# The cylindrical plate is subjected to the three 2

The cylindrical plate is subjected to the three cable forces which are concurrent at point D. Express each force which the cables exert on the plate as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force.

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 write the locations of points A, B, C, and D in Cartesian vector form.

Using the diagram, the points are at the following locations:

$A:(0.75i+0j+0k)$ m

$B:(-0.75\sin30^0i+0.75\cos30^0j+0k)\,=\,(-0.375i+0.65j+0k)$ m

$C:(-0.75\cos45^0i-0.75\sin45^0j+0k)\,=\,(-0.53i-0.53j+0k)$ m

$D:(0i+0j+3k)$ m

(Note that the radius of the circle is 0.75 m. Using our sine and cosine functions, we can then figure out the x and y coordinates of each point)

We will now write position vectors for points from A to D, B to D, and C to D.

$r_{AD}\,=\,\left\{(0-0.75)i+(0-0)j+(3-0)k\right\}$

$r_{AD}\,=\,\left\{-0.75i+0j+3k\right\}$ m

$r_{BD}\,=\,\left\{(0-(-0.375))i+(0-0.65)j+(3-0)k\right\}$

$r_{BD}\,=\,\left\{0.375i-0.65j+3k\right\}$ m

$r_{CD}\,=\,\left\{(0-(-0.53))i+(0-(-0.53))j+(3-0)k\right\}$

$r_{CD}\,=\,\left\{0.53i+0.53j+3k\right\}$ m

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

We will now find the magnitude of each position vector.

magnitude of $r_{AD}\,=\,\sqrt{(-0.75)^2+(0)^2+(3)^2}\,=\,3.09$ m

magnitude of $r_{BD}\,=\,\sqrt{(0.375)^2+(-0.65)^2+(3)^2}\,=\,3.09$ m

magnitude of $r_{CD}\,=\,\sqrt{(0.53)^2+(0.53)^2+(3)^2}\,=\,3.09$ m

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.

Let us now write the unit vector for each position vector.

$u_{AD}\,=\,\left(-\dfrac{0.75}{3.09}i+0j+\dfrac{3}{3.09}k\right)$

$u_{BD}\,=\,\left(\dfrac{0.375}{3.09}i-\dfrac{0.65}{3.09}j+\dfrac{3}{3.09}k\right)$

$u_{CD}\,=\,\left(\dfrac{0.53}{3.09}i+\dfrac{0.53}{3.09}j+\dfrac{3}{3.09}k\right)$

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

We can now write each force in Cartesian form by multiplying the given magnitudes of each force by the unit vector.

$F_A\,=\,6\left(-\dfrac{0.75}{3.09}i+0j+\dfrac{3}{3.09}k\right)$

$F_A\,=\,\left\{-1.46i+0j+5.82k\right\}$ kN

$F_B\,=\,8\left(\dfrac{0.375}{3.09}i-\dfrac{0.65}{3.09}j+\dfrac{3}{3.09}k\right)$

$F_B\,=\,\left\{0.97i-1.68j+7.77k\right\}$ kN

$F_C\,=\,5\left(\dfrac{0.53}{3.09}i+\dfrac{0.53}{3.09}j+\dfrac{3}{3.09}k\right)$

$F_C\,=\,\left\{0.86i+0.86j+4.85k\right\}$ kN

Next, we will find the resultant force.

$F_R\,=\,F_A+F_B+F_C$

$F_R\,=\,(-1.46+0.97+0.86)i+(0-1.68+0.86)j+(5.82+7.77+4.85)k$

$F_R\,=\,\left\{0.37i-0.82j+18.44k\right\}$ kN

To figure out the coordinate direction angles, we need to find the magnitude of the resultant force.

magnitude of $F_R\,=\,\sqrt{(0.37)^2+(-0.82)^2+(18.44)^2}\,=\,18.46$ kN

We can now write the coordinate direction angles by taking the cosine inverse of each component of the resultant force divided by it’s magnitude.

$\alpha\,=\,\cos^{-1}\left(\dfrac{0.37}{18.46}\right)\,=\,88.8^0$

$\beta\,=\,\cos^{-1}\left(\dfrac{-0.82}{18.46}\right)\,=\,92.5^0$

$\gamma\,=\,\cos^{-1}\left(\dfrac{18.44}{18.46}\right)\,=\,2.66^0$

$F_A\,=\,\left\{-1.46i+0j+5.82k\right\}$ kN

$F_B\,=\,\left\{0.97i-1.68j+7.77k\right\}$ kN

$F_C\,=\,\left\{0.86i+0.86j+4.85k\right\}$ kN

$F_R\,=\,\left\{0.37i-0.82j+18.44k\right\}$ kN

magnitude of $F_R\,=\,18.46$ kN

$\alpha\,=\,88.8^0$

$\beta\,=\,92.5^0$

$\gamma\,=\,2.66^0$

## 2 thoughts on “The cylindrical plate is subjected to the three”

• questionsolutions Post author

Sorry for the late reply. We use sin30 and cos30 to figure out the exact point where B lies with respective to the x and y axes. Using sin30 we can figure out the x length, and using cos30 we can figure out the y length. Remember that sin is opposite over hypotenuse while cos is adjacent over hypotenuse.