09.1 Moments (Couples) Concept Overview
MOMENTS (COUPLES) | CONCEPT OVERVIEW The TOPIC of MOMENTS (COUPLES) can be referenced on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
CONCEPT INTRO: A FORCE has TWO EFFECTS when acting on a RIGID BODY, or SOLID OBJECT, in a TWO or THREE DIMENSIONAL space, they are: 1. A FORCE will MOVE the CENTER OF MASS of a RIGID BODY creating ACCELERATION along the LINE OF ACTION of the FORCE, as well as an associated DISPLACEMENT. 2. A FORCE will cause a RIGID BODY, or SOLID OBJECT, to ROTATE about an AXIS. The tendency for a FORCE to ROTATE, TURN, or TWIST a RIGID BODY about an actual or assumed pivot point is referred to as a MOMENT. When a MOMENT acts on an UNRESTRAINED RIGID BODY, that body will ROTATE. On the other hand, when a MOMENT acts on a RESTRAINED RIGID BODY, there will be NO ROTATION.
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The MOMENT of a FORCE about some specified point QUANTIFIES the TENDENCY of that FORCE to ROTATE the RIGID BODY about that point. The MAGNITUDE of a MOMENT identifies the AXIS OF ROTATION associated with the rotational force. If the LINE OF ACTION of a force PASSES THROUGH the CENTER OF ROTATION, the NET RESULTANT MOMENT on the body will be ZERO, thus, there will be NO ROTATION. The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ You may also be familiar with a MOMENT being presented in the form: π = π Γ πΉ = π πΉ π πππ Where: β’ The variable r is the RADIUS VECTOR, or DISTANCE, from the point at which the MOMENT is being calculated to the point at which the force is being applied.
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This distance is typically given in units of meters, inches, feet, or another unit of length. β’ The variable F represents the FORCE being applied within the system that is creating the MOMENT. This FORCE is typically given in units of pounds or Newtons. β’ π is the angle between the FORCE and the AXIS that we are pulling our RADIUS VECTOR, or DISTANCE, back to. This MOMENT can also be broken up in to COMPONENT MOMENTS about a particular AXIS using the following formulas: π+ = π¦πΉ- β π§πΉ0 π0 = π§πΉ+ β π₯πΉπ- = π₯πΉ0 β π¦πΉ+ When we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM.
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Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. We may also commonly hear a MOMENT referred to as a TORQUE when we are working with applications such as shafts and other power-transmitting machines. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics. A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. Whereas a FORCE has a unique POINT OF APPLICATION (location in space) and a defined LINE OF ACTION (line in space on which the vector is acting), a MOMENT equivalently and uniquely defined by the AXIS OF ROTATION, which is the AXIS within the FORCE system that the FORCE ROTATES the RIGID BODY.
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Illustrating this further, we can state that:
The AXIS OF ROTATION can either be WITHIN the RIGID BODY, which results in a rotational behavior known as PIVOTING, or OUTSIDE the RIGID BODY, resulting in a rotational behavior known as ORBITING. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your
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answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system.
AXIS OF ROTATION: In TWO DIMENSIONS, the X and Y CARTESIAN AXES are usually reside in the plane of the page, which results in the THIRD AXIS, the Z-AXIS, protruding out of the page as it runs perpendicular to the TWO DIMENSIONAL AXES. A TWO DIMENSIONAL AXES MOMENT of a FORCE located in the XY CARTESIAN PLANE is always rotating about the Z-AXIS (the AXIS OF ROTATION is PARALLEL to the Z-AXIS). In vector terms, this fact means that the MOMENT in the XY CARTESIAN PLANE has a UNIT VECTOR DIRECTION of π (either positive, or negative depending on the SENSE of the moment). The AXIS OF ROTATION of the MOMENT is NORMAL to the PLANE containing the FORCE VECTOR and the POSITION VECTOR.
THE RIGHT HAND RULE: The SENSE of a MOMENT can quickly be determined using the RIGHT HAND RULE.
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The RIGHT HAND RULE only works with the right hand! REMEMBER that come exam day and under timed conditions, if you use your left hand by mistake, the direction of your Z-AXIS will be flipped backwards. Using the RIGHT HAND RILE, we will: β’ ALIGN our fingers with the CARTESIAN AXIS for which our FORCE and POSITION VECTOR reside. β’ DEFINE the AXIS OF ROTATION by looking at the DIRECTION that your THUMB is pointing. β’ DEFINE the SENSE of the MOMENT by looking at the DIRECTION that your THUMB rotates when you curl your fingers in to your body. In THREE DIMENSIONS, a MOMENT is POSITIVE about the X-AXIS if itβs acting COUNTERCLOCKWISE when youβre looking at the ROTATION of the tip of your thumb. The same applies to the Y-AXIS and your forefinger and to the Z-AXIS and your middle finger.
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Illustrating the AXIS OF ROTATION versus the standard of POSITIVE MOMENTS, we have:
To solidify your understanding and ensure that you are comfortable using the RIGHT HAND RULE, grab a pencil and pen and practice using the following procedure. β’ Designate the left end of the pencil as the pivot, and hold it rigidly with your finger. β’ The pen will be the force. o If the pen represents an upward force, you will be applying the force below the pencil, but directed upwards. o If the pen represents a downward force, you will be applying the force above the pencil, but directed downwards.
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Run through the following scenarios, noticing which DIRECTION the pencil is ROTATING, keeping the following in mind:
Note on a piece of PAPER whether you are seeing CLOCKWISE, COUNTERCLOCKWIS, or if it is NOT ROTATING at all: 1. Apply an upward force through the center of the pencil 2. Apply an upward force through the pivot 3. Apply an upward force through the edge farthest from the pivot 4. Apply a downward force through the center of the pencil 5. Apply a downward force through the pivot 6. Apply a downward force through the edge farthest from the pivot Here is what you should have observed: 1. COUNTERCLOCKWISE 2. NOT ROTATING 3. COUNTERCLOCKWISE
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4. CLOCKWISE 5. NOT ROTATING 6. COUNTERCLOCKWISE
COUNTERACTING A MOMENT (OR COUPLE): Consider the beam as illustrated:
This beam is ROTATING at some PIVOT POINT due to a FORCE being applied at some unspecified distance r away. We donβt know the self weight of this BEAM, nor do we know the location of the PIVOT point, however, we do know that the FORCE applied to this beam will cause it to ROTATE about that PIVOT POINT. This can be confirmed and quantified had we had more definition of the FORCE, F, and that DISTANCE, r, illustrated for us.
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However, based on the limited information we do have, we can conclude that with the FORCE acting to the LEFT of the PIVOT POINT, and with a DOWNWARD direction, that the ROTATION experience at the PIVOT POINT will be COUNTERCLOCKWISE. In ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want these ROTATIONS to occur, we need to counteract them in some form or fashion. This can be done in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. The LOCATION of the FORCE can be either on the LEFT or RIGHT of the PIVOT POINT as long as it is creating that MOMENT we need to establish EQUILIBRIUM. The GEOMETRY of the BEAM along with the LOCATION of the PIVOT POINT will also drive the location of this additional FORCE, as a POSITION VECTOR, along with the FORCE, are the TWO COMPONENTS that define a MOMENT. The following scenarios illustrate the GEOMETRICAL dependency of equalizing FORCES and their subsequent MOMENTS.
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PIVOT LOCATED AT CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at the very CENTER of itβs GEOMETRY, so that the DISTANCE from the CENTER POINT to the LEFT END and the DISTANCE from the CENTER POINT to the RIGHT END are EQUAL, or r:
Adding a FORCE, F, at an equal distance, r, to the RIGHT of the PIVOT POINT and acting in the SAME DIRECTION, DOWNWARD, would create a MOMENT equal in MAGNITUDE and an OPPOSITE SENSE, CLOCKWISE.
PIVOT LOCATED OFFSET AND TO THE RIGHT OF THE CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at a DISTANCE that is to the RIGHT of the CENTER of its GEOMETRY.
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The DISTANCE from the CENTER POINT and to the RIGHT END, as illustrated, is half that of the DISTANCE from the CENTER POINT and to the LEFT END. The MOMENT in this case would be the PRODUCT of the FORCE TIMES the DISTANCE of the FORCE from the PIVOT POINT, or: π = ππΉ To create EQUILIBRIUM, we need to NEUTRALIZE the effects of this MOMENT with a FORCE placed elsewhere on the BEAM, such that: π34 = π35 Or: π34 β π35 = 0
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We can do one of two things here: β’ Add a FORCE, F, that is EQUAL in MAGNITUDE but OPPOSITE in DIRECTION at the SAME LOCATION of the ORIGINAL FORCE. β’ Add a FORCE, F, that is DOUBLE the MAGNITUDE of the FORCE on the LEFT and ACTING in the SAME DIRECTION, DOWNWARD. Because the DISTANCE to the RIGHT END is half that of the DISTANCE to the LEFT END, this has to be accounted for by DOUBLING the counteracting FORCE. Laying this out in FORMULAIC TERMS, we would have:
ππΉ4 =
π 2πΉ 2 4
PIVOT LOCATED OFFSET AND TO THE LEFT OF THE CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at a DISTANCE that is to the RIGHT of the CENTER of its GEOMETRY.
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The DISTANCE from the CENTER POINT and to the RIGHT END, as illustrated, is double that of the DISTANCE from the CENTER POINT and to the LEFT END. The MOMENT in this case would be the PRODUCT of the FORCE TIMES the DISTANCE of the FORCE from the PIVOT POINT, or: π = ππΉ To create EQUILIBRIUM, we need to NEUTRALIZE the effects of this MOMENT with a FORCE placed elsewhere on the BEAM, such that: π34 = π35 Or: π34 β π35 = 0 We can do one of two things here: β’ Add a FORCE, F, that is EQUAL in MAGNITUDE but OPPOSITE in DIRECTION at the SAME LOCATION of the ORIGINAL FORCE. β’ Add a FORCE, F, that is HALF the MAGNITUDE of the FORCE on the LEFT and ACTING in the SAME DIRECTION, DOWNWARD. Because the DISTANCE to the RIGHT END is double that of the DISTANCE to the LEFT END, this has to be accounted for by HALVING the counteracting FORCE.
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Laying this out in FORMULAIC TERMS, we would have:
ππΉ4 = 2π
πΉ4 2
UNITS: The TOPIC of UNITS of FORCE can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The most common UNITS we will encounter in our studies for a MOMENT, or TORQUE, will be the SI UNIT known as the NEWTON-METER, denoted as N-m. Though, we are probably more familiar with and comfortable using SI UNITS, we should be prepared to work problems come exam day using the USCS UNITS as well. The standard USCS UNIT for MOMENTS, or TORQUE, is the FOOT-POUND, denoted using the symbol ft-lb. The GENERAL FORMULA that will help is CONVERT the UNITS OF A MOMENT can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. To CONVERT from SI UNITS to USCS UNITS, we can use the following CONVERSION FACTORS: β’ 1 ππ = 0.738 ππ‘ ππ β’ 1 ππ‘ β ππ = 1.356 π β π
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To fully solidify the existence, and our understanding, of MOMENTS, or TORQUES, it is important that we visualize practical examples as they would occur in engineering systems: To further assist in developing your understanding, and where these MOMENTS and TOURQUES, can occur, here are a few examples: β’ MOMENTS can emerge as REACTION FORCES. One example of this is the RESISTANCE you encounter when TURNING THE STEERING WHEEL of a car. The FORCE you feel is created by MOMENTS acting on the wheels where they touch the ground. β’ MOMENTS can also appear as INTERNAL FORCES in structural members or components. One example of this can be seen when a BEAM BENDS because of an INTERNAL MOMENT with a DIRECTION TRANSVERSE to the direction of the beam. An internal moment causes points to rotate relative to each other just as an internal force causes points in a solid to move relative to each other.
MOMENT-COUPLE SYSTEMS: A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force.
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The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, as expressed: π4 = π5 Or: πΉ4 π4 = πΉ5 π5 A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements. One example of a COUPLE is the forces from your fingers on to a doorknob. As you turn the knob, one finger is pushing up on the side of the doorknob and the other is pushing down. The CUMULATIVE AFFECT of these TWO FORCES TOGETHER cause the knob to ROTATE.
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CONCENTRATED MOMENTS: You may also hear a COUPLE referred to as a PURE MOMENT or CONCENTRATED MOMENT. When developing a FREE BODY DIAGRAM, CONCENTRATED MOMENTS are easy to apply and account for. MOMENTS are vectors, and because of this, they too will display MAGNITUDE, SENSE, and a POINT OF APPLICATION. For both TWO and THREE DIMENSIONAL objects, we will need to IDENTIFY the MAGNITUDE of a CONCENTRATED MOMENT as a NUMERICAL VALUE, if known, or a LABEL it as a VECTOR if itβs not known.
THE TRANSMISSION OF MOMENTS: The TRANSMISSION of MOMENTS can occur through multiple objects. For example, this can be observed in the operation of the drive train in your car, or even when you are turning the steering wheel, as illustrated earlier. As far as the drive train example, the engine of your car causes the transmission to rotate, which in turn causes the axles to rotate, which in turn cause the wheels to rotate.
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MOMENTS (COUPLES) | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The magnitude of the moment about the base point, O, created by the 600 π force as applied, is best represented as:
A. 2,610 N β m (counterclockwise) B. 2,610 N β m (clockwise) C. 2,980 N β m (counterclockwise) D. 2,980 N β m (clockwise)
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SOLUTION: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics.
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A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE In this problem, we have the POINT OF APPLICATION, but need to define the MAGNITUDE as well as the SENSE acting at POINT O. The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. When we are dealing with a FORCE that is at an ANGLE relative to the COORDINATE SYSTEM in which it resides, we will need to use our knowledge of RESOLVING a FORCE in to COMPONENTS so that we can assess each one INDIVIDUALLY and how it affects the ROTATION about our POINT OF APPLICATION.
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In this problem, we are given the illustration:
The 600 π force generates a MOMENT about the base at point βπβ. At point βπβ, this MOMENT will generate rotation, a SENSE, in either the clockwise or counterclockwise direction. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your
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answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system. We will assume the COUNTERCLOCKWISE direction as POSITIVE. In this problem, we have all the GEOMETRY nicely illustrated for us, so this affords us an opportunity to attempt to do one of two things, we can: β’ Break the FORCE VECTOR in to itβs COMPONENTS and define a PAIR of MOMENTS to determine the CUMULATIVE AFFECT at POINT O. β’ We can define the LINE OF ACTION of the FORCE and an associated LEVER ARM that is PERPENDICULAR between the LINE OF ACTION and AXIS OF ROTATION. The FACT that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATION is IMPORTANTβ¦many fail right off the bat by creating a LEVER ARM from the POINT OF APPLICATION back to the AXIS OF ROTATION. This second approach, if possible, is a HACK of sorts, if you get it dialed, you are going to DOMINATE these types of problems, so letβs take this routeβ¦at least to see if we can make it happen.
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If not, we just fall back to our initial approach, which at the end of the day, unfolds as we progress through the attempted HACK. Illustrating the creation of a LEVER ARM from the AXIS OF ROTATION PERPENDICULAR back to the AXIS OF ROTATION for this problem, we have:
The next step is to define a COORDINATE SYSTEM relative to the AXIS OF ROTATION.
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This will allow us to use the GEOMETRY defined, if possible, to determine the PERPENDICULAR DISTANCE between the AXIS OF ROTATION and the LINE OF ACTION of the FORCE, doing so we have:
We know by the GENERAL definition of a MOMENT, that we need TWO COMPONENTS defined, the PERPENDICULAR DISTANCE and the MAGNITUDE of the FORCE.
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As we are given the MAGNITUDE of the FORCE, 600 N, we need to see if we can get the LENGTH of the LEVEL ARM as illustrated:
Adding in the DIMENSIONS that we are given, we have:
Unfortunately, there is no quick way to determine the LENGTH of our LEVER ARM from what we are given, which means we will need to fall back to calculating the HORIZONTAL and VERTICAL COMPONENTS based on the provided GEOMETRY.
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As you see, no time is lost up to this point, however, if we were able to define that LEVER ARM as illustrated, it would of cut down massively on the amount it takes to solve this problemβ¦but letβs move on. The illustration above provides us a clear representation of the HORIZONTAL and VERTICAL PERPENDICULAR DISTANCES between the AXIS OF ROTATION and the LINE OF ACTION for the FORCE, which affords us the ability to break the force into its COMPONENTS. The GENERAL FORMULAS revolving around the COMPONENTS of a VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For this reason, we must memorize this process and the various formulas, understanding fully its application independent of the NCEES Supplied Reference Handbook. Knowing where our ANGLE of 40Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉ+ = πΉ\ cos π β’ The VERTICAL COMPONENT of the force: πΉ0 = πΉ\ sin π Letβs start with calculating the HORIZONTAL COMPONENT.
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Pulling the data we have and plugging it in to the general formula established, we get: πΉ+]^^ = (600 π) cos(40Β°) Or: πΉ+]^^ = 460 π Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉ0]^^ = (600 π) sin(40Β°) Or: πΉ0]^^ = 386 π Revisiting and pulling back down the GENERAL FORMULA for a MOMENT in our TWO DIMENSIONAL space, we have: π = ππΉ Since we have broken our single FORCE in to TWO COMPONENT forces, we are going to have to create a LINE OF ACTION for each, define the associated DISTANCES and add the AFFECTS of each COMPONENT FORCE to determine the CUMULATIVE AFFECT of the SINGLE FORCE on the ROTATION about the AXIS OF ROTATION.
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In FORMULAIC TERMS, this will resolve to: π_ = πΉ+ π+ + πΉ0 π0 The GEOMETRY is already laid out perfectly for us, revisiting it, we have:
Remember that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATION, this is IMPORTANT! This allows us to define: π+ = 4 π π0 = 2 π And already having defined: πΉ+ = 460 π
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πΉ0 = 386 π We are now at a point where we can just plug and chug. Plugging in the data that we have defined up to this point, we can calculate the CUMULATIVE MOMENT generated about point βπβ as: π_ = 460 π 4.00 π + (386 π) 2.00 π Or: π_ = 2,610 π β π The last thing we need to do is determine the SENSE of the MOMENT about POINT O. We can do this using the RIGHT HAND RULE or visually looking at the FREE BODY DIAGRAM along with the MAGNITUDE of each individual COMPONENT FORCE and the MOMENT it creates about POINT O. Taking this route, we can conclude that: β’ πΉ+ will create a MOMENT that is CLOCKWISE about POINT O β’ πΉ0 will create a MOMENT that is CLOCKWISE about POINT O Therefore, because both MOMENTS create a CLOCKWISE MOMENT about POINT O, the CUMULATIVE MOMENT will be CLOCKWISE as well.
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However, had each COMPONENT FORCE created a MOMENT that is OPPOSITE from the other, then we would have to ASSESS the MAGNITUDE of the MOMENT created by each COMPONENT. The correct answer choice is B. π, πππ π β π¦ (πππππππππ)
HOW IT ALL PLAYS OUT: The FOLLOWING STEPS are paramount when starting any problem that involves finding REACTIONS or MOMENTS. 1. ESTABLISH a COORDINATE AXIS SYSTEM. Drawing a small coordinate system in the corner of the problem will help! 2. ESTABLISH whether CLOCKWISE, or COUNTER-CLOCKWISE MOMENT direction will be POSITIVE. It doesnβt matter which route you take, stick to your choice, CONSISTENCY is key! 3. USE the EQUATIONS OF EQUILBRIUM to determine the REACTIONS or MOMENTS of any given problem. If the structure, or beam, is said to be in EQUILIBRIUM, then all of these MUST be trueβ¦everything must balance. β’ Sum of the forces in the βπ₯β direction:
πΉ+ = 0
β’ Sum of the forces in the βπ¦β direction:
πΉ0 = 0
β’ Sum of the moment about a chosen point:
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CONCEPT INTRO: A FORCE has TWO EFFECTS when acting on a RIGID BODY, or SOLID OBJECT, in a TWO or THREE DIMENSIONAL space, they are: 1. A FORCE will MOVE the CENTER OF MASS of a RIGID BODY creating ACCELERATION along the LINE OF ACTION of the FORCE, as well as an associated DISPLACEMENT. 2. A FORCE will cause a RIGID BODY, or SOLID OBJECT, to ROTATE about an AXIS. The tendency for a FORCE to ROTATE, TURN, or TWIST a RIGID BODY about an actual or assumed pivot point is referred to as a MOMENT. When a MOMENT acts on an UNRESTRAINED RIGID BODY, that body will ROTATE. On the other hand, when a MOMENT acts on a RESTRAINED RIGID BODY, there will be NO ROTATION.
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The MOMENT of a FORCE about some specified point QUANTIFIES the TENDENCY of that FORCE to ROTATE the RIGID BODY about that point. The MAGNITUDE of a MOMENT identifies the AXIS OF ROTATION associated with the rotational force. If the LINE OF ACTION of a force PASSES THROUGH the CENTER OF ROTATION, the NET RESULTANT MOMENT on the body will be ZERO, thus, there will be NO ROTATION. The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ You may also be familiar with a MOMENT being presented in the form: π = π Γ πΉ = π πΉ π πππ Where: β’ The variable r is the RADIUS VECTOR, or DISTANCE, from the point at which the MOMENT is being calculated to the point at which the force is being applied.
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This distance is typically given in units of meters, inches, feet, or another unit of length. β’ The variable F represents the FORCE being applied within the system that is creating the MOMENT. This FORCE is typically given in units of pounds or Newtons. β’ π is the angle between the FORCE and the AXIS that we are pulling our RADIUS VECTOR, or DISTANCE, back to. This MOMENT can also be broken up in to COMPONENT MOMENTS about a particular AXIS using the following formulas: π+ = π¦πΉ- β π§πΉ0 π0 = π§πΉ+ β π₯πΉπ- = π₯πΉ0 β π¦πΉ+ When we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM.
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Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. We may also commonly hear a MOMENT referred to as a TORQUE when we are working with applications such as shafts and other power-transmitting machines. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics. A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. Whereas a FORCE has a unique POINT OF APPLICATION (location in space) and a defined LINE OF ACTION (line in space on which the vector is acting), a MOMENT equivalently and uniquely defined by the AXIS OF ROTATION, which is the AXIS within the FORCE system that the FORCE ROTATES the RIGID BODY.
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Illustrating this further, we can state that:
The AXIS OF ROTATION can either be WITHIN the RIGID BODY, which results in a rotational behavior known as PIVOTING, or OUTSIDE the RIGID BODY, resulting in a rotational behavior known as ORBITING. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your
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answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system.
AXIS OF ROTATION: In TWO DIMENSIONS, the X and Y CARTESIAN AXES are usually reside in the plane of the page, which results in the THIRD AXIS, the Z-AXIS, protruding out of the page as it runs perpendicular to the TWO DIMENSIONAL AXES. A TWO DIMENSIONAL AXES MOMENT of a FORCE located in the XY CARTESIAN PLANE is always rotating about the Z-AXIS (the AXIS OF ROTATION is PARALLEL to the Z-AXIS). In vector terms, this fact means that the MOMENT in the XY CARTESIAN PLANE has a UNIT VECTOR DIRECTION of π (either positive, or negative depending on the SENSE of the moment). The AXIS OF ROTATION of the MOMENT is NORMAL to the PLANE containing the FORCE VECTOR and the POSITION VECTOR.
THE RIGHT HAND RULE: The SENSE of a MOMENT can quickly be determined using the RIGHT HAND RULE.
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The RIGHT HAND RULE only works with the right hand! REMEMBER that come exam day and under timed conditions, if you use your left hand by mistake, the direction of your Z-AXIS will be flipped backwards. Using the RIGHT HAND RILE, we will: β’ ALIGN our fingers with the CARTESIAN AXIS for which our FORCE and POSITION VECTOR reside. β’ DEFINE the AXIS OF ROTATION by looking at the DIRECTION that your THUMB is pointing. β’ DEFINE the SENSE of the MOMENT by looking at the DIRECTION that your THUMB rotates when you curl your fingers in to your body. In THREE DIMENSIONS, a MOMENT is POSITIVE about the X-AXIS if itβs acting COUNTERCLOCKWISE when youβre looking at the ROTATION of the tip of your thumb. The same applies to the Y-AXIS and your forefinger and to the Z-AXIS and your middle finger.
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Illustrating the AXIS OF ROTATION versus the standard of POSITIVE MOMENTS, we have:
To solidify your understanding and ensure that you are comfortable using the RIGHT HAND RULE, grab a pencil and pen and practice using the following procedure. β’ Designate the left end of the pencil as the pivot, and hold it rigidly with your finger. β’ The pen will be the force. o If the pen represents an upward force, you will be applying the force below the pencil, but directed upwards. o If the pen represents a downward force, you will be applying the force above the pencil, but directed downwards.
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Run through the following scenarios, noticing which DIRECTION the pencil is ROTATING, keeping the following in mind:
Note on a piece of PAPER whether you are seeing CLOCKWISE, COUNTERCLOCKWIS, or if it is NOT ROTATING at all: 1. Apply an upward force through the center of the pencil 2. Apply an upward force through the pivot 3. Apply an upward force through the edge farthest from the pivot 4. Apply a downward force through the center of the pencil 5. Apply a downward force through the pivot 6. Apply a downward force through the edge farthest from the pivot Here is what you should have observed: 1. COUNTERCLOCKWISE 2. NOT ROTATING 3. COUNTERCLOCKWISE
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4. CLOCKWISE 5. NOT ROTATING 6. COUNTERCLOCKWISE
COUNTERACTING A MOMENT (OR COUPLE): Consider the beam as illustrated:
This beam is ROTATING at some PIVOT POINT due to a FORCE being applied at some unspecified distance r away. We donβt know the self weight of this BEAM, nor do we know the location of the PIVOT point, however, we do know that the FORCE applied to this beam will cause it to ROTATE about that PIVOT POINT. This can be confirmed and quantified had we had more definition of the FORCE, F, and that DISTANCE, r, illustrated for us.
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However, based on the limited information we do have, we can conclude that with the FORCE acting to the LEFT of the PIVOT POINT, and with a DOWNWARD direction, that the ROTATION experience at the PIVOT POINT will be COUNTERCLOCKWISE. In ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want these ROTATIONS to occur, we need to counteract them in some form or fashion. This can be done in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. The LOCATION of the FORCE can be either on the LEFT or RIGHT of the PIVOT POINT as long as it is creating that MOMENT we need to establish EQUILIBRIUM. The GEOMETRY of the BEAM along with the LOCATION of the PIVOT POINT will also drive the location of this additional FORCE, as a POSITION VECTOR, along with the FORCE, are the TWO COMPONENTS that define a MOMENT. The following scenarios illustrate the GEOMETRICAL dependency of equalizing FORCES and their subsequent MOMENTS.
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PIVOT LOCATED AT CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at the very CENTER of itβs GEOMETRY, so that the DISTANCE from the CENTER POINT to the LEFT END and the DISTANCE from the CENTER POINT to the RIGHT END are EQUAL, or r:
Adding a FORCE, F, at an equal distance, r, to the RIGHT of the PIVOT POINT and acting in the SAME DIRECTION, DOWNWARD, would create a MOMENT equal in MAGNITUDE and an OPPOSITE SENSE, CLOCKWISE.
PIVOT LOCATED OFFSET AND TO THE RIGHT OF THE CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at a DISTANCE that is to the RIGHT of the CENTER of its GEOMETRY.
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The DISTANCE from the CENTER POINT and to the RIGHT END, as illustrated, is half that of the DISTANCE from the CENTER POINT and to the LEFT END. The MOMENT in this case would be the PRODUCT of the FORCE TIMES the DISTANCE of the FORCE from the PIVOT POINT, or: π = ππΉ To create EQUILIBRIUM, we need to NEUTRALIZE the effects of this MOMENT with a FORCE placed elsewhere on the BEAM, such that: π34 = π35 Or: π34 β π35 = 0
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We can do one of two things here: β’ Add a FORCE, F, that is EQUAL in MAGNITUDE but OPPOSITE in DIRECTION at the SAME LOCATION of the ORIGINAL FORCE. β’ Add a FORCE, F, that is DOUBLE the MAGNITUDE of the FORCE on the LEFT and ACTING in the SAME DIRECTION, DOWNWARD. Because the DISTANCE to the RIGHT END is half that of the DISTANCE to the LEFT END, this has to be accounted for by DOUBLING the counteracting FORCE. Laying this out in FORMULAIC TERMS, we would have:
ππΉ4 =
π 2πΉ 2 4
PIVOT LOCATED OFFSET AND TO THE LEFT OF THE CENTER OF UNIFORM BEAM: In this scenario, we have a UNIFORM BEAM with a PIVOT POINT placed at a DISTANCE that is to the RIGHT of the CENTER of its GEOMETRY.
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The DISTANCE from the CENTER POINT and to the RIGHT END, as illustrated, is double that of the DISTANCE from the CENTER POINT and to the LEFT END. The MOMENT in this case would be the PRODUCT of the FORCE TIMES the DISTANCE of the FORCE from the PIVOT POINT, or: π = ππΉ To create EQUILIBRIUM, we need to NEUTRALIZE the effects of this MOMENT with a FORCE placed elsewhere on the BEAM, such that: π34 = π35 Or: π34 β π35 = 0 We can do one of two things here: β’ Add a FORCE, F, that is EQUAL in MAGNITUDE but OPPOSITE in DIRECTION at the SAME LOCATION of the ORIGINAL FORCE. β’ Add a FORCE, F, that is HALF the MAGNITUDE of the FORCE on the LEFT and ACTING in the SAME DIRECTION, DOWNWARD. Because the DISTANCE to the RIGHT END is double that of the DISTANCE to the LEFT END, this has to be accounted for by HALVING the counteracting FORCE.
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Laying this out in FORMULAIC TERMS, we would have:
ππΉ4 = 2π
πΉ4 2
UNITS: The TOPIC of UNITS of FORCE can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The most common UNITS we will encounter in our studies for a MOMENT, or TORQUE, will be the SI UNIT known as the NEWTON-METER, denoted as N-m. Though, we are probably more familiar with and comfortable using SI UNITS, we should be prepared to work problems come exam day using the USCS UNITS as well. The standard USCS UNIT for MOMENTS, or TORQUE, is the FOOT-POUND, denoted using the symbol ft-lb. The GENERAL FORMULA that will help is CONVERT the UNITS OF A MOMENT can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. To CONVERT from SI UNITS to USCS UNITS, we can use the following CONVERSION FACTORS: β’ 1 ππ = 0.738 ππ‘ ππ β’ 1 ππ‘ β ππ = 1.356 π β π
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To fully solidify the existence, and our understanding, of MOMENTS, or TORQUES, it is important that we visualize practical examples as they would occur in engineering systems: To further assist in developing your understanding, and where these MOMENTS and TOURQUES, can occur, here are a few examples: β’ MOMENTS can emerge as REACTION FORCES. One example of this is the RESISTANCE you encounter when TURNING THE STEERING WHEEL of a car. The FORCE you feel is created by MOMENTS acting on the wheels where they touch the ground. β’ MOMENTS can also appear as INTERNAL FORCES in structural members or components. One example of this can be seen when a BEAM BENDS because of an INTERNAL MOMENT with a DIRECTION TRANSVERSE to the direction of the beam. An internal moment causes points to rotate relative to each other just as an internal force causes points in a solid to move relative to each other.
MOMENT-COUPLE SYSTEMS: A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force.
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The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, as expressed: π4 = π5 Or: πΉ4 π4 = πΉ5 π5 A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements. One example of a COUPLE is the forces from your fingers on to a doorknob. As you turn the knob, one finger is pushing up on the side of the doorknob and the other is pushing down. The CUMULATIVE AFFECT of these TWO FORCES TOGETHER cause the knob to ROTATE.
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CONCENTRATED MOMENTS: You may also hear a COUPLE referred to as a PURE MOMENT or CONCENTRATED MOMENT. When developing a FREE BODY DIAGRAM, CONCENTRATED MOMENTS are easy to apply and account for. MOMENTS are vectors, and because of this, they too will display MAGNITUDE, SENSE, and a POINT OF APPLICATION. For both TWO and THREE DIMENSIONAL objects, we will need to IDENTIFY the MAGNITUDE of a CONCENTRATED MOMENT as a NUMERICAL VALUE, if known, or a LABEL it as a VECTOR if itβs not known.
THE TRANSMISSION OF MOMENTS: The TRANSMISSION of MOMENTS can occur through multiple objects. For example, this can be observed in the operation of the drive train in your car, or even when you are turning the steering wheel, as illustrated earlier. As far as the drive train example, the engine of your car causes the transmission to rotate, which in turn causes the axles to rotate, which in turn cause the wheels to rotate.
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MOMENTS (COUPLES) | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The magnitude of the moment about the base point, O, created by the 600 π force as applied, is best represented as:
A. 2,610 N β m (counterclockwise) B. 2,610 N β m (clockwise) C. 2,980 N β m (counterclockwise) D. 2,980 N β m (clockwise)
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SOLUTION: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics.
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A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE In this problem, we have the POINT OF APPLICATION, but need to define the MAGNITUDE as well as the SENSE acting at POINT O. The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. When we are dealing with a FORCE that is at an ANGLE relative to the COORDINATE SYSTEM in which it resides, we will need to use our knowledge of RESOLVING a FORCE in to COMPONENTS so that we can assess each one INDIVIDUALLY and how it affects the ROTATION about our POINT OF APPLICATION.
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In this problem, we are given the illustration:
The 600 π force generates a MOMENT about the base at point βπβ. At point βπβ, this MOMENT will generate rotation, a SENSE, in either the clockwise or counterclockwise direction. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your
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answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system. We will assume the COUNTERCLOCKWISE direction as POSITIVE. In this problem, we have all the GEOMETRY nicely illustrated for us, so this affords us an opportunity to attempt to do one of two things, we can: β’ Break the FORCE VECTOR in to itβs COMPONENTS and define a PAIR of MOMENTS to determine the CUMULATIVE AFFECT at POINT O. β’ We can define the LINE OF ACTION of the FORCE and an associated LEVER ARM that is PERPENDICULAR between the LINE OF ACTION and AXIS OF ROTATION. The FACT that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATION is IMPORTANTβ¦many fail right off the bat by creating a LEVER ARM from the POINT OF APPLICATION back to the AXIS OF ROTATION. This second approach, if possible, is a HACK of sorts, if you get it dialed, you are going to DOMINATE these types of problems, so letβs take this routeβ¦at least to see if we can make it happen.
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If not, we just fall back to our initial approach, which at the end of the day, unfolds as we progress through the attempted HACK. Illustrating the creation of a LEVER ARM from the AXIS OF ROTATION PERPENDICULAR back to the AXIS OF ROTATION for this problem, we have:
The next step is to define a COORDINATE SYSTEM relative to the AXIS OF ROTATION.
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This will allow us to use the GEOMETRY defined, if possible, to determine the PERPENDICULAR DISTANCE between the AXIS OF ROTATION and the LINE OF ACTION of the FORCE, doing so we have:
We know by the GENERAL definition of a MOMENT, that we need TWO COMPONENTS defined, the PERPENDICULAR DISTANCE and the MAGNITUDE of the FORCE.
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As we are given the MAGNITUDE of the FORCE, 600 N, we need to see if we can get the LENGTH of the LEVEL ARM as illustrated:
Adding in the DIMENSIONS that we are given, we have:
Unfortunately, there is no quick way to determine the LENGTH of our LEVER ARM from what we are given, which means we will need to fall back to calculating the HORIZONTAL and VERTICAL COMPONENTS based on the provided GEOMETRY.
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As you see, no time is lost up to this point, however, if we were able to define that LEVER ARM as illustrated, it would of cut down massively on the amount it takes to solve this problemβ¦but letβs move on. The illustration above provides us a clear representation of the HORIZONTAL and VERTICAL PERPENDICULAR DISTANCES between the AXIS OF ROTATION and the LINE OF ACTION for the FORCE, which affords us the ability to break the force into its COMPONENTS. The GENERAL FORMULAS revolving around the COMPONENTS of a VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For this reason, we must memorize this process and the various formulas, understanding fully its application independent of the NCEES Supplied Reference Handbook. Knowing where our ANGLE of 40Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉ+ = πΉ\ cos π β’ The VERTICAL COMPONENT of the force: πΉ0 = πΉ\ sin π Letβs start with calculating the HORIZONTAL COMPONENT.
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Pulling the data we have and plugging it in to the general formula established, we get: πΉ+]^^ = (600 π) cos(40Β°) Or: πΉ+]^^ = 460 π Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉ0]^^ = (600 π) sin(40Β°) Or: πΉ0]^^ = 386 π Revisiting and pulling back down the GENERAL FORMULA for a MOMENT in our TWO DIMENSIONAL space, we have: π = ππΉ Since we have broken our single FORCE in to TWO COMPONENT forces, we are going to have to create a LINE OF ACTION for each, define the associated DISTANCES and add the AFFECTS of each COMPONENT FORCE to determine the CUMULATIVE AFFECT of the SINGLE FORCE on the ROTATION about the AXIS OF ROTATION.
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In FORMULAIC TERMS, this will resolve to: π_ = πΉ+ π+ + πΉ0 π0 The GEOMETRY is already laid out perfectly for us, revisiting it, we have:
Remember that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATION, this is IMPORTANT! This allows us to define: π+ = 4 π π0 = 2 π And already having defined: πΉ+ = 460 π
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πΉ0 = 386 π We are now at a point where we can just plug and chug. Plugging in the data that we have defined up to this point, we can calculate the CUMULATIVE MOMENT generated about point βπβ as: π_ = 460 π 4.00 π + (386 π) 2.00 π Or: π_ = 2,610 π β π The last thing we need to do is determine the SENSE of the MOMENT about POINT O. We can do this using the RIGHT HAND RULE or visually looking at the FREE BODY DIAGRAM along with the MAGNITUDE of each individual COMPONENT FORCE and the MOMENT it creates about POINT O. Taking this route, we can conclude that: β’ πΉ+ will create a MOMENT that is CLOCKWISE about POINT O β’ πΉ0 will create a MOMENT that is CLOCKWISE about POINT O Therefore, because both MOMENTS create a CLOCKWISE MOMENT about POINT O, the CUMULATIVE MOMENT will be CLOCKWISE as well.
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However, had each COMPONENT FORCE created a MOMENT that is OPPOSITE from the other, then we would have to ASSESS the MAGNITUDE of the MOMENT created by each COMPONENT. The correct answer choice is B. π, πππ π β π¦ (πππππππππ)
HOW IT ALL PLAYS OUT: The FOLLOWING STEPS are paramount when starting any problem that involves finding REACTIONS or MOMENTS. 1. ESTABLISH a COORDINATE AXIS SYSTEM. Drawing a small coordinate system in the corner of the problem will help! 2. ESTABLISH whether CLOCKWISE, or COUNTER-CLOCKWISE MOMENT direction will be POSITIVE. It doesnβt matter which route you take, stick to your choice, CONSISTENCY is key! 3. USE the EQUATIONS OF EQUILBRIUM to determine the REACTIONS or MOMENTS of any given problem. If the structure, or beam, is said to be in EQUILIBRIUM, then all of these MUST be trueβ¦everything must balance. β’ Sum of the forces in the βπ₯β direction:
πΉ+ = 0
β’ Sum of the forces in the βπ¦β direction:
πΉ0 = 0
β’ Sum of the moment about a chosen point:
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