05 Resultant of Force Concept Overview
RESULTANT OF A FORCE | CONCEPT OVERVIEW The TOPICS OF RESULTANT OF A FORCE can be referenced on page 67 of the NCEES Supplied Reference Handbook, 9.4 version for Computer Based Testing.
CONCEPT INTRO: A force can be broken down into components that represent the force in a particular direction or along a particular axis. The vector form of a three-dimensional force is represented as: πΉ = πΉ# π + πΉ& π + πΉ( π Any vector can be broken down into two or more individual vectors. These are called the COMPONENT VECTORS of the original vector. When asked to resolve a vector into its components, this is the answer being asked for. The process of creating a vector component is known as the RESOLUTION OF A VECTOR. In working with vectors in statics, you always resolve two-dimensional actions such as forces and displacements into exactly two pieces for two-dimensional problems and into exactly three pieces for three-dimensional problems.
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You can also depict two-dimensional actions with three-dimensional components by making a third component that has zero magnitude. To accomplish this task, you need to decide which type of component is actually required. These types of components are often separated into two categories: Cartesian components and non-Cartesian components. As the name Cartesian components implies, all of the resolved vector components are aligned with the Cartesian π₯β, π¦β, πππ π§ β ππ₯ππ of your coordinate system. Sometimes you see them referred to as rectangular components. Rectangular components are probably the most common components you calculate, and are fortunately usually the easiest to compute. In three-dimensional problems, the rectangular components of a vector can be expressed in terms of π, π, πππ π unit vectors in the π₯, π¦, πππ π§ directions, respectively.
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Non-Cartesian components of a vector arenβt necessarily aligned with the Cartesianaxes. One or more components may be aligned with the Cartesian axes, but at least one is not. If all were aligned with the axes, theyβd be Cartesian components. The FORMULAS FOR THE RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Forces acting at some angle from the coordinates axes can be resolved into mutually perpendicular forces called COMPONENTS. The component of a force parallel to the π₯ β ππ₯ππ is called the X-COMPONENT, and the component of a force parallel to the π¦ β ππ₯ππ is the called the Y-COMPONENT. We typically resolve forces along the π₯ β πππ π¦ β ππ₯ππ , as the Cartesian Coordinate System is typically used when working with problems in two dimensions.
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We can use basic trigonometric functions to break up the force into components, by multiplying the magnitude or resultant of the force by the appropriate trigonometric function. You should remember that the sine and cosine of an angle in a right triangle are defined as the ratio of the opposite side and the adjacent side to the hypotenuse. In two dimensions, you need to represent two components for every resultant vector, and in three dimensions you actually have three components that you need to determine. However, for both two- and three-dimensional cases, these components must always result in the same original combined behavior when separated. A component of a resultant vector must also be the same type of vector as the resultant itself. If youβre finding the components of a force vector, the components are also force vectors. Components of a force are not necessary equal to the projections of the force unless the axes on which the forces are projected are orthogonal, meaning they are perpendicular to each other. The angle between the point of application of the force and the axes will determine the component of a force. The FORMULAS FOR THE COMPONENTS OF A VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The π₯ β ππ₯ππ typically represents the horizontal component of the force: π»ππππ§πππ‘ππ πΆππππππππ‘: πΉ# = πΉ cos π The π¦ β ππ₯ππ typically represents the vertical component of the force: ππππ‘ππππ πΆππππππππ‘: πΉ& = πΉ sin π It is common to use the sine function to represent the vertical component, and cosine function to represent the horizontal component. We can express the horizontal and vertical components using trigonometric functions.
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The formula for the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT or magnitude of the force, (πΉ), of βπβ forces with components πΉ# ,F πππ πΉ& has the magnitude of: ,F
J
G
πΉ=
πΉ# ,F
J
G
+
FHI
πΉ& FHI
=
,F
J
G
πΉ# ,F FHI
J
G
+
πΉ& FHI
I J
,F
The formula for the RESULTANT DIRECTION OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT DIRECTION with respect to the π₯ β ππ₯ππ is calculated as the ππππ‘ππππππ‘ of the ratio of the vertical component to the horizontal component:
π = arctan
G FHI πΉ& ,F G FHI πΉ# ,F
It is important to remember that the component reflects the relationship of a component of the force using the trigonometry of a provided angle. Depending on the geometry given in the problem, it is possible to use either the π πππ or πππ πππ function to reflect either the π πππ or πππ πππ function. Below are two examples when trigonometry was used to solve for the components a force given two different angles.
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In the figure below, we are given an angle that is a common scenario where the π πππ function is used to calculate the vertical component, and the πππ πππ function is used to calculate horizontal component.
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In the figure below, we are given a different angle where the πππ πππ function is used to calculate the vertical component, and the π πππ function is used to calculate the horizontal component. It is important to realize that you need to make sure the correct angle, with respect to your coordinate system, is being used when breaking a force into components with respect to a particular axis.
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FREE-BODY DIAGRAMS: A FREE BODY DIAGRAM represents the physical conditions of the rigid object you want to analyze, including dimension data, and the forces acting on the system. Freebody diagrams can be complex pictures of multiple objects and systems, or diagrams of a small subcomponent of a larger piece within a system. We use Free Body Diagrams (FBD) to assess the interaction of forces on bodies of mass. A FBD is essentially a sketch of a body separated from its surrounding and in complete equilibrium. All forces acting on a body at any specific moment in time can be represented on a FBD.
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CONCEPT EXAMPLE: The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. A hook has three cables attached to it, as illustrated with πΌ = 35Β°. What is the resultant magnitude and resultant direction of the three forces?
A. 137 π; 86.6Β° B. 217 π; 176.6Β° C. 309 π; 86.6Β° D. 217 π; 176.6Β°
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SOLUTION: The topic of the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. This problem requires us to break each individual force up into components, sum them together, and then plug those components in the formula to calculate the resultant. The first step in this problem is to define our Coordinate System. When working with two-dimensional problems, it is typical to use a Cartesian Coordinate System such that the π₯ β ππ₯ππ represents the horizontal component, and the π¦ β ππ₯ππ represents the vertical component.
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The next step is to cycle through each individual force, one at a time, and break the force into its horizontal and vertical components. We will start with the 100 π force highlighted in red.
The line of action of the force is provided, as well as the sense of the angle relative to the origin, given as πΌ = 35Β°. We can then use this angle to determine the components.
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We will use trigonometric identities to solve for each of the components. We can identify the π₯ β ππ₯ππ as our adjacent side, π¦ β ππ₯ππ as our opposite side, and the force vector as our hypotenuse.
Calculating the horizontal component, we use the π₯ β ππ₯ππ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: πΉ#]^^ = πΉ cos π = (+100 π)(cos 35Β°) = +81.92 π NOTE: Make sure that your calculator is in degree mode not in radian mode.
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Moving on to the vertical component, we use the π¦ β ππ₯ππ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: πΉ&]^^ = πΉ sin π = (β100)(sin 35Β°) = β57.36 π NOTE: Make sure that your calculator is in degree mode and not radian mode. We now have both the horizontal and vertical components for the 100 β π force. We will now do the same process for the other two forces acting on the hook. The 150 π force acts on the hook at an angle that is not given in the problem statement.
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We need to realize that we can add the given angles of 30Β° and πΌ = 30Β° to calculate the angle at which the 150 π forces acts relative to the π₯ β ππ₯ππ .
Solving for the horizontal component of the force, we find: πΉ#]a^ = πΉ cos π = (+150 π)(cos 65Β°) = +63.39 π Solving for the vertical component of the force, we find: πΉ&]a^ = πΉ sin π = (β150)(sin 65Β°) = β135.95 π NOTE: Make sure that your calculator is in degree mode and not radian mode.
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Next, the 200 β π force acts with an opposite sense, but similar angle relative to the origin, given as πΌ = 35Β°. We can then use this angle to determine the components.
Solving for the horizontal component of the force, we find: πΉ#b^^ = πΉ cos π = (β200 π)(cos 35Β°) = β163.83 π Solving for the vertical component of the force, we find: πΉ&b^^ = πΉ sin π = (β200)(sin 35Β°) = β114.72 π NOTE: Make sure that your calculator is in degree mode. You can also use 145Β° if you continue to use the same point of reference from the first round of force calculations.
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As we now have the horizontal and vertical components of all the forces, we will sum the forces to calculate the resultant of the forces in each component. Once we have the resultant of each component, we can use the formula to calculate the net resultant for the force for a 2-dimensional force system. Summing the horizontal component of each force, we find the net resultant force the horizontal component is:
πΉ# = πΉ#]^^ + πΉ#]a^ + πΉ#b^^ = 81.92 + 63.39 β 163.83 = β18.52 π
Summing the vertical component of each force, we find the net resultant force the vertical component is:
πΉ& = πΉ&]^^ + πΉ&]a^ + πΉ&b^^ = β57.36 β 135.95 β 114.72 = β308.03 π The formula for the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT or magnitude of the force, πΉ, of π forces with components πΉ# ,F πππ πΉ& has the magnitude of:
J
G
πΉ=
πΉ# ,F FHI
J
G
+
πΉ& FHI
Made with
,F
J
G
=
πΉ# ,F FHI
J
G
+
πΉ& FHI
I J
,F
by Prepineer | Prepineer.com
,F
Plugging in our calculated values for the resultant of the horizontal and vertical components, we find the net resultant of the forces acting on the hook to be:
πΉ=
β18.52
J
+ β308.03
J
= 309 π
The formula for the RESULTANT DIRECTION OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We then need to determine the angle that the force is acting at relative to the x-axis. The RESULTANT DIRECTION with respect to the x-axis is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
π = arctan
G FHI πΉ& ,F G FHI πΉ# ,F
Plugging in our calculated values for the resultant of the horizontal and vertical components, we find the resultant direction of the forces acting on the hook to be:
π = arctan β
308.03 = 86.6Β° β18.52
Therefore, the correct answer choice is C. πππ π΅; ππ. πΒ°
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CONCEPT INTRO: A force can be broken down into components that represent the force in a particular direction or along a particular axis. The vector form of a three-dimensional force is represented as: πΉ = πΉ# π + πΉ& π + πΉ( π Any vector can be broken down into two or more individual vectors. These are called the COMPONENT VECTORS of the original vector. When asked to resolve a vector into its components, this is the answer being asked for. The process of creating a vector component is known as the RESOLUTION OF A VECTOR. In working with vectors in statics, you always resolve two-dimensional actions such as forces and displacements into exactly two pieces for two-dimensional problems and into exactly three pieces for three-dimensional problems.
Made with
by Prepineer | Prepineer.com
You can also depict two-dimensional actions with three-dimensional components by making a third component that has zero magnitude. To accomplish this task, you need to decide which type of component is actually required. These types of components are often separated into two categories: Cartesian components and non-Cartesian components. As the name Cartesian components implies, all of the resolved vector components are aligned with the Cartesian π₯β, π¦β, πππ π§ β ππ₯ππ of your coordinate system. Sometimes you see them referred to as rectangular components. Rectangular components are probably the most common components you calculate, and are fortunately usually the easiest to compute. In three-dimensional problems, the rectangular components of a vector can be expressed in terms of π, π, πππ π unit vectors in the π₯, π¦, πππ π§ directions, respectively.
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Non-Cartesian components of a vector arenβt necessarily aligned with the Cartesianaxes. One or more components may be aligned with the Cartesian axes, but at least one is not. If all were aligned with the axes, theyβd be Cartesian components. The FORMULAS FOR THE RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Forces acting at some angle from the coordinates axes can be resolved into mutually perpendicular forces called COMPONENTS. The component of a force parallel to the π₯ β ππ₯ππ is called the X-COMPONENT, and the component of a force parallel to the π¦ β ππ₯ππ is the called the Y-COMPONENT. We typically resolve forces along the π₯ β πππ π¦ β ππ₯ππ , as the Cartesian Coordinate System is typically used when working with problems in two dimensions.
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by Prepineer | Prepineer.com
We can use basic trigonometric functions to break up the force into components, by multiplying the magnitude or resultant of the force by the appropriate trigonometric function. You should remember that the sine and cosine of an angle in a right triangle are defined as the ratio of the opposite side and the adjacent side to the hypotenuse. In two dimensions, you need to represent two components for every resultant vector, and in three dimensions you actually have three components that you need to determine. However, for both two- and three-dimensional cases, these components must always result in the same original combined behavior when separated. A component of a resultant vector must also be the same type of vector as the resultant itself. If youβre finding the components of a force vector, the components are also force vectors. Components of a force are not necessary equal to the projections of the force unless the axes on which the forces are projected are orthogonal, meaning they are perpendicular to each other. The angle between the point of application of the force and the axes will determine the component of a force. The FORMULAS FOR THE COMPONENTS OF A VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The π₯ β ππ₯ππ typically represents the horizontal component of the force: π»ππππ§πππ‘ππ πΆππππππππ‘: πΉ# = πΉ cos π The π¦ β ππ₯ππ typically represents the vertical component of the force: ππππ‘ππππ πΆππππππππ‘: πΉ& = πΉ sin π It is common to use the sine function to represent the vertical component, and cosine function to represent the horizontal component. We can express the horizontal and vertical components using trigonometric functions.
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by Prepineer | Prepineer.com
The formula for the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT or magnitude of the force, (πΉ), of βπβ forces with components πΉ# ,F πππ πΉ& has the magnitude of: ,F
J
G
πΉ=
πΉ# ,F
J
G
+
FHI
πΉ& FHI
=
,F
J
G
πΉ# ,F FHI
J
G
+
πΉ& FHI
I J
,F
The formula for the RESULTANT DIRECTION OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT DIRECTION with respect to the π₯ β ππ₯ππ is calculated as the ππππ‘ππππππ‘ of the ratio of the vertical component to the horizontal component:
π = arctan
G FHI πΉ& ,F G FHI πΉ# ,F
It is important to remember that the component reflects the relationship of a component of the force using the trigonometry of a provided angle. Depending on the geometry given in the problem, it is possible to use either the π πππ or πππ πππ function to reflect either the π πππ or πππ πππ function. Below are two examples when trigonometry was used to solve for the components a force given two different angles.
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In the figure below, we are given an angle that is a common scenario where the π πππ function is used to calculate the vertical component, and the πππ πππ function is used to calculate horizontal component.
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In the figure below, we are given a different angle where the πππ πππ function is used to calculate the vertical component, and the π πππ function is used to calculate the horizontal component. It is important to realize that you need to make sure the correct angle, with respect to your coordinate system, is being used when breaking a force into components with respect to a particular axis.
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by Prepineer | Prepineer.com
FREE-BODY DIAGRAMS: A FREE BODY DIAGRAM represents the physical conditions of the rigid object you want to analyze, including dimension data, and the forces acting on the system. Freebody diagrams can be complex pictures of multiple objects and systems, or diagrams of a small subcomponent of a larger piece within a system. We use Free Body Diagrams (FBD) to assess the interaction of forces on bodies of mass. A FBD is essentially a sketch of a body separated from its surrounding and in complete equilibrium. All forces acting on a body at any specific moment in time can be represented on a FBD.
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CONCEPT EXAMPLE: The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. A hook has three cables attached to it, as illustrated with πΌ = 35Β°. What is the resultant magnitude and resultant direction of the three forces?
A. 137 π; 86.6Β° B. 217 π; 176.6Β° C. 309 π; 86.6Β° D. 217 π; 176.6Β°
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SOLUTION: The topic of the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. This problem requires us to break each individual force up into components, sum them together, and then plug those components in the formula to calculate the resultant. The first step in this problem is to define our Coordinate System. When working with two-dimensional problems, it is typical to use a Cartesian Coordinate System such that the π₯ β ππ₯ππ represents the horizontal component, and the π¦ β ππ₯ππ represents the vertical component.
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The next step is to cycle through each individual force, one at a time, and break the force into its horizontal and vertical components. We will start with the 100 π force highlighted in red.
The line of action of the force is provided, as well as the sense of the angle relative to the origin, given as πΌ = 35Β°. We can then use this angle to determine the components.
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We will use trigonometric identities to solve for each of the components. We can identify the π₯ β ππ₯ππ as our adjacent side, π¦ β ππ₯ππ as our opposite side, and the force vector as our hypotenuse.
Calculating the horizontal component, we use the π₯ β ππ₯ππ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: πΉ#]^^ = πΉ cos π = (+100 π)(cos 35Β°) = +81.92 π NOTE: Make sure that your calculator is in degree mode not in radian mode.
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Moving on to the vertical component, we use the π¦ β ππ₯ππ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: πΉ&]^^ = πΉ sin π = (β100)(sin 35Β°) = β57.36 π NOTE: Make sure that your calculator is in degree mode and not radian mode. We now have both the horizontal and vertical components for the 100 β π force. We will now do the same process for the other two forces acting on the hook. The 150 π force acts on the hook at an angle that is not given in the problem statement.
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We need to realize that we can add the given angles of 30Β° and πΌ = 30Β° to calculate the angle at which the 150 π forces acts relative to the π₯ β ππ₯ππ .
Solving for the horizontal component of the force, we find: πΉ#]a^ = πΉ cos π = (+150 π)(cos 65Β°) = +63.39 π Solving for the vertical component of the force, we find: πΉ&]a^ = πΉ sin π = (β150)(sin 65Β°) = β135.95 π NOTE: Make sure that your calculator is in degree mode and not radian mode.
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Next, the 200 β π force acts with an opposite sense, but similar angle relative to the origin, given as πΌ = 35Β°. We can then use this angle to determine the components.
Solving for the horizontal component of the force, we find: πΉ#b^^ = πΉ cos π = (β200 π)(cos 35Β°) = β163.83 π Solving for the vertical component of the force, we find: πΉ&b^^ = πΉ sin π = (β200)(sin 35Β°) = β114.72 π NOTE: Make sure that your calculator is in degree mode. You can also use 145Β° if you continue to use the same point of reference from the first round of force calculations.
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As we now have the horizontal and vertical components of all the forces, we will sum the forces to calculate the resultant of the forces in each component. Once we have the resultant of each component, we can use the formula to calculate the net resultant for the force for a 2-dimensional force system. Summing the horizontal component of each force, we find the net resultant force the horizontal component is:
πΉ# = πΉ#]^^ + πΉ#]a^ + πΉ#b^^ = 81.92 + 63.39 β 163.83 = β18.52 π
Summing the vertical component of each force, we find the net resultant force the vertical component is:
πΉ& = πΉ&]^^ + πΉ&]a^ + πΉ&b^^ = β57.36 β 135.95 β 114.72 = β308.03 π The formula for the RESULTANT OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT or magnitude of the force, πΉ, of π forces with components πΉ# ,F πππ πΉ& has the magnitude of:
J
G
πΉ=
πΉ# ,F FHI
J
G
+
πΉ& FHI
Made with
,F
J
G
=
πΉ# ,F FHI
J
G
+
πΉ& FHI
I J
,F
by Prepineer | Prepineer.com
,F
Plugging in our calculated values for the resultant of the horizontal and vertical components, we find the net resultant of the forces acting on the hook to be:
πΉ=
β18.52
J
+ β308.03
J
= 309 π
The formula for the RESULTANT DIRECTION OF A FORCE can be referenced under the topic of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We then need to determine the angle that the force is acting at relative to the x-axis. The RESULTANT DIRECTION with respect to the x-axis is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
π = arctan
G FHI πΉ& ,F G FHI πΉ# ,F
Plugging in our calculated values for the resultant of the horizontal and vertical components, we find the resultant direction of the forces acting on the hook to be:
π = arctan β
308.03 = 86.6Β° β18.52
Therefore, the correct answer choice is C. πππ π΅; ππ. πΒ°
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