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.

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.

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.

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

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

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

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

$\alpha=37.8^0$

$\beta=67.1^0$

$\gamma=118^0$