06 Resultant of Force Problem Set
RESULTANT OF A FORCE | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: A force system is shown below. What is the resultant magnitude and resultant direction of the three forces?
A. 19.83 đđ; 14.09° B. 19.83 đđ; 24.09° C. 29.83 đđ; 14.09° D. 29.83 đđ; 24.09°
Made with
by Prepineer | Prepineer.com
SOLUTION 1: 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
Made with
by Prepineer | Prepineer.com
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 đ force highlighted in red.
The line of action of the force is provided, as well as the direction of the angle relative to the origin, given as 30°. We can then use this angle to determine the components.
Made with
by Prepineer | Prepineer.com
We will use trigonometric identities to solve for each of the components. We can identify the đĨ â đđĨđđ as our adjacent side, đĻ â đđĨđđ as our opposite side, and the force vector as our hypotenuse.
Calculating the horizontal component, we use the đĨ â đđĨđđ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: đš6788 = (100 đđ)(cos 30°) = 86.6 đđ
Made with
by Prepineer | Prepineer.com
Moving on to the vertical component, we use the đĻ â đđĨđđ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: đš@788 = (100 đđ)(sin 30°) = 50 đđ We now have both the horizontal and vertical components for the 100 â đ force. We will now do the same process for the other two forces acting on the system.
Made with
by Prepineer | Prepineer.com
The next force we will break up into components is the 50 đđ force. The angle at which the force is acting is not given, but we are given the side of a triangles. We can use trigonometric ratios to calculate each of the components.
Solving for the horizontal component of the force, we find:
đš6D8 = đš cos đ = (â50 đđ)
4 = â40 đđ 5
Solving for the vertical component of the force, we find:
đš@D8 = đš sin đ = (50 đđ)
3 = 30 đđ 5
Made with
by Prepineer | Prepineer.com
Next, the 80 đđ force acts with an opposite sense but similar angle relative to the origin, given as đ = 20°. We can then use this angle to determine the components.
Solving for the horizontal component of the force, we find: đš6F8 = đš sin đ = (â80 đđ)(sin 20°) = â27.36 đđ Solving for the vertical component of the force, we find: đš@F8 = đš cos đ = (â80 đđ)(cos 20°) = â75.17 đđ
Made with
by Prepineer | Prepineer.com
As we know have the horizontal and vertical components of all the forces, we will sum the components 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: đš6 = đš6788 + đš6D8 + đš6F8 = 86.6 â 40 â 27.36 = 19.24 đđ Summing the vertical component of each force, we find the net resultant force the vertical component is:
đš@ = đš@788 + đš@D8 + đš@F8 = 50 + 30 â 75.17 = 4.83 đđ
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 (đš) of âđâ forces with components đš6 ,K đđđ đš@ has the magnitude of: ,K
P
M
đš=
đš6 ,K KNO
P
M
+
đš@ KNO
Made with
,K
P
M
=
đš6 ,K KNO
P
M
+
đš@ KNO
O P
,K
by Prepineer | Prepineer.com
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:
đš=
19.24
P
+ 4.83
P
= 19.83 đđ
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 đĨ â đđĨđđ . The RESULTANT DIRECTION with respect to the đĨ â đđĨđđ is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
đ = arctan
M KNO đš@ ,K M KNO đš6 ,K
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:
đ = arctan
4.83 = 14.09° 19.24
Therefore, the correct answer choice is A. đđ. đđ đđ; đđ. đđ°
Made with
by Prepineer | Prepineer.com
PROBLEM 2: A force system is shown below, what is the resultant magnitude and resultant direction of the three forces?
A. 45.29 đđ; 22.22° B. 45.29 đđ; 67.76° C. 53.10 đđ; 33.09° D. 53.10 đđ; 43.09°
Made with
by Prepineer | Prepineer.com
SOLUTION 2: 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 into 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
Made with
by Prepineer | Prepineer.com
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 120 đđ force highlighted in red.
Made with
by Prepineer | Prepineer.com
The line of action of the force is provided, as well as the sense of the angle relative to the origin, and is given as 40°. We can then use this angle to determine the components.
We will use trigonometric identities to solve for each of the components. We can identify the đĨ â đđĨđđ as our adjacent side, đĻ â đđĨđđ as our opposite side, and the force vector as our hypotenuse.
Made with
by Prepineer | Prepineer.com
Calculating the horizontal component, we use the đĨ â đđĨđđ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: đš67\8 = (120 đđ)(cos 40°) = 91.93 đđ Moving on to the vertical component, we use the đĻ â đđĨđđ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: đš@7\8 = (120 đđ)(sin 40°) = 77.13 đđ We now have both the horizontal and vertical components for the 120 đđ force. We will now do the same process for the other two forces acting in the system.
Made with
by Prepineer | Prepineer.com
The next force we will break up into components is the 60 đđ force. The angle at which the force is acting, is not given explicitly, but since the force is acting in only one direction we can determine its components easily. As the force acts at an angle on the đĻ â đđĨđđ , the force has a horizontal component of 0, as the force has only a vertical component.
Solving for the horizontal component of the force, we find: đš6]8 = 0 đđ Solving for the vertical component of the force, we find: đš@]8 = 60 đđ
Made with
by Prepineer | Prepineer.com
The last force we will break up into components is the 130 đđ force. The angle at which the force is acting is not given, but we are given the side of a triangles. We can use trigonometric ratios to calculate each of the components.
Solving for the horizontal component of the force, we find:
đš67^8 = đš cos đ = (â130 đđ)
5 = â50 đđ 13
Solving for the vertical component of the force, we find:
đš@7^8 = đš sin đ = (â130 đđ)
Made with
12 = â120 đđ 13
by Prepineer | Prepineer.com
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:
đš6 = đš67\8 + đš6]8 + đš67^8 = 91.93 + 0 â 50 = 41.93 đđ
Summing the vertical component of each force, we find the net resultant force the vertical component is:
đš@ = đš@7\8 + đš@]8 + đš@7^8 = 77.13 + 60 â 120 = 17.13 đđ 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, đš, of đ forces with components đš6 ,K đđđ đš@ has the magnitude of:
P
M
đš=
đš6 ,K KNO
P
M
+
đš@ KNO
Made with
,K
P
M
=
đš6 ,K KNO
P
M
+
đš@ KNO
O P
,K
by Prepineer | Prepineer.com
,K
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:
đš=
41.93
P
+ 17.13
P
= 45.29 đđ
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 đĨ â đđĨđđ . The RESULTANT DIRECTION with respect to the đĨ â đđĨđđ is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
đ = arctan
M KNO đš@ ,K M KNO đš6 ,K
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:
đ = arctan
17.13 = 22.22° 41.93
Therefore, the correct answer choice is A. đđ. đđ đđ; đđ. đđ°
Made with
by Prepineer | Prepineer.com
PROBLEM 3: Using the free-body diagram of point C provided below, the sum of the forces along the đĨ â đđĨđđ is represented by which of the following expressions?
A. đšP sin 50° â 20 = 0 B. đšP cos 50° â 20 = 0 C. đšP sin 50° â đšO = 0 D. đšP cos 50° + 20 = 0
Made with
by Prepineer | Prepineer.com
SOLUTION 3: 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. 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
As we are simply looking for an expression to represent the sum of the forces about the x-axis, we can sum the force to write the expression: âđš6 = 0: đšP cos 50° â 20 = 0
Therefore, the correct answer choice is B. đđ đđđ đđ° â đđ = đ
Made with
by Prepineer | Prepineer.com
PROBLEM 4: Using the free-body diagram of point C provided below, the sum of the forces along the đĨ â đđĨđđ is represented by which of the following expressions?
A. đšP sin 50° â 20 = 0 B. đšP cos 50° â 20 = 0 C. đšP sin 50° â đšO = 0 D. đšP cos 50° + 20 = 0
Made with
by Prepineer | Prepineer.com
SOLUTION 4: 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. 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
As we are simply looking for an expression to represent the sum of the forces about the y-axis, we can sum the force to write the expression: âđš@ = 0: đšP sin 50° â đšO = 0
Therefore, the correct answer choice is C. đđ đđđ đđ° â đđ = đ
Made with
by Prepineer | Prepineer.com
PROBLEM 1: A force system is shown below. What is the resultant magnitude and resultant direction of the three forces?
A. 19.83 đđ; 14.09° B. 19.83 đđ; 24.09° C. 29.83 đđ; 14.09° D. 29.83 đđ; 24.09°
Made with
by Prepineer | Prepineer.com
SOLUTION 1: 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
Made with
by Prepineer | Prepineer.com
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 đ force highlighted in red.
The line of action of the force is provided, as well as the direction of the angle relative to the origin, given as 30°. We can then use this angle to determine the components.
Made with
by Prepineer | Prepineer.com
We will use trigonometric identities to solve for each of the components. We can identify the đĨ â đđĨđđ as our adjacent side, đĻ â đđĨđđ as our opposite side, and the force vector as our hypotenuse.
Calculating the horizontal component, we use the đĨ â đđĨđđ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: đš6788 = (100 đđ)(cos 30°) = 86.6 đđ
Made with
by Prepineer | Prepineer.com
Moving on to the vertical component, we use the đĻ â đđĨđđ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: đš@788 = (100 đđ)(sin 30°) = 50 đđ We now have both the horizontal and vertical components for the 100 â đ force. We will now do the same process for the other two forces acting on the system.
Made with
by Prepineer | Prepineer.com
The next force we will break up into components is the 50 đđ force. The angle at which the force is acting is not given, but we are given the side of a triangles. We can use trigonometric ratios to calculate each of the components.
Solving for the horizontal component of the force, we find:
đš6D8 = đš cos đ = (â50 đđ)
4 = â40 đđ 5
Solving for the vertical component of the force, we find:
đš@D8 = đš sin đ = (50 đđ)
3 = 30 đđ 5
Made with
by Prepineer | Prepineer.com
Next, the 80 đđ force acts with an opposite sense but similar angle relative to the origin, given as đ = 20°. We can then use this angle to determine the components.
Solving for the horizontal component of the force, we find: đš6F8 = đš sin đ = (â80 đđ)(sin 20°) = â27.36 đđ Solving for the vertical component of the force, we find: đš@F8 = đš cos đ = (â80 đđ)(cos 20°) = â75.17 đđ
Made with
by Prepineer | Prepineer.com
As we know have the horizontal and vertical components of all the forces, we will sum the components 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: đš6 = đš6788 + đš6D8 + đš6F8 = 86.6 â 40 â 27.36 = 19.24 đđ Summing the vertical component of each force, we find the net resultant force the vertical component is:
đš@ = đš@788 + đš@D8 + đš@F8 = 50 + 30 â 75.17 = 4.83 đđ
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 (đš) of âđâ forces with components đš6 ,K đđđ đš@ has the magnitude of: ,K
P
M
đš=
đš6 ,K KNO
P
M
+
đš@ KNO
Made with
,K
P
M
=
đš6 ,K KNO
P
M
+
đš@ KNO
O P
,K
by Prepineer | Prepineer.com
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:
đš=
19.24
P
+ 4.83
P
= 19.83 đđ
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 đĨ â đđĨđđ . The RESULTANT DIRECTION with respect to the đĨ â đđĨđđ is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
đ = arctan
M KNO đš@ ,K M KNO đš6 ,K
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:
đ = arctan
4.83 = 14.09° 19.24
Therefore, the correct answer choice is A. đđ. đđ đđ; đđ. đđ°
Made with
by Prepineer | Prepineer.com
PROBLEM 2: A force system is shown below, what is the resultant magnitude and resultant direction of the three forces?
A. 45.29 đđ; 22.22° B. 45.29 đđ; 67.76° C. 53.10 đđ; 33.09° D. 53.10 đđ; 43.09°
Made with
by Prepineer | Prepineer.com
SOLUTION 2: 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 into 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
Made with
by Prepineer | Prepineer.com
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 120 đđ force highlighted in red.
Made with
by Prepineer | Prepineer.com
The line of action of the force is provided, as well as the sense of the angle relative to the origin, and is given as 40°. We can then use this angle to determine the components.
We will use trigonometric identities to solve for each of the components. We can identify the đĨ â đđĨđđ as our adjacent side, đĻ â đđĨđđ as our opposite side, and the force vector as our hypotenuse.
Made with
by Prepineer | Prepineer.com
Calculating the horizontal component, we use the đĨ â đđĨđđ to represent the horizontal component of the force. Solving for the horizontal component of the force, we find: đš67\8 = (120 đđ)(cos 40°) = 91.93 đđ Moving on to the vertical component, we use the đĻ â đđĨđđ to represent the vertical component of the force.
Solving for the vertical component of the force, we find: đš@7\8 = (120 đđ)(sin 40°) = 77.13 đđ We now have both the horizontal and vertical components for the 120 đđ force. We will now do the same process for the other two forces acting in the system.
Made with
by Prepineer | Prepineer.com
The next force we will break up into components is the 60 đđ force. The angle at which the force is acting, is not given explicitly, but since the force is acting in only one direction we can determine its components easily. As the force acts at an angle on the đĻ â đđĨđđ , the force has a horizontal component of 0, as the force has only a vertical component.
Solving for the horizontal component of the force, we find: đš6]8 = 0 đđ Solving for the vertical component of the force, we find: đš@]8 = 60 đđ
Made with
by Prepineer | Prepineer.com
The last force we will break up into components is the 130 đđ force. The angle at which the force is acting is not given, but we are given the side of a triangles. We can use trigonometric ratios to calculate each of the components.
Solving for the horizontal component of the force, we find:
đš67^8 = đš cos đ = (â130 đđ)
5 = â50 đđ 13
Solving for the vertical component of the force, we find:
đš@7^8 = đš sin đ = (â130 đđ)
Made with
12 = â120 đđ 13
by Prepineer | Prepineer.com
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:
đš6 = đš67\8 + đš6]8 + đš67^8 = 91.93 + 0 â 50 = 41.93 đđ
Summing the vertical component of each force, we find the net resultant force the vertical component is:
đš@ = đš@7\8 + đš@]8 + đš@7^8 = 77.13 + 60 â 120 = 17.13 đđ 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, đš, of đ forces with components đš6 ,K đđđ đš@ has the magnitude of:
P
M
đš=
đš6 ,K KNO
P
M
+
đš@ KNO
Made with
,K
P
M
=
đš6 ,K KNO
P
M
+
đš@ KNO
O P
,K
by Prepineer | Prepineer.com
,K
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:
đš=
41.93
P
+ 17.13
P
= 45.29 đđ
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 đĨ â đđĨđđ . The RESULTANT DIRECTION with respect to the đĨ â đđĨđđ is calculated as the arctangent of the ratio of the vertical component to the horizontal component:
đ = arctan
M KNO đš@ ,K M KNO đš6 ,K
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:
đ = arctan
17.13 = 22.22° 41.93
Therefore, the correct answer choice is A. đđ. đđ đđ; đđ. đđ°
Made with
by Prepineer | Prepineer.com
PROBLEM 3: Using the free-body diagram of point C provided below, the sum of the forces along the đĨ â đđĨđđ is represented by which of the following expressions?
A. đšP sin 50° â 20 = 0 B. đšP cos 50° â 20 = 0 C. đšP sin 50° â đšO = 0 D. đšP cos 50° + 20 = 0
Made with
by Prepineer | Prepineer.com
SOLUTION 3: 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. 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
As we are simply looking for an expression to represent the sum of the forces about the x-axis, we can sum the force to write the expression: âđš6 = 0: đšP cos 50° â 20 = 0
Therefore, the correct answer choice is B. đđ đđđ đđ° â đđ = đ
Made with
by Prepineer | Prepineer.com
PROBLEM 4: Using the free-body diagram of point C provided below, the sum of the forces along the đĨ â đđĨđđ is represented by which of the following expressions?
A. đšP sin 50° â 20 = 0 B. đšP cos 50° â 20 = 0 C. đšP sin 50° â đšO = 0 D. đšP cos 50° + 20 = 0
Made with
by Prepineer | Prepineer.com
SOLUTION 4: 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. 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 đĨ â đđĨđđ represents the horizontal component, and the đĻ â đđĨđđ represents the vertical component.
As we are simply looking for an expression to represent the sum of the forces about the y-axis, we can sum the force to write the expression: âđš@ = 0: đšP sin 50° â đšO = 0
Therefore, the correct answer choice is C. đđ đđđ đđ° â đđ = đ
Made with
by Prepineer | Prepineer.com
