle.

Fz = 0   F1 sin 30O – 2.8 = 0   Then:  F1 = 5.6 KN

Fy = 0   - F2 (24/25) + (8.5) cos 30 = 0  Gives: F2 = 8.55  k

F3 – 5.6 cos 30O – 8.55 (7/25) – 8.5 sin 15O = 0    Then:   F3 = 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

FB = FB i

FC = FC cos 120 i + FC cos 135 j + FC cos 60 k

= -0.5 FC  i - 0.707 FC j + 0.5 FC k

FD  = FD ( -1 i + 2j + 2k) / SQRT (-12 +22 + 22) = -0.333 FD i + 0.667 FD j + 0.667 FD k

W = - 981 k

Σ F = 0    Then:

FB i  -  0.5 FC i – 0.707 FC j + 0.5 FD i + 0.667 FD j + 0.667 FD k – 981 k = 0

Setting  Σ Fx = 0 ,  Σ Fy = 0 , Σ Fz = 0  We have three unknowns and three equations, then:

FC = 813 N

FD = 862 N

FB = 693.7 N

The stretch of the spring is F = 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 Fx=0 , (4.5–1.5)/SQRT[(4.5-1.5)2+(1.5)2+(3)2] FDA+(-1.5) /SQRT[(-1.5)2+(2.5-1.5)2+(3)2] FDC

= 3 / 4.5 FDA – 1.5 / 3.5 FDC  = 0.667 FDA - 0.429 FDC = 0

Σ Fy=0 , (–1.5)/SQRT[(4.5-1.5)2+(-1.5)2+(3)2] FDA+(2.5-1.5) /SQRT[(1.5)2+(3)2+(2.5-1.5)2] FDC

+ (-1.5)/ SQRT[(1.5)2 +0 + 0)] FDB

= -1.5/ 4.5 FDA + 1 / 3.5 FDC - FDB = - 0.33 FDA + 0.286 FDC  -  FDB = 0

Σ Fz = 0 then:

(3)/SQRT[(4.5-1.5)2+(1.5)2+(3)2] FDA+(3) /SQRT[(1.5)2+(3)2+(2.5-1.5)2] FDC - W

= 3 / 4.5 FDA + 3 / 3.5 FDC  - 200 = 0.667 FDA + 0.857 FDC – 200 = 0

The above are three equations and three unknowns FDA, FDB and FDC. Solving yields:

FDA = 100 lb

FDB = 11.1 lb , and’

FDC = 155.6 lb

 Lecture 10 Force System Resultant Moment of a Force-Scalar 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 right-hand rule defined earlier.

If F is a force along A-A line, the magnitude of the moment of the force around point O is:

Mo = │F│ d = │F││r│sin θ

Where d is the “moment arm” which is the perpendicular distance from O to the line A-A.

F is parallel to x-y axis           F is parallel to z-y axis               F passes through 0

Note that if the force does not lie in a plane perpendicular to the moment axis, resolve the force into its two components: one parallel to the moment axis and he other in a plane perpendicular to the moment axis. Since the parallel force lie on the same plane as the moment’s axis, this force would have no tendency to rotate the body. The moment of the second component is the only moment of the force with respect to the axis.

EXAMPLE (From Hibbeler)

Determine the magnitude and the directional sense of the resultant moment of the force about point P.

Solution:

+

MP = (260) (5/13) [3] +(260) (12/13) [2] – (400) (sin 30o) [4-2] + (400) (cos 30o) [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 MO

a) d = 24 cos 60 = 12 in

MO = (100 lb) (12 in) = 1200 lb-in Perpendicular to the paper and pointed into the paper.

b) d = (24 in) sin 60 = 20.8 in Then: MO = 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 30o m) – 40 N (4 m + 3 Cos 30o m)

=    - 334 N.m

=     334 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 right-hand 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 = (Px i + Py j + Pz k ) X ( Qx i + Qy j + Qz k)

= Px Qx (i X i) + Px Qy (i X j) + Px Qz (i X k) + Py Qx (j X i) + Py Qy (j X j) + Py Qz (j X k)

+ Pz Qx (k X i) + Pz Qy (k X j) + Pz Qz (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 = (Py Qz- Pz Qy) i – (Px Qz – Pz Qx) j + (Px Qy – Py Qx) ki  =    Px  Py  Pz

Where the symbol is known as the determinant                 Qx Qy  Qz

EXAMPLE:

A 300 lb force acts on bracket as shown. Find the moment MA about point A

The moment is the vector MA = Δr X F where Δr is the vector pointing from A to C

Mx = 300 cos 30Oi(6)  =1558.8 in-lb   My =300 sin 30O (7) = 1050

Then MA = 1558-1050 =  508.8 in-lb  Pointing into paper

i        j        k

MA = Δr X F = | 7       6       0 | = (7x150 – 6x259.8) k = – 508.8 k  (-) Pointing into the paper

259.8  150

Example:

Determine the direction θ for  0O ≤ θ ≥ 180O 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

MA = 400 [ ( 32 + 22) ] = 1442 N-m   Since: Ф = tan-1 (2/3) = 33.69O Then θ = 90O – 33.69=  56.3O

b) MA = 0  Then:  θ = 180O – 33.69O = 146O

 Lecture 12 Force System Resultant Moment of a Force-Vector Formulation (4.3)

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 r1 = F x r2 = ׀ 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:

(MR)O = ∑(r X F)

 Lecture 13 Force System Resultant Principle of Moments (4.4)

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 = │ rx   ry   rz │= [ry Fz – rzFy] i - [rx Fz – rzFx] j + [rx Fy – ryFx] k Ξ r X F

Fx Fy Fz

│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

MO = r X F =     │ rx   ry   rz │= [ry Fz – rzFy] i - [rx Fz – rz(-Fx)] j + [rx Fy – ry(-Fx)] k

-Fx Fy Fz

Where: r is the position vector connecting point O to any point on the line of action of F.

F

z

(MO)z

r

rz                                                                                                   y

rx                                                              (MO)y

ry

x       (MO)x

The i component of the moment is the result of three moments: one due to  Fx about the x axis (but producing no moment since Fx is parallel to x-axis), the second is due to Fz about the x-axis (Ξ +ry Fz acts in the positive x-axis), the third is due to Fy about the x-axis (Ξ - rzFy acts in the negative x-axis).

Similarly:

The j component of the moment is the result of three moments: one due to Fy about the y axis (but producing no moment since Fy is parallel to y-axis), the second is due to Fz about the y-axis (Ξ - rx Fz acts in the negative y-axis), the third is due to Fx about the y-axis (Ξ - rzFx acts in the negative y-axis). Also: The k component of the moment is the result of three moments: one due to Fz about the z axis (but producing no moment since Fz is parallel to z-axis), the second is due to Fy about the z-axis (Ξ + rx Fy acts in the positive z-axis), the third is due to Fx about the z-axis (Ξ + ryFx acts in the positive y-axis).

EXAMPLE: (Higdon & Williams)

Determine the moment of the 400 lb force with respect to the horizontal line A-A

The length of CD is L = [42 + 52 + 72]1/2 = 9.49

uCD = 7/9.49 i +4/9.49 j – 5/9.49 k

F = 400 [7/9.49 i +4/9.49 j – 5/9.49 k] = 295 i + 168.6 j + 210.7 k       This produces zero moment

The z component of F is parallel to A-A line therefore it produces no moment with respect to A-A. The other two components are:

Fx = 295 i      and:  Fy = 168.6 j

MAB = the vector sum of moments of the components

i            j          k

MA-A = ( 5           6         0   )    = [(6)(210.7)] i – [(5)(210.7)] j + [(0) (5)(168.6) – (6)(295)

295     168.6   210.7   = 1264.2 i – 1053.5 j

= 1645.6 in-lb

 Lecture 14 Force System Resultant Moment of a Force about Specified Axes (4.5)

Approaches Used to Obtain Moments about Specified Axis:

Consider a force F acting on a rigid body containing point O as shown. The moment of this force about a given axis, say a-a, passing through point O, can be found  by applying the cross product between the position vector OA with the force F producing moment MO perpendicular to the plane formed by F and OA and projecting the resulting moment  into axis a-a.

1)       Scalar Approach:

Alternatively, the moment about the axis can be found by resolving F into its three components, finding the individual moments about O by multiplying the force components with shortest distance between the force and the point in question. Finally, the moment about the axis is computed by projecting the moments into the axis in question and adding them algebraically.