Particle Equilibrium

• Discusses the equilibrium of a particle, which is small enough so that size does not affect equilibrium.
• There is no rotation in these analyses because they involve infinitesimal particles.
• Equilibrium means that a particle or body is not moving and all forces are balanced.

Introduction to Particle Equilibrium

• With a small particle on the left, size does not affect equilibrium.
• On the right, a particle is pictured where size affects equilibrium because rotations could occur.

Balanced Forces

• Balanced forces mean:
• ΣFx = 0: All vector components of the Cartesian î axis.
• ΣFy = 0: All vector components of the Cartesian ĵ axis.
• ΣFz = 0: All vector components of the Cartesian k axis.
• There are multiple equations: 3 in space, 2 in the plane.
• In 2D: ΣFx = 0, ΣFy = 0
• In 3D: ΣFx = 0, ΣFy = 0, ΣFz = 0
• Only these equations are relevant because of dealing with particles. If particle size is large enough, equilibrium would have to include pairs or moments (torques in physics).

Free Body Diagram

• A free-body diagram is a starting concept.
• Draw a diagram with the forces involved, examining whether all forces should be considered.
• Determine the point of interest.
• Isolate to observe the forces involved.
• Draw the system that should be in equilibrium, noting that ropes, cables, and chains are always in tension and cannot withstand compression.
• Label all elements in the free-body diagram.
• Resolve all forces that have an angle in the system as Cartesian components.
• If the angled force moves away from the point, the components move away from that point, and vice versa if they move away.
• Write the equilibrium equations: ΣFx = 0, ΣFy = 0
• Solve for unknown forces using the equilibrium equations.

System Resolution Example

• The following system is presented as an example to resolve:
• A structure with points A, B, and C.
• A line runs from points A to B and A to C, both anchored at the top.
• Element A is connected to a block that weighs 50 Newtons

Steps 1-4

• Steps 1 to 4: Determine the point of interest, isolate the body to be studied, and add the forces involved.
• Step 5: Label forces.
• Step 6: Resolve the forces into Cartesian components.

Steps 7 & 8

• Step 7: Write the equilibrium equations.
• Step 8: Solve for the unknowns to the equation ΣFx = 0 = TABcos(20°) – TAccos(30°) and ΣFy = 0 = TABsin(20°) + TAcsin(30°) – 50.
• TAB = 56.5 N
• TAC = 61.3 N

Angle Example

• Exercise: Find the angle provided in the image.

Cable Tension Example

• Exercise: Find the tension in the cable provided in the image.

Cable and Pulley Example

• Exercise: Find the tension in the cable and the force on the pulley shown in the diagram

Coplanar Equilibrium Problems

• Coplanar force equilibrium problems for a particle can be solved using the following procedure:

Free-Body Diagram Steps

• Establish the x and y axes in any suitable orientation.
• Mark on the diagram all the magnitudes and directions of known and unknown forces.
• Assume the direction of a force with an unknown magnitude.

Equilibrium Equation Steps

• Apply the equilibrium equations ΣFx = 0 and ΣFy = 0.
• The components are positive if directed along a positive axis; they are negative if directed along a negative axis.
• Springs, apply F = ks to relate the spring force to the spring's deformation, s. Since the magnitude of a force is always positive, a negative result indicates that the force's direction is opposite to that shown on the free-body diagram.

Original Angle

• If the original angle was 45°, consider calculating the weight of each block in the diagram.

Maximum Load Problems

• Maximum load problems, considerations:
• Step 1: Choose a point to start, draw the free-body diagram, and solve for the unknowns in that part of the problem. (Follow the same analysis steps as for particle equilibrium.)
• Step 2: Choose a member under maximum load.
• Step 3: Write and solve the equilibrium equations.
• Step 4: If many free-body diagrams exist, draw the next free-body diagram, usually the contiguous one.
  • Note: With multiple free-body diagrams, force vectors in both diagrams (contiguous diagrams) act in opposite directions (interaction between bodies should be in equilibrium, therefore, forces are opposite).
• Step 5: Write the equilibrium equations and solve.
• Step 6: Repeat the procedure to find all unknowns.
• Recommendation: For problems that provide a minimum and maximum condition, never condition more than one vector under maximum load simultaneously when solving the system.

Find Max Weight

• Find the maximum weight of the lamp without exceeding 100# in the cord.

Step 1: Select Starting

• Step 1: The starting point is selected, generally as far as possible from the point of interest (the lamp's weight); begin at point A.

Step 2: Select Cord

• Step 2: Select a cord with a maximum load of 100# and solve for the remaining tensions.
• If there are three unknowns, one is given (100), and there are two equations to find the remaining tensions.
• Selecting the correct cord will result in the other tensions being less than 100#.

Step 3: Equilibrium Equations

• Step 3: The equilibrium equations are written and observations are made regarding calculated tensions.
• ΣFx = 0 = 100cos(45°) + ABcos(75°) – AD
• ΣFy = 0 = 100sin(45°) – ABsin(75°)
• The result is AD = 89.66 lbs and AB = 73.19 lbs.

Step 4: Free Diagram

• Step 4: Diagram the following equilibrium point.
• ΣFx = 0 = BCcos(30°) – 73.19cos(75°)
• ΣFy = 0 = BCsin(30°) + 73.19sin(75°) – Wlight
• The result is BC = 21.87 lbs and Wlight = 81.63 lbs max.

System Example

• Task: Solve the system.

Cable Weights

• For the problem below, weights are connected by a cable system through an ideal pulley with no friction; find forces DB, AE, and AB

Max Cable Weight

• If the weight of 175# is unknown and cable AB's maximum tension is 500#, the maximum weight in A must be determined.

Static Rigid Bodies

• The object is non-deformable, this is not a real situation, but sufficient for practical statics problems.
• Dimensions of the object intervene in rigid-body equilibrium.
• Equilibrium requires ∑F (forces) and ∑M (moments).

Equilibrium Question

• Determine if the following figure is in equilibrium.
• The terms torque, moment, and pair are introduced.
• All three terms can be defined as a tendency to rotate.
• Moment is a general term; torque indicates torsion.

Moments & Torque

• External moments are represented as a curved double arrow or a single curved arrow according to the right-hand rule. The course uses a single curved arrow to indicate the moment.
• A pair is two parallel forces of the same magnitude in opposite directions.
• There's a demonstration of torque involving twisting a pipe as it causes torsion to appear.

Figure Analysis

• The question is posed to identify which of the figures is an example of a moment
• Also the state of equilibrium is investigated
• Furthermore, the additional product produced by torsion is inquired about

Calculating Moment

• The equations for calculating a moment are M= F * d, where
• F is Force
• d is the perpendicular distance to the force vector.
• Additionally M = r x F, where
• r represents the position vector

Position Vector

• The location of the tail of a vector, from where the moment needs to be known, and whose point falls on the line of action of the force creating the moment.
• It is clarified that r is a position vector, not a unit vector, and is not related to Lamda (λ).

Moment Magnitude

• The equation for calculating a moment is M = F * d, where:
• F: is force, and
• d: is the perpendicular distance to the force vector Also M = r x F
• Where r is a positional vector, which is the location of the tail of a vector from where it is required to know the moment and whose point falls into the line of action of the force
• It's also clarified that r is a positional vector and NOT a unit vector, in addition to not being equal to Lamda(λ), and it's stated that the magnitude of any Lambda (λ) is one, in effect, lacking to show the space to discover the moments, from which the distance required to know it can be extracted.

Moment Characteristics

• Characteristics of the moment:
• A torque wrench is presented as an example
• One way to increase a moment is by increasing its arm or the distance d, or the position vector in a vectorial manner

Increasing Torque

• There's a demonstration of the capacity to make the nut rotate increasing when the key has its length also boosted.

2D Problems

• The resolution of 2D problems is performed in two manners, first M = r x F needs to be written in a Cartesian format from which both the vector of position and the force.
• Then obtaining a cross product using the determinant method or Cramer's rule.

Moment Calculation

• The other method to find the moment would be through the components of force multiplied by their perpendicular distance.

Determinant Methodology

• Calculate the cross-product
• Create a matrix with vectorial components î,ĵ, and k in the first fila
• In the second fila, vectorial components of the first vector are collated
• In the third fila, the components of the second vector are placed.

Vectorial Quantities

• Cover the î column and by means of cross-multiplication multiply the coefficients of the first vector with the coefficients of the ĥ vector, then subtract the * k* components of the second vector.

Cross Product

• Vectorial quantities (Cont.):
• Next, the j column will be removed to commence its multiplication and find the factor of vector for the coefficient of vector by the result, but subtracting the resulting coefficients and vector by the result.
• The outcome for this term has a subtraction sign before the operation and the cofactor for the situation is negative.

Cross Product Multiplying

• In the next step, the coefficient (k) for the first vector requires a product and the coefficient for the unit (i) and its position need to be the position vector from which the unit vector is also necessary for the vectorial coefficients and its result must be also used.
• Finally, the cross product result for the position in each unit needs to occur.

Matrix Steps

• An example for calculating axb is laid out
• The first step is building the matrix
  • a = 5î - 7ĵ + ak
  • b = 6î + 8ĵ - 7k

Cover J

• In second, “j” gets covered while multiplying "j" to "k" for the first vector.

Negative Sing

• Negative Sing:
• Step A: "î" gets covered, and it will be "I" x " K"
• The product must precede a negative sign

Vector Calculation

• Step #4: Cover k and multiply “i” times “j” for the second vector and then reverse to the first, This term MUST be positive
• In step #5 the equation, which yields: [(49) - (16)] i - [(−35) – (12)]j + [(40) – (-42)] k = 33i + 47j + 82k

Examples to Practice

• Find axb.

Moment Example

• Example: Calculate the moment using two methods.
• A pole attached at the bottom makes a 30 degree angle with the ground.
• There is a force of 100 # a the top of the pole.

Moments Around a Point

• Take into account knowing the momentum by vector method requires performing the product in given order M =řxF
• Vectorial Methods should obtain a accurate result. Component Method; F* d requires only for the distance to be the force perpendicular. The method F*d should be for 2D scenarios

Recommendable Products

• The method M =řxF and is recommended for 3-dimensions as its is on axis (3D) *

Moment Considerations

• In cases one were to use M=F"d in (3D), ensure a consideration
• A force in 2 units doesn't occur in other ones or otherwise.
• A final note about a difference in momentum

Moments Around a Point: Right-Hand Rule

• Moments Around Point Rule (Right-Hand):
• Pointing fingers (on your right-hand) at position vector

Right Hand Vectors

• The thumb in (R.H.) shows what direction the Force Vector is pushing (it pushes).
• The First vector (i) that pushes the Vector's arm for (positive or negative) Moments.

Moment Problems

• Find both ways the Vector that applies the Momentum to point A. (Two examples are cited for this Point).

Class Problems

• Examples of calculating the moment at point using M=Fd* and M = r x F

Torque & Force

• If the Force to arm are perpendicular then the results are max!

Tensile Cord Question

• What happens as the Tension is 2500lbs?
• And what way does the Tensile pull?

Moment Around Axis

• The amount that occurs around any given area, like air flow!
• Vectors have used for known parts for a test

Procedure for Calculation

• Procedure for Calculation:
• A: Place the first Unit in (Cartesian) form.
• B: PinPoint the right location for the Axle (360 spin).
• C: Align vector tail at the Rotation Axis.
• D: Draw+ (Cartesian) force action.
• E: Vector forces (Cart).

Matrical Methods

• Matrical:
• a) (this generates momentum: cartesian);
• 2. Write the Unit Vector (Cartesian) for the Arbritary Axle;
• 3. dot point, vector (Moment) the Unit's Axe. Note: scalar, it also measures the (Arbertary) movement and is (negative or positive*
• The axis moment is the arbritary anti-clock.

Torque Findings

• Solve Vector to tube

Vectory Point

• Follow vectors to there points

Cartesian Expression

• Vector is already expressed

Calculating Torque

• Solve for products

Vectory findings for point A through B for Axles

• Vectors must also have a Torque/Vector A.

Finding Point Products

• Step by point from now on

Axis Torques

• Find axis the tube M=F*d, M = ř x F, solución a los problemas de clase.

Moment Findings

Calculations need what to do with max and mins

• to offer max; Vector forces have perpendicular angles -to offer min: Vectors must be close in length.

Rigid Cord Notes

• At Point A, the Pull occurs while the String is at (2500lbs force for Max. Points).

Areas of All Arcs

• The Axial is the (Rotation/Torque Point Vectors = The Vector that flows to point by the arbritary

Summary of Finding

• Find where the axis is at the torque -

Equivalent Systems

• In finding areas and what connects them it can cause (one) Force at (same) point.
• Redact (Re-State/Sub) you can re work parts to cause basic patterns

Equivalent Notes

• System types may vary! (1+ A and C will be easy finds)!

Equivalent Systems

The systems should have force in place while the other does not require a multiplier and sum. to that degree.

Moment Observations

If there are systems that have observations that occur for multipliers then the system is (Horizontally) positive and visa versa.

Equal Forms Rules

• You must find an equal version and follow equal rules or steps
• Find (X) Force for first System

Make (system) find Y power * equal systems And finally, make (S) to see what equal

Simplified Step

• Step 4 must show each force equal

Equal System

• Step= Be Cautios, force must meet/greet

Beam Load

• Replace this by : A Par or Force with a Force at all

Task to March

Equivalent force, in task. A force in action.