# The shear leg derrick is used to haul 1

The shear leg derrick is used to haul the 200-kg net of fish onto the dock. Determine the compressive force along each of the legs AB and CB and the tension in the winch cable DB. Assume the force in each leg acts along its axis.

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

#### Solution:

We will first find the position vectors for the two legs of AB, AC and the cable DB. Note that while the cable will be in tension (a force going from B to D), the two legs will be in compression (forces going from A to B and C to B).

Looking at the diagram, the locations of the points A, B, C, and D are:

$A:(2i+0j+0k)$

$B:(0i+4j+4k)$

$C:(-2i+0j+0k)$

$D:(0i-5.6j+0k)$

Now we can write the position vectors. Again, remember that in the cable, the force will be going from B to D as it is in tension while the two legs will carry the force upwards because they are in compression.

$r_{AB}\,=\,\left\{(0-2)i+(4-0)j+(4-0)k\right\}\,=\,\left\{-2i+4j+4k\right\}$

$r_{CB}\,=\,\left\{(0-(-2))i+(4-0)j+(4-0)k\right\}\,=\,\left\{2i+4j+4k\right\}$

$r_{BD}\,=\,\left\{(0-0)i+(-5.6-4)j+(0-4)k\right\}\,=\,\left\{0i-9.6j-4k\right\}$

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Let us now find the magnitude of each position vector:

magnitude of $r_{AB}\,=\,\sqrt{(-2)^2+(4)^2+(4)^2}\,=\,6$

magnitude of $r_{CB}\,=\,\sqrt{(2)^2+(4)^2+(4)^2}\,=\,6$

magnitude of $r_{BD}\,=\,\sqrt{(0)^2+(-9.6)^2+(-4)^2}\,=\,10.4$

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 write our unit vectors for each position vector.

$u_{AB}\,=\,\left(-\dfrac{2}{6}i+\dfrac{4}{6}j+\dfrac{4}{6}k\right)$

$u_{CB}\,=\,\left(\dfrac{2}{6}i+\dfrac{4}{6}j+\dfrac{4}{6}k\right)$

$u_{BD}\,=\,\left(0i-\dfrac{9.6}{10.4}j-\dfrac{4}{10.4}k\right)$

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, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

The next step is to express each force in Cartesian vector form.

$F_{AB}\,=\,F_{AB}\left(-\dfrac{2}{6}i+\dfrac{4}{6}j+\dfrac{4}{6}k\right)$

$F_{CB}\,=\,F_{CB}\left(\dfrac{2}{6}i+\dfrac{4}{6}j+\dfrac{4}{6}k\right)$

$F_{BD}\,=\,F_{BD}\left(0i-\dfrac{9.6}{10.4}j-\dfrac{4}{10.4}k\right)$

(simplify by expanding the brackets and writing the fractions in decimal form)

$F_{AB}\,=\,\left\{-0.333F_{AB}i+0.667F_{AB}j+0.667F_{AB}k\right\}$

$F_{CB}\,=\,\left\{0.333F_{CB}i+0.667F_{CB}j+0.667F_{CB}k\right\}$

$F_{BD}\,=\,\left\{0i-0.923F_{BD}j-0.385F_{BD}k\right\}$

$W\,=\,\left\{0i+0j-1962k\right\}$

(Force W is the weight of the weight of the fish. The weight is equal to (200)(9.81)=1962 N)

As the system is in equilibrium, all forces added together must equal zero.

$\sum \text{F}\,=\,0$

$F_{AB}+F_{AC}+F_{BD}+W\,=\,0$

Since all forces added together must equal zero, each component (x, y, z-components) added together must also equal zero.

x-components:

$-0.333F_{AB}+0.333F_{CB}\,=\,0$

y-components:

$0.667F_{AB}+0.667F_{CB}-0.923F_{BD}\,=\,0$

z-components:

$0.667F_{AB}+0.667F_{CB}-0.385F_{BD}-1962\,=\,0$

Solving the three equations gives us:

$F_{AB}\,=\,2523$ N

$F_{CB}\,=\,2523$ N

$F_{BD}\,=\,3647$ N