03.1 Forces Concept Overview
FORCES | CONCEPT OVERVIEW The TOPIC of FORCES can be referenced on page 67 of the NCEES Supplied Reference Handbook, 9.4 version for Computer Based Testing.
CONCEPT INTRO: A FORCE represents the ACTION of one body exerted on to another. This FORCE can be applied through CONTACT, which is the most obvious means, but can also be applied from a DISTANCE, as in the case of gravitational and magnetic forces. The TOPIC of FORCES can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A FORCE, precisely in the same way as a VECTOR, can be generally illustrated as:
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All FORCES are VECTOR QUANTITIES uniquely defined by the following CHARACTERISTICS: • MAGNITUDE – The magnitude of a vector represents its LENGTH, specified by a single numerical value having appropriate units for the realm in which it acts. In the instance that we are dealing with pure numbers, no units are used. • LINE OF ACTION – The line of action of a vector is a hypothetical straight line that runs COLLINEAR with the vector. This exaggerated straight line demonstrates in greater context the direction at which the force is acting. • POINT OF ACTION – The point of action, or point of application, of a vector is the EXACT LOCATION, or POINT, which the vector force is acting from, or on, in a defined SYSTEM or BODY of analysis. • SENSE – The sense is the ORIENTATION about which the force acts on the body at the POINT OF ACTION, relative to a defined coordinate system and origin. • HEAD – The HEAD of a vector is specified by the order of two points on a line parallel to the vector. You can think of the HEAD of a VECTOR as the DIRECTION in which that vector is moving relative to a defined coordinate system. When working with MUTLIPLE FORCES, there will be instances where we will need to combine these FORCES to determine a single force, otherwise referred to as the RESULTANT FORCE. Carrying out this process of combination, the
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CHARACTERISTICS of each individual FORCE will make an impact individually in defining what the hybrid RESULTANT will be. For this reason, we can’t afford to mischaracterize any of the individual FORCES we are working with. Using the CARTESIAN COORDINATE SYSTEM, we can lay out the basic sign convention as such: • HORIZONTAL TRANSLATION to the RIGHT along the X-AXIS and VERTICAL TRANSLATION UPWARDS along the Y-AXIS is defined as POSITIVE. • HORIZONTAL TRANSLATION to the LEFT along the X-AXIS and VERTICAL TRANSLATION DOWNWARD along the Y-AXIS is defined as NEGATIVE. This CONVENTION can be generically illustrated as:
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UNITS OF FORCE: The TOPIC of UNITS of a 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 commonly used unit for FORCE is the NEWTON. The NEWTON is an SI unit that is denoted using the symbol “N”. This unit is derived from NEWTON’S SECOND LAW OF MOTION which states that a &
FORCE of 1 N causes a mass of 1 kg to accelerate 1 ( . '
With this general definition, a more conventional statement of a NEWTON can be defined as 1
)*∙& '(
. Remember this fact, for it can be a simple conceptual problem given
to you come exam day, and further, a simple point in your direction. Though we are probably more familiar and comfortable using SI units, the exam specifies that it is fair game to see either SI or USCS units. For this reason, we should be prepared and comfortable working problems in both worlds. The standard US unit for a FORCE is the POUND denoted with a symbol 𝑙𝑏& or 𝑙𝑏. . 25
Similar to the NEWTON, a FORCE of 1 lb2 causes a mass of 1 slug to accelerate at 1 ( . 6
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USCS units can be a confusing way of representing mass because universally the mass of an object is reported as a weight in 𝑙𝑏& or 𝑙𝑏. . However, the weight of an object in pounds is not a mass, but actually the force of gravity acting on the mass. Consequently, an extra step to determine an object’s mass in SLUGS, given it’s WEIGHT, must be completed using the following conversion:
𝑚'89*' =
𝑙𝑏. 𝑔
Where:
𝑔 = 32.17
.? '(
the acceleration due to gravity
NEWTON’S THREE LAWS OF MOTION: The topic of NEWTON’S THREE LAWS OF MOTION is not provided directly in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of utmost importance that we understand these LAWS front to back, therefore, we will present and study them so that we can fully understand their application independent of the NCEES Supplied Reference Handbook. NEWTON'S THREE LAWS OF MOTION are physical laws that form the basis for classical mechanics and can be summarized as such: 1. NEWTON’S FIRST LAW:
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a. NEWTON’S FIRST LAW OF MOTION, also commonly referred to as the LAW OF INERTIA, states that an object at rest tends to stay at rest until acted upon by an unbalanced force. b. If the RESULTANT FORCE ACTING on a PARTICLE IS ZERO, the particle WILL REMAIN AT REST (if originally at rest) or will move with CONSTANT SPEED in a straight line (if originally in motion). 2. NEWTON’S SECOND LAW: a. NEWTON’S SECOND LAW OF MOTION states that when a force acts upon an object that has a specified mass (𝑚), a corresponding acceleration is produced. b. If the RESULTANT FORCE ACTING on a PARTICLE IS NOT ZERO, the particle will have an ACCELERATION PROPORTIONAL to the MAGNITUDE of the resultant in the direction of this resultant force. c. A body of MASS (𝑚), subject to a NET FORCE (𝐹), undergoes an ACCELERATION (𝑎) that has the SAME DIRECTION as the force and a MAGNITUDE that is DIRECTLY PROPORTIONAL to the force and inversely proportional to the mass, or in more familiar formulaic terms: 𝐹 = 𝑚𝑎 Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.
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3. NEWTON’S THIRD LAW: a. NEWTON’S THIRD LAW OF MOTION states that for EVERY ACTION, there is an EQUAL and OPPOSITE REACTION. b. The FORCES of an ACTION and REACTION between bodies in contact have the same magnitude, same line of action, and opposite sense, or in other terms, they are equal, opposite, and collinear. c. This means that whenever a first body exerts a force (𝐹) on a second body, the second body exerts a force (−𝐹) on the first body. These two FORCES are EQUAL IN MAGNITUDE and OPPOSITE IN DIRECTION. d. This law is sometimes referred to as the action-reaction law, with “𝐹” called the "action" and (−𝐹) the "reaction". The ACTION and the REACTION are simultaneous.
TYPES OF FORCES: Forces can be classified in to SPECIFIC CATEGORIES based on how they are applied to another body. These categories that generally be stated as: • EXTERNAL FORCES: These FORCES are created by the INTERACTION between the SYSTEM OF INTEREST and its SURROUNDINGS. These forces include GRAVITATIONAL forces, DRAG or LIFT forces, and BUOYANCY forces, amongst others.
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• CONSTRAINT FORCES: Forces exerted by one part of a structure on to another part are called CONSTRAINT FORCES, or SUPPORT REACTIONS. We will see these forces occur in places such as JOINTS and CONNECTIONS between components, and will play a vital role in the analysis of statics problems. • INTERNAL FORCES: Forces that occur inside a given structure, or component, as a response to an applied EXTERNAL LOAD or FORCE, are referred to as INTERNAL FORCES. An example of an INTERNAL FORCE is that TENSION that is experiences within a rope stretched by an EXTERNAL FORCE. Most solid objects have a very complex distribution of INTERNAL FORCES. Due to the complex nature of INTERNAL FORCE DISTRIBUTIONS, it is IMPOSSIBLE to DESIGN a component that is GUARANTEED NOT TO FAIL OR DEFORM, and this is why we design using a FACTOR OF SAFETY defined based on a very specific set of loading circumstances.
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FORCES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. If a net force of 20 N causes a chair to accelerate at a rate of 4 m/s G , the mass of the chair (in lbs) is most close to: A. 0.44 B. 176 C. 5 D. 11
SOLUTION: As in most STATICS problems we can expect to see on the exam, we will need to use the EQUATIONS OF EQUILIBRIUM to solve for any UNKNOWNS that aren’t originally stated for us. These UNKNOWNS, although not directly defined in the problem statement, will be driven by data that is defined within the problem statement. Generically, the process will include identifying the loads and external forces applied to a structure and using the EQUATIONS OF EQUILIBRIUM to solve for the desired UNKNOWNS.
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The topic of NEWTON’S THREE LAWS OF MOTION is not provided directly in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of utmost importance that we understand these LAWS front to back, therefore, we will present and study them so that we can fully understand their application independent of the NCEES Supplied Reference Handbook. The goal of this problem is to determine the mass of a chair, in pounds, pushed with a net force of 20 N and accelerating at a rate of 4 m/s G . We will employ our knowledge of NEWTON’S SECOND LAW OF MOTION to derive our result. NEWTON’S SECOND LAW OF MOTION can generally be broken down like this: • NEWTON’S SECOND LAW OF MOTION states that when a force acts upon an object that has a specified mass (𝑚), a corresponding acceleration is produced. • If the RESULTANT FORCE ACTING on a PARTICLE IS NOT ZERO, the particle will have an ACCELERATION PROPORTIONAL to the MAGNITUDE of the resultant force and in the direction of this resultant force. • A body of MASS (𝑚), subject to a NET FORCE (𝐹), undergoes an ACCELERATION (𝑎) that has the SAME DIRECTION as the force and a MAGNITUDE that is DIRECTLY PROPORTIONAL to the force and inversely proportional to the mass, or in more familiar formulaic terms: 𝐹 = 𝑚𝑎
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Where: • F is The RESULTANT FORCE acting on the ELEMENT • m is the MASS of the ELEMENT • a is the ACCELERATION of the ELEMENT In this problem, we are given: 𝐹 = 20 𝑁 𝑎 = 4 𝑚 ∙ 𝑠 LG At this point, we have all the data we need to simply plug and chug. Recall that NEWTON’S SECOND LAW OF MOTION states that a FORCE of 1 N causes &
a mass of 1 kg to accelerate 1 ( . '
Therefore, we can equivalently write our FORCE as:
𝐹 = 20
𝑘𝑔 ∙ 𝑚 𝑠G
Restating the GENERAL FORMULA for NEWTON’S SECOND LAW OF MOTION, we have: 𝐹 = 𝑚𝑎
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We have the FORCE and the ACCELERATION and we are after the MASS…therefore, rearranging to isolate and solve for the MASS we get:
𝑚=
𝐹 𝑎
Plugging in our data, we have:
𝑚=
𝑘𝑔 ∙ 𝑚 𝑠G 𝑚 4 G 𝑠
20
Or: 𝑚 = 5 𝑘𝑔 Many will stop here, 5 is given as an answer option, and under timed conditions, we may be quick to select it and move on…this will be catastrophic as it will give us the wrong answer after getting so far. We need to take just one more minor step, that being, converting from KILOGRAMS to POUNDS which the problem requests. The most COMMON CONVERSION UNITS can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We are able to convert KILOGRAMS to POUNDS using the following relationship:
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• 1 kg = 2.2 lbs CONVERTING our value, we get:
𝑚 = 5 𝑘𝑔
2.2 𝑙𝑏𝑠 1 𝑘𝑔
Or: m = 11 lbs The correct answer choice is D. 𝟏𝟏, as the mass of the chair is 𝟏𝟏 𝒍𝒃𝒔.
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CONCEPT INTRO: A FORCE represents the ACTION of one body exerted on to another. This FORCE can be applied through CONTACT, which is the most obvious means, but can also be applied from a DISTANCE, as in the case of gravitational and magnetic forces. The TOPIC of FORCES can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A FORCE, precisely in the same way as a VECTOR, can be generally illustrated as:
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All FORCES are VECTOR QUANTITIES uniquely defined by the following CHARACTERISTICS: • MAGNITUDE – The magnitude of a vector represents its LENGTH, specified by a single numerical value having appropriate units for the realm in which it acts. In the instance that we are dealing with pure numbers, no units are used. • LINE OF ACTION – The line of action of a vector is a hypothetical straight line that runs COLLINEAR with the vector. This exaggerated straight line demonstrates in greater context the direction at which the force is acting. • POINT OF ACTION – The point of action, or point of application, of a vector is the EXACT LOCATION, or POINT, which the vector force is acting from, or on, in a defined SYSTEM or BODY of analysis. • SENSE – The sense is the ORIENTATION about which the force acts on the body at the POINT OF ACTION, relative to a defined coordinate system and origin. • HEAD – The HEAD of a vector is specified by the order of two points on a line parallel to the vector. You can think of the HEAD of a VECTOR as the DIRECTION in which that vector is moving relative to a defined coordinate system. When working with MUTLIPLE FORCES, there will be instances where we will need to combine these FORCES to determine a single force, otherwise referred to as the RESULTANT FORCE. Carrying out this process of combination, the
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CHARACTERISTICS of each individual FORCE will make an impact individually in defining what the hybrid RESULTANT will be. For this reason, we can’t afford to mischaracterize any of the individual FORCES we are working with. Using the CARTESIAN COORDINATE SYSTEM, we can lay out the basic sign convention as such: • HORIZONTAL TRANSLATION to the RIGHT along the X-AXIS and VERTICAL TRANSLATION UPWARDS along the Y-AXIS is defined as POSITIVE. • HORIZONTAL TRANSLATION to the LEFT along the X-AXIS and VERTICAL TRANSLATION DOWNWARD along the Y-AXIS is defined as NEGATIVE. This CONVENTION can be generically illustrated as:
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UNITS OF FORCE: The TOPIC of UNITS of a 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 commonly used unit for FORCE is the NEWTON. The NEWTON is an SI unit that is denoted using the symbol “N”. This unit is derived from NEWTON’S SECOND LAW OF MOTION which states that a &
FORCE of 1 N causes a mass of 1 kg to accelerate 1 ( . '
With this general definition, a more conventional statement of a NEWTON can be defined as 1
)*∙& '(
. Remember this fact, for it can be a simple conceptual problem given
to you come exam day, and further, a simple point in your direction. Though we are probably more familiar and comfortable using SI units, the exam specifies that it is fair game to see either SI or USCS units. For this reason, we should be prepared and comfortable working problems in both worlds. The standard US unit for a FORCE is the POUND denoted with a symbol 𝑙𝑏& or 𝑙𝑏. . 25
Similar to the NEWTON, a FORCE of 1 lb2 causes a mass of 1 slug to accelerate at 1 ( . 6
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USCS units can be a confusing way of representing mass because universally the mass of an object is reported as a weight in 𝑙𝑏& or 𝑙𝑏. . However, the weight of an object in pounds is not a mass, but actually the force of gravity acting on the mass. Consequently, an extra step to determine an object’s mass in SLUGS, given it’s WEIGHT, must be completed using the following conversion:
𝑚'89*' =
𝑙𝑏. 𝑔
Where:
𝑔 = 32.17
.? '(
the acceleration due to gravity
NEWTON’S THREE LAWS OF MOTION: The topic of NEWTON’S THREE LAWS OF MOTION is not provided directly in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of utmost importance that we understand these LAWS front to back, therefore, we will present and study them so that we can fully understand their application independent of the NCEES Supplied Reference Handbook. NEWTON'S THREE LAWS OF MOTION are physical laws that form the basis for classical mechanics and can be summarized as such: 1. NEWTON’S FIRST LAW:
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a. NEWTON’S FIRST LAW OF MOTION, also commonly referred to as the LAW OF INERTIA, states that an object at rest tends to stay at rest until acted upon by an unbalanced force. b. If the RESULTANT FORCE ACTING on a PARTICLE IS ZERO, the particle WILL REMAIN AT REST (if originally at rest) or will move with CONSTANT SPEED in a straight line (if originally in motion). 2. NEWTON’S SECOND LAW: a. NEWTON’S SECOND LAW OF MOTION states that when a force acts upon an object that has a specified mass (𝑚), a corresponding acceleration is produced. b. If the RESULTANT FORCE ACTING on a PARTICLE IS NOT ZERO, the particle will have an ACCELERATION PROPORTIONAL to the MAGNITUDE of the resultant in the direction of this resultant force. c. A body of MASS (𝑚), subject to a NET FORCE (𝐹), undergoes an ACCELERATION (𝑎) that has the SAME DIRECTION as the force and a MAGNITUDE that is DIRECTLY PROPORTIONAL to the force and inversely proportional to the mass, or in more familiar formulaic terms: 𝐹 = 𝑚𝑎 Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.
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3. NEWTON’S THIRD LAW: a. NEWTON’S THIRD LAW OF MOTION states that for EVERY ACTION, there is an EQUAL and OPPOSITE REACTION. b. The FORCES of an ACTION and REACTION between bodies in contact have the same magnitude, same line of action, and opposite sense, or in other terms, they are equal, opposite, and collinear. c. This means that whenever a first body exerts a force (𝐹) on a second body, the second body exerts a force (−𝐹) on the first body. These two FORCES are EQUAL IN MAGNITUDE and OPPOSITE IN DIRECTION. d. This law is sometimes referred to as the action-reaction law, with “𝐹” called the "action" and (−𝐹) the "reaction". The ACTION and the REACTION are simultaneous.
TYPES OF FORCES: Forces can be classified in to SPECIFIC CATEGORIES based on how they are applied to another body. These categories that generally be stated as: • EXTERNAL FORCES: These FORCES are created by the INTERACTION between the SYSTEM OF INTEREST and its SURROUNDINGS. These forces include GRAVITATIONAL forces, DRAG or LIFT forces, and BUOYANCY forces, amongst others.
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• CONSTRAINT FORCES: Forces exerted by one part of a structure on to another part are called CONSTRAINT FORCES, or SUPPORT REACTIONS. We will see these forces occur in places such as JOINTS and CONNECTIONS between components, and will play a vital role in the analysis of statics problems. • INTERNAL FORCES: Forces that occur inside a given structure, or component, as a response to an applied EXTERNAL LOAD or FORCE, are referred to as INTERNAL FORCES. An example of an INTERNAL FORCE is that TENSION that is experiences within a rope stretched by an EXTERNAL FORCE. Most solid objects have a very complex distribution of INTERNAL FORCES. Due to the complex nature of INTERNAL FORCE DISTRIBUTIONS, it is IMPOSSIBLE to DESIGN a component that is GUARANTEED NOT TO FAIL OR DEFORM, and this is why we design using a FACTOR OF SAFETY defined based on a very specific set of loading circumstances.
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FORCES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. If a net force of 20 N causes a chair to accelerate at a rate of 4 m/s G , the mass of the chair (in lbs) is most close to: A. 0.44 B. 176 C. 5 D. 11
SOLUTION: As in most STATICS problems we can expect to see on the exam, we will need to use the EQUATIONS OF EQUILIBRIUM to solve for any UNKNOWNS that aren’t originally stated for us. These UNKNOWNS, although not directly defined in the problem statement, will be driven by data that is defined within the problem statement. Generically, the process will include identifying the loads and external forces applied to a structure and using the EQUATIONS OF EQUILIBRIUM to solve for the desired UNKNOWNS.
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The topic of NEWTON’S THREE LAWS OF MOTION is not provided directly in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of utmost importance that we understand these LAWS front to back, therefore, we will present and study them so that we can fully understand their application independent of the NCEES Supplied Reference Handbook. The goal of this problem is to determine the mass of a chair, in pounds, pushed with a net force of 20 N and accelerating at a rate of 4 m/s G . We will employ our knowledge of NEWTON’S SECOND LAW OF MOTION to derive our result. NEWTON’S SECOND LAW OF MOTION can generally be broken down like this: • NEWTON’S SECOND LAW OF MOTION states that when a force acts upon an object that has a specified mass (𝑚), a corresponding acceleration is produced. • If the RESULTANT FORCE ACTING on a PARTICLE IS NOT ZERO, the particle will have an ACCELERATION PROPORTIONAL to the MAGNITUDE of the resultant force and in the direction of this resultant force. • A body of MASS (𝑚), subject to a NET FORCE (𝐹), undergoes an ACCELERATION (𝑎) that has the SAME DIRECTION as the force and a MAGNITUDE that is DIRECTLY PROPORTIONAL to the force and inversely proportional to the mass, or in more familiar formulaic terms: 𝐹 = 𝑚𝑎
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Where: • F is The RESULTANT FORCE acting on the ELEMENT • m is the MASS of the ELEMENT • a is the ACCELERATION of the ELEMENT In this problem, we are given: 𝐹 = 20 𝑁 𝑎 = 4 𝑚 ∙ 𝑠 LG At this point, we have all the data we need to simply plug and chug. Recall that NEWTON’S SECOND LAW OF MOTION states that a FORCE of 1 N causes &
a mass of 1 kg to accelerate 1 ( . '
Therefore, we can equivalently write our FORCE as:
𝐹 = 20
𝑘𝑔 ∙ 𝑚 𝑠G
Restating the GENERAL FORMULA for NEWTON’S SECOND LAW OF MOTION, we have: 𝐹 = 𝑚𝑎
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We have the FORCE and the ACCELERATION and we are after the MASS…therefore, rearranging to isolate and solve for the MASS we get:
𝑚=
𝐹 𝑎
Plugging in our data, we have:
𝑚=
𝑘𝑔 ∙ 𝑚 𝑠G 𝑚 4 G 𝑠
20
Or: 𝑚 = 5 𝑘𝑔 Many will stop here, 5 is given as an answer option, and under timed conditions, we may be quick to select it and move on…this will be catastrophic as it will give us the wrong answer after getting so far. We need to take just one more minor step, that being, converting from KILOGRAMS to POUNDS which the problem requests. The most COMMON CONVERSION UNITS can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We are able to convert KILOGRAMS to POUNDS using the following relationship:
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• 1 kg = 2.2 lbs CONVERTING our value, we get:
𝑚 = 5 𝑘𝑔
2.2 𝑙𝑏𝑠 1 𝑘𝑔
Or: m = 11 lbs The correct answer choice is D. 𝟏𝟏, as the mass of the chair is 𝟏𝟏 𝒍𝒃𝒔.
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