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.

#### Solution:

Show me the final answers↓

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:

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\{-0.75i+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

We will now find the magnitude of each position vector.

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

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

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)

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

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\,=\,(-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.

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.

\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

#### Final Answers:

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

Can you please explain to me why the point B you use sin30i and cos30j

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.