The two mooring cables exert forces


The two mooring cables exert forces on the stern of a ship as shown. Represent each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant.

The two mooring cables exert forces

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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We will first write the position vectors for each cable. To do so, we will write the locations of points A, B and C in Cartesian vector form.

The two mooring cables exert forces

Using the diagram, the locations of the points are:

A:(50i+10j-30k) ft

B:(50i+50j-30k) ft

C:(0i+0j+0k) ft

 

Let us now write our position vectors:

r_{CA}\,=\,\left\{(50-0)i+(10-0)j+(-30-0)k\right\}=\left\{50i+10j-30k\right\} ft

r_{CB}\,=\,\left\{(50-0)i+(50-0)j+(-30-0)k\right\}=\left\{50i+50j-30k\right\} ft

A position vector, denoted \mathbf{r} is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was (x_A,y_A,z_A) and the coordinates of point B was(x_B,y_B,z_B), then r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

 

The next step is to find the magnitude of each position vector.

magnitude of r_{CA}\,=\,\sqrt{(50)^2+(10)^2+(-30)^2}=59.16

magnitude of r_{CB}\,=\,\sqrt{(50)^2+(50)^2+(-30)^2}=76.81

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r\,=\,ai+bj+ck, then the magnitude would be, r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

 

We can now find the unit vector for each position vector as follows:

u_{CA}\,=\,\left(\dfrac{50}{59.16}i+\dfrac{10}{59.16}j-\dfrac{30}{59.16}k\right)

u_{CB}\,=\,\left(\dfrac{50}{76.81}i+\dfrac{50}{76.81}j-\dfrac{30}{76.81}k\right)

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was r\,=\,ai+bj+ck, then unit vector, u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}

 

Let us now express each force in Cartesian vector form. From the question, we know F_A=200 lb and F_B=150 lb.

F_{CA}=200\left(\dfrac{50}{59.16}i+\dfrac{10}{59.16}j-\dfrac{30}{59.16}k\right)

F_{CA}=\left\{169i+33.8j-101.4k\right\} lb

 

F_{CB}=150\left(\dfrac{50}{76.81}i+\dfrac{50}{76.81}j-\dfrac{30}{76.81}k\right)

F_{CB}=\left\{97.6i+97.6j-58.6k\right\} lb

 

We can now find the resultant force by adding the two forces together.

F_R=F_{CA}+F_{CB}

F_R=\left\{169i+33.8j-101.4k\right\}+\left\{97.6i+97.6j-58.6k\right\}

F_R=\left\{266.6i+131.4j-160k\right\} lb

 

The magnitude of this resultant force is:

magnitude of F_R\,=\,\sqrt{(266.6)^2+(131.4)^2+(-160)^2}=337.5 lb

 

The coordinate direction angles can be calculated by taking the cosine inverse of each component of the resultant force divided by the magnitude.

\alpha=\cos^{-1}\left(\dfrac{266.6}{337.5}\right)=37.8^0

\beta=\cos^{-1}\left(\dfrac{131.4}{337.5}\right)=67.1^0

\gamma=\cos^{-1}\left(\dfrac{-160}{337.5}\right)=118^0

 

Final Answers:

F_{CA}=\left\{169i+33.8j-101.4k\right\} lb

F_{CB}=\left\{97.6i+97.6j-58.6k\right\} lb

F_R=\left\{266.6i+131.4j-160k\right\} lb

magnitude of F_R=337.5 lb

\alpha=37.8^0

\beta=67.1^0

\gamma=118^0

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-101.

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