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Determining Position Vectors and
Normal Vectors for a Mechanical
System
Solutions Manual for Engineering Mechanics:
Statics (2nd Edition)
Overview
The provided text discusses the Solutions Manual for the 2nd edition of the textbook "Engineering Mechanics: Statics" by Michael E. Plesha, Gary L. Gray, and Francesco Costanzo. The manual was created with the assistance of several individuals, including Chris Punshon, Andrew J. Miller, Justin High, Chris O'Brien, Chandan Kumar, Joseph Wyne, and Jonathan Fleischmann.
Content and Structure
The Solutions Manual provides detailed solutions and explanations for the problems and exercises presented in the 2nd edition of the "Engineering Mechanics: Statics" textbook. It is designed to aid students in understanding the concepts and applying the principles covered in the course.
The manual includes solutions for a wide range of topics, such as: - Vector mechanics - Equilibrium of particles and rigid bodies - Structural analysis - Distributed forces - Internal forces and moments - Friction - Virtual work and potential energy
Each solution is presented in a clear and comprehensive manner, with step- by-step explanations and relevant diagrams or illustrations to help students follow the problem-solving process.
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Additional Information
The text also mentions that the Solutions Manual is for the 2nd edition of the "Engineering Mechanics: Statics" textbook, which is used in a course at the Korea Advanced Institute of Science and Technology (KAIST), as indicated by the Korean title " ( )".
Solutions Manual for Engineering Mechanics
Statics and Dynamics 2nd Edition by Plesha
Important Information about this Solutions Manual
We encourage you to occasionally visit http://www.mhhe.com/pgc2e to obtain the most up-to-date version of this solutions manual. If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the authors and editors via email at: plesha@engr.wisc.edu, and stat_solns@email.esm.psu.edu.
Accuracy of Numbers in Calculations
Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 or 4 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.
Problem 2.
For each vector, write two expressions using polar vector representations, one using a positive value of θ and the other a negative value, where θ is measured counterclockwise from the right-hand horizontal direction.
Part (a): E r = 12 in. @ 90° or E r = 12 in. @ -270°
Part (b): E F = 23 N @ 135° or E F = 23 N @ -225°
Part (c): E v = 15 m/s @ 240° or E v = 15 m/s @ -120°
Problem 2.
Add the two vectors shown to form a resultant vector E R, and report your result using polar vector representation.
The law of cosines is used to determine R: R = √((101 mm)^2 + ( mm)^2 - 2(101 mm)(183 mm)cos 125°) = 254.7 mm
The law of sines is used to determine the angle β: R/sin α = 183 mm/sin β β = sin^-1((183 mm)/(254.7 mm)sin 125°) = 36.05°
Using these results, the vector E R is reported using polar vector representation as: E R = 255 mm @ 36.05°
Using polar vector representation, the resultant is: E R = 456.4 lb @ 159.7°
Problem 2.
Add the two vectors shown to form a resultant vector E R, and report your result using polar vector representation.
Part (a): R = √((139 lb)^2 + (200 lb)^2) = 243.6 lb α = tan^-1(200 lb/ lb) = 55.20°
Using polar vector representation, the resultant is: E R = 243.6 lb @ -34.80°
Part (b): The law of cosines is used to determine R: R = √((6 in.)^2 + ( in.)^2 - 2(6 in.)(8 in.)cos 80°) = 9.129 in.
The law of sines is used to determine the angle α: 8 in./sin α = R/sin 80° α = sin^-1((8 in.)/(9.129 in.)sin 80°) = 59.66°
Using polar vector representation, the resultant is: E R = 9.129 in. @ 100.3°
Problem 2.
Add the two vectors shown to form a resultant vector E R, and report your result using polar vector representation.
Part (a): R = √((35 kN)^2 + (18 kN)^2) = 39.36 kN β = tan^-1(35 kN/ kN) = 62.78°
The direction for E R measured from the right-hand horizontal direction is 180° - 62.78° = 117.2°. Therefore, the polar vector representation for E R is: E R = 39.4 kN @ 117°
Part (b): We observe from the vector polygon that α = 180° - 60° - 45° = 75°.
Using the Law of Cosines
The given information can be used to determine the magnitude and direction of the vector $\vec{R}$ representing the position from boat A to boat C.
Step 1: Calculate the magnitude of $\vec{R}$
Using the law of cosines, we have:
$R = \sqrt{(3 \text{ km})^2 + (4 \text{ km})^2 - 2(3 \text{ km})(4 \text{ km}) \cos 75^\circ} = 1.970 \text{ km}$
Step 2: Calculate the angle $\beta$
Using the law of sines, we have:
$\sin \beta = \frac{4 \text{ km}}{1.970 \text{ km}} \sin 75^\circ$ $\beta = \sin^{-1}\left(\frac{4 \text{ km}}{1.970 \text{ km}} \sin 75^\circ\right) = 67.91^\circ$
Step 3: Determine the direction of $\vec{R}$
The direction of $\vec{R}$ measured from the right-hand horizontal direction is $67.91^\circ - 45^\circ = 22.91^\circ$.
Step 4: Represent the vector $\vec{R}$ in polar form
The polar vector representation of $\vec{R}$ is: $\vec{R} = 1.97 \text{ km} @ 22.9^\circ$
Vector Polygon Sketches and Evaluations
Part (a): $\vec{R} = \vec{A} + \vec{B}$
Where $\vec{A} = 2 \text{ m} @ 0^\circ$ and $\vec{B} = 6 \text{ m} @ 90^\circ$
The magnitude of $\vec{R}$ is given by: $R = \sqrt{A^2 + B^2} = \sqrt{( \text{ m})^2 + (6 \text{ m})^2} = 6.32 \text{ m}$
The angle $\beta$ is found using the law of sines: $\sin \beta = \frac{6 \text{ m}}{6.32 \text{ m}} \sin 90^\circ$ $\beta = 71.6^\circ$
The polar vector representation of $\vec{R}$ is: $\vec{R} = 6.32 \text{ m} @ 71.6^\circ$
Part (b): $\vec{R} = 2\vec{A} - \vec{B}$
The magnitude and angle of $\vec{R}$ are calculated as: $R = \sqrt{( \cdot 2 \text{ m} - 6 \text{ m})^2 + (6 \text{ m})^2} = 7.211 \text{ m}$ $ \beta = \sin^{-1}\left(\frac{6 \text{ m}}{7.211 \text{ m}}\right) = -56.3^\circ$
The polar vector representation of $\vec{R}$ is: $\vec{R} = 7.21 \text{ m} @ -56.3^\circ$
Part (c): $\vec{R} = |\vec{A}|\vec{B} + |\vec{B}|\vec{A}$
The magnitude and angle of $\vec{R}$ are calculated as: $R = \sqrt{( \text{ m} \cdot 6 \text{ m})^2 + (2 \text{ m} \cdot 6 \text{ m})^2} = 16. \text{ m}^2$ $\beta = \tan^{-1}\left(\frac{2 \text{ m} \cdot 6 \text{ m}}{ \text{ m} \cdot 6 \text{ m}}\right) = 45^\circ$
The polar vector representation of $\vec{R}$ is: $\vec{R} = 17.0 \text{ m} ^2 @ 45^\circ$
Remarks
The text notes that Equations (2) and (6) are identical, and hence $\alpha = \beta = 24.15^\circ$. This means that the law of cosines has multiple solutions, two of which are $\alpha = \pm 24.15^\circ$. Using this result, both possible position vectors from boat A to boat C could have been obtained.
Force Polygon and Resultant Force
Force Polygon
The force polygon shown in the image corresponds to the addition of the forces applied by the three ropes to the ship. The known force vectors, 2 kN and 3 kN, are sketched first. There are many possible choices of the force F such that the resultant force will be parallel to line a. The smallest value of F occurs when the force F is perpendicular to line a, i.e., when the angle θ is 90°.
Magnitude of Force F
The magnitude of the force F is found by using the force polygon to write:
F = (3 kN) sin 60° - (2 kN) sin 30°
This equation represents the solution to the problem.
Shortest-Distance Route
Path Walked by the Surveyor
The vector polygon shown in the image corresponds to the addition of the four position vectors corresponding to the path walked by the surveyor. The first three position vectors take the surveyor to the point at which she begins to travel back to the line that is directly northeast of her starting position (shown as a dashed line in the vector polygon). The path she takes to reach this line has distance d, and several possibilities are shown.
Determining the Shortest Distance
By examining the vector polygon, the smallest value of d results when she travels directly southeast, in which case d is given by:
d = (400 m) sin 45° - (200 m) sin 45° = 141 m
Final Distance from Starting Point
To summarize, the surveyor should walk 141 m in the southeast direction. The distance R from her starting point to her final position is given by:
R = (200 m) cos 45° + (400 m) cos 45° + 300 m = 724 m
Resultant Force on Utility Pole
Case 1: P = 0
If the force P is 0, the resultant force applied by the wires to the pole can be determined using polar vector representation.
Case 2: P = 500 N and α = 60°
If the force P is 500 N and the angle α is 60°, the resultant force applied by the wires to the pole can be determined using polar vector representation.
Force Polygon Analysis
Part (a)
The force polygon shown at the right can be used to determine the resultant force Q. Regardless of which force polygon is used, the law of cosines provides:
Q = √[(400 N)^2 + (650 N)^2 - 2(400 N)(650 N)cos 30°] = 363.5 N
Using the first force polygon, the law of sines is used to determine the angle θ1 as:
400 N / sin θ1 = Q / sin 30° θ1 = sin^-1(400 / 363.5) = 33.38°
The orientation of Q to be used for its polar vector representation is 180° - θ1 = 146.6°, and hence the vector representation of Q is:
Q = 364 N @ 147°
Alternatively, the second force polygon could be used, which would also yield Q = 363.3 N. The orientation of Q would be 30° + θ2 = 146.6°, and hence the vector representation would be the same:
Q = 364 N @ 147°
Part (b)
Our strategy is to add the force vector P to the result for Q obtained in Part (a). The force polygon is shown, where Q from Eq. (1) and θ1 from Eq. (2) are used, such that:
θ4 = 60° - θ1 = 26.62°
The law of cosines may be used to find R:
R = √[(500 N)^2 - 2(500 N)(363.3 N)cos θ4] = 239.1 N
$\vec{F}_a = 105.7 N @ 0^\circ$ (4) $\vec{F}_b = 226.6 N @ 90^\circ$ (5)
Oblique Force Addition
Force Polygon
The force polygon corresponding to the addition of $\vec{F}_a$ and $ \vec{F}_b$ to produce the resultant $\vec{F}$ is shown in the figure.
Angle Calculations
Since $\vec{F}_a$ and $\vec{F}_b$ are not perpendicular, the laws of sines and cosines must be used. The angles $\theta_1$, $\theta_2$, and $\theta_3$ are easily determined as:
$\theta_1 = 90^\circ - 65^\circ = 25^\circ$ (2) $\theta_2 = 90^\circ - 50^\circ = 40^\circ$ (3) $\theta_3 = 180^\circ - \theta_1 - \theta_2 = 115^\circ$ (4)
Magnitude Calculations
The law of sines provides:
$\frac{250 N}{\sin \theta_2} = \frac{|\vec{F}_a|}{\sin \theta_1} = \frac{| \vec{F}_b|}{\sin \theta_3}$ (5)
Solving for the magnitudes:
$|\vec{F}_a| = (250 N) \sin 25^\circ / \sin 40^\circ = 164.4 N$ (6) $| \vec{F}_b| = (250 N) \sin 115^\circ / \sin 40^\circ = 352.5 N$ (7)
Polar Vector Representation
Using polar vector representation, the forces are:
$\vec{F}_a = 164.4 N @ -50^\circ$ (8) $\vec{F}_b = 352.5 N @ 90^\circ$ (8)
Wind Force on a Sailing Canoe
Force Decomposition
Let the force perpendicular to the keel be denoted by $F_\perp$ and the force parallel to the keel be denoted by $F_\parallel$. The sketch shows the addition of these two forces to yield the 40 lb force applied to the sail.
$F_\perp = (40 lb) \sin 20^\circ = 13.7 lb$ (1) $F_\parallel = (40 lb) \cos 20^\circ = 37.6 lb$ (2)
Optimization of Force Polygon
Relationship for $F_{OC}'$
A relationship for $F_{OC}'$ in terms of $F_\parallel$ and $\beta$ is needed, and this may be obtained using the force polygon and the law of sines:
$F_{OC}' = (400 lb) / (\sin 30^\circ / \sin \beta)$ (1)
Minimizing $F_{OC}'$
To determine the minimum value of $F_{OC}'$ as a function of $\beta$, we make $F_{OC}'$ stationary by setting its derivative with respect to $\beta$ equal to zero:
$\frac{dF_{OC}'}{d\beta} = 0 = \frac{(400 lb)(\sin 30^\circ)(\cos \beta)} {(\sin \beta)^2}$ (2)
Satisfying this equation requires $\cos \beta = 0$, which gives $\beta = 90^\circ$. From Eq. (1), this corresponds to the minimum value of $F_{OC}'$.
Vector Representations
Part (a): Position vector from A to B
The position vector from A to B, denoted as E rAB, can be represented in Cartesian coordinates as:
E rAB = (4 m) i - (3 m) j
Part (b): Position vector from B to A
The position vector from B to A, denoted as E rBA, can be represented in Cartesian coordinates as:
E rBA = -(4 m) i + (3 m) j
Part (c): Unit vector from A to B
The unit vector in the direction from A to B, denoted as O uAB, can be calculated as:
O uAB = E rAB / |E rAB| = (4/5) i - (3/5) j
Part (d): Unit vector from B to A
The unit vector in the direction from B to A, denoted as O uBA, can be calculated as:
Part (e)
The force vector from point A to point B is given by:
$\vec{\mathbf{F}}_{AB} = (8 \text{ lb}) \left(\cos 30^\circ \mathbf{i} + \sin 30^\circ \mathbf{j}\right)$
Part (f)
The force vector from point B to point A is related to the force vector from point A to point B by:
$\vec{\mathbf{F}} {BA} = -\vec{\mathbf{F}} {AB}$
Therefore, the force vector from point B to point A is:
$\vec{\mathbf{F}}_{BA} = -(6.93 \mathbf{i} + 4.00 \mathbf{j}) \text{ lb}$
Problem 2.
Solution
In Cartesian vector representation, the resultant force vector $ \vec{\mathbf{R}}$ is expressed as:
$\vec{\mathbf{R}} = \vec{\mathbf{F}}_1 + \vec{\mathbf{F}}_2 + \vec{\mathbf{F}}_3$
$\vec{\mathbf{R}} = (400 \text{ N})(-\sin 40^\circ \mathbf{i} + \cos 40^\circ \mathbf{j}) + (200 \text{ N})\left(\frac{5}{13} \mathbf{i} + \frac{12}{13} \mathbf{j}\right) + (100 \text{ N})(\cos 30^\circ \mathbf{i}
- \sin 30^\circ \mathbf{j})$
$\vec{\mathbf{R}} = (-93.6 \mathbf{i} + 541 \mathbf{j}) \text{ N}$
The magnitude of the resultant force is:
$R = \sqrt{(-93.59)^2 + (541.0)^2} = 549 \text{ N}$
The orientation of the resultant force, measured from the vertical direction, is:
$\beta = \tan^{-1}\left(\frac{541.0 \text{ N}}{93.59 \text{ N}}\right) = 80.18^\circ$
The orientation of the resultant force, measured from the right-hand horizontal direction, is:
$\theta = 90^\circ + 9.814^\circ = 99.81^\circ$
Therefore, the resultant force in polar vector representation is:
$\vec{\mathbf{R}} = 549 \text{ N} @ 99.8^\circ$
Problem 2.
Solution
The resultant force is:
$\vec{\mathbf{R}} = \vec{\mathbf{F}}_1 + \vec{\mathbf{F}}_2 + \vec{\mathbf{F}}_3$
$\vec{\mathbf{R}} = 2000 \text{ lb}\left(\frac{15}{\sqrt{15^2 + (-6)^2}} \mathbf{i} - \frac{6}{\sqrt{15^2 + (-6)^2}} \mathbf{j}\right) + 1600 \text{ lb}(-\sin 50^\circ \mathbf{i} - \cos 50^\circ \mathbf{j}) + 1100 \text{ lb}(\cos 15^\circ \mathbf{i} - \sin 15^\circ \mathbf{j})$
$\vec{\mathbf{R}} = (1694 \mathbf{i} - 2056 \mathbf{j}) \text{ lb}$
The magnitude of the resultant force is:
$R = \sqrt{(1694 \text{ lb})^2 + (-2056 \text{ lb})^2} = 2664 \text{ lb}$
The orientation of the resultant force, measured from the horizontal direction, is:
$\theta = \tan^{-1}\left(\frac{-2056 \text{ lb}}{1694 \text{ lb}}\right) = 50.52^\circ$
Therefore, the resultant force in polar vector representation is:
$\vec{\mathbf{R}} = 2664 \text{ lb} @ -50.52^\circ$
or, using a positive angle:
$\vec{\mathbf{R}} = 2664 \text{ lb} @ 309.5^\circ$
Problem 2.
Solution
The resultant force is:
$\vec{\mathbf{R}} = \vec{\mathbf{F}}_1 + \vec{\mathbf{F}}_2 + \vec{\mathbf{F}}_3 + \vec{\mathbf{F}}_4$
$\vec{\mathbf{R}} = 600 \text{ N}(-\cos 55^\circ \mathbf{i} - \sin 55^\circ \mathbf{j}) + 700 \text{ N}(-\sin 10^\circ \mathbf{i} - \cos 10^\circ \mathbf{j}) + 800 \text{ N}\left(\frac{13}{p13^2 + 15^2} \mathbf{i} + \frac{15}{p13^2 + 15^2} \mathbf{j}\right) + 900 \text{ N}\left(\frac{20} {p20^2 + 9^2} \mathbf{i} + \frac{9}{p20^2 + 9^2} \mathbf{j}\right)$
$\vec{\mathbf{R}} = (-168.9 \mathbf{i} - 207.0 \mathbf{j}) \text{ N}$
The magnitude of the resultant force is:
Part (c)
The unit vector in the direction from point A to point D is:
$\vec{\mathbf{u}} {AD} = \frac{\vec{\mathbf{r}} {AD}}{r_{AD}} = \frac{5.481 \mathbf{i} + 57.12 \mathbf{j}}{57.38 \text{ cm}} = 0. \mathbf{i} + 0.995 \mathbf{j}$
Problem 2.
Solution
The position vectors are:
$\vec{\mathbf{r}}_{AB} = 12.5 \text{ m}(\cos 75^\circ \mathbf{i} + 0.2225 \mathbf{j})$
$\vec{\mathbf{r}}_{BC} = 2.8 \text{ m}\left(\frac{36}{p36^2 + 15^2} \mathbf{i} + \frac{15}{p36^2 + 15^2} \mathbf{j}\right)$
$\vec{\mathbf{r}}_{CD} = 7 \text{ m}(\sin 55^\circ \mathbf{i} - \cos 55^\circ \mathbf{j})$
$\vec{\mathbf{r}}_{DE} = 2.5 \text{ m}\left(\frac{2}{p2^2 + 1^2} \mathbf{i} - \frac{1}{p2^2 + 1^2} \mathbf{j}\right)$
The position vector from point A to point E is:
$\vec{\mathbf{r}} {AE} = \vec{\mathbf{r}} {AB} + \vec{\mathbf{r}} _{BC}
- \vec{\mathbf{r}}_ {CD} + \vec{\mathbf{r}}_{DE}$
$\vec{\mathbf{r}}_{AE} = (13.79 \mathbf{i} + 8.018 \mathbf{j}) \text{ m} $
Problem 2.
Solution
The position vectors are:
$\vec{\mathbf{r}}_{OA} = 100 \text{ mm} \mathbf{j}$
$\vec{\mathbf{r}}_{AB} = 400 \text{ mm}(\cos \alpha \mathbf{i} + \sin \alpha \mathbf{j})$
$\vec{\mathbf{r}}_{BC} = 300 \text{ mm}(\cos(\alpha - \beta) \mathbf{i}
- \sin(\alpha - \beta) \mathbf{j})$
The position vector from point O to point C is:
$\vec{\mathbf{r}} {OC} = \vec{\mathbf{r}} {OA} + \vec{\mathbf{r}} _{AB}
$\vec{\mathbf{r}}_{OC} = (400 \text{ mm} \cos \alpha + 300 \text{ mm} \cos(\alpha - \beta)) \mathbf{i} + (100 \text{ mm} + 400 \text{ mm} \sin \alpha + 300 \text{ mm} \sin(\alpha - \beta)) \mathbf{j}$
This expression for $\vec{\mathbf{r}}_{OC}$ in terms of the angles $ \alpha$ and $\beta$ can be used to determine the position of the actuator at point C.
Determining the Ratio of Forces F2/F1 for a
Vertical Resultant
Problem 2.
Two ropes are used to lift a pipe in a congested region. The goal is to determine the ratio of the forces F2/F1 so that the resultant of the forces F and F2 is vertical.
The resultant force R is given by:
R = F1(-cos 15° i + sin 15° j) + F2(sin 40° i + cos 40° j)
(1) To make the x-component of the resultant force zero, we have:
0 = -F1 cos 15° + F2 sin 40°
(2) Solving for the ratio F2/F1, we get:
F2/F1 = cos 15° / sin 40° = 1.
Therefore, the ratio of the forces F2/F1 that will make the resultant force vertical is 1.50.
Determining the Maximum Value of Force P
Problem 2.
A welded steel tab is subjected to forces F and P. The goal is to determine the largest value P may have if F = 1000 lb and the magnitude of the resultant force cannot exceed 1500 lb.
The resultant force R is given by:
R = F(-4/5 i - 3/5 j) + P(12/13 i - 5/13 j)
(1) Setting the magnitude of the resultant force R equal to 1500 lb, we get a quadratic equation in P:
P^2 - 1.32×10^4/13 P - 1.25×10^6 = 0
(2) Solving this equation, we obtain two values for P:
Determining the Angle α to Minimize the
Resultant Force Magnitude
Problem 2.
A short cantilever beam is subjected to three forces. If F = 8 kN, the goal is to determine the value of α that minimizes the magnitude of the resultant of the three forces.
(1) Examining the force polygon, the resultant force R of smallest magnitude is obtained when the force F has a direction from point A to point O. This corresponds to:
α = tan^-1(8 kN/12 kN) = 33.7°
(2) The magnitude of the resultant force R is then calculated as:
R = √((5.344 kN)^2 + (3.562 kN)^2) = 6.35 kN
Determining the Value of Force F to Minimize
the Resultant Force Magnitude
Problem 2.
A short cantilever beam is subjected to three forces. If α = 45°, the goal is to determine the value of F that will make the magnitude of the resultant of the three forces smallest.
(1) Examining the force polygon, the resultant force R has the smallest magnitude when its direction is perpendicular to the force F.
(2) Equating the x and y components of the resultant force, we obtain:
F = 12 kN - (8 kN)/√2 = 8 kN
This value of F will make the magnitude of the resultant force smallest.
Transforming Vectors from One Coordinate
System to Another
Derivation of Transformation Equations
The transformation equations that allow the conversion of vector components from one Cartesian coordinate system to another are derived as follows:
Consider a vector v that has components vx and vy in the x-y coordinate system. This vector can also be expressed in terms of its components vt and
vn in the t-n coordinate system, where the t axis is rotated counterclockwise by an angle θ from the positive x axis.
Using the sketch provided, we can resolve the vector vx acting in the x direction into its t and n components:
vt = vx cos θ vn = -vx sin θ
Similarly, we can resolve the vector vy acting in the y direction into its t and n components:
vt = vy sin θ vn = vy cos θ
By observing that v = vx i + vy j = vt t + vn n, we can derive the transformation equations:
vt = vx cos θ + vy sin θ vn = -vx sin θ + vy cos θ
These equations are identical to Eqs. (2) and (3) in the problem statement.
Application to Example 2.
For the eyebolt and post problem (Example 2.7), the x and y components of the resultant force were found to be:
Rx = -809.9 lb Ry = 373.9 lb
Since the t axis is rotated 40° counterclockwise from the x axis, θ = 40°, and we can use Eqs. (2) and (3) to find the t and n components of the resultant force:
Rt = (-809.9 lb) cos 40° + (373.9 lb) sin 40° = -380.0 lb Rn = (-809.9 lb) sin 40° + (373.9 lb) cos 40° = 807.0 lb
These results agree with the values provided in Eq. (9) of the problem statement.
Importance of t and n Components
The t and n components of the resultant force vector are useful to know because they provide information about the force acting parallel (t) and perpendicular (n) to the surface or axis of interest. This information is important for determining if the object will slide or tip under the applied forces, using concepts such as Coulomb's law of friction.
Determining the Resultant Force Components
Part (a)
To determine the x and y components of the resultant force vector at point A due to forces P1 and P2, we can use the following equations:
