Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

1.1 , 1.2, 1.3, 1.4) 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:

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/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

Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Force Vectors

2.1, 2.2) 2.3) Vectors, Vector Operations and Vector Addition of Forces:

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.

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 (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:

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 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 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

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 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

2.3) Vector Addition of Forces

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

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

2.4) Resultant of Coplanar Forces:

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

F₁ F₂

F_3y F_3x

F₃

F₁ = - F_1x + F_1y , F₂ = F_2x + F_2y , F₃ = F_3x - F_3y

The resultant is: F_R = F₁ + F₂ + 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_Rx² + F_Ry² ) θ = Tan^-1 (ІF_Ry І / ІF_Rx І)

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

R_x = +199.9 R_y = +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) = 4.1^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 = F₁ + F₂ + F₃

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 Cos² α + Cos² β + Cos² α = 1 (0.42857)² + (0.28571)² + (0.8571)² = 1 OK

2.5, 2.6) Cartesian Vectors & Position 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 + Cos γ k

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

Cos α² i + Cos² β j + Cos² γ k = 1

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 │

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 u_BA:

r_BA = SQRT [(9)² + (12)² + (7)²] = 16.553

u_BA = (-9 i + 12 j + 7 k) / 16.553 = -0.5437 i + 0.7249 j + 0.4229 k

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 │ = ( 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 F_R = 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 Cos α = 0.2617 α = 74.8^o and: Cos β = - 0.2094 β = 102^o

and: Cos γ = 0.9422 γ= 19.6^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.

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_AB = (120 N) [ r_AC / ׀r_AC׀ ] = (120 N) [ (4 / 6) i + (2 / 6) – (4 / 6) k ]

F_AB = [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

2.7, 2.8) Position Vector Along a Line:

We have shown that the position vector along a line AB is:

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 1: (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: 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

2.9) Dot Product:

Dot product of two vectors P and Q (otherwise known as scalar product) is defined as the product of their two magnitudes and the cosine of the angle formed by P & Q. Dot product of two vectors 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_yQ_y + P_z Q_z

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

u_{P = P /│P│}

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

Since │P_a-a│ = │P│cos θ Then: │P_a-a│ = P . u_a-a and: P_a-a = [ P . u_a-a ] u_aa

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

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 ( 6² + 4.5² ) = 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^‘

1) 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 the forces acting on a body from its surroundings and drawing the forces acting on the body.

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 spring has stiffness of 500 N/m and outstretched length of 6 m. 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:

AC = 3.3541 m

Σ F_x = 0 If the tension at each spring is T Then: T_x = (1.5/3.3541) T

Then, 2(1.5 / 3.3541) (T) – F = 0 = 3.3541-3=0.3541 m

T = K S = (500)( 0.3541) = 177.05 N Then: F = 158 N

2) Coplanar Force System:

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 ∑ F_x = 0 and ∑ F_x = 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:

∑ 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 unreformed length of the spring is L^’_AB = 0.4 m and its stiffness is 300 N/m

Y

T_AC

T_AB

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

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.

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

∑ 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: 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 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