The door is held opened by means of two chains. If the tension in AB and CD is F_A = 300 N and F_C = 250 N, respectively, express each of these forces in Cartesian vector form.

#### Solution:

Show me the final answer↓

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

Notice how we can figure our the z-coordinates of point A and C using the triangle we formed. Using sine function (opposite over hypotenuse), we can figure out h which is the height (z-component). The height of points A and C is:

h\,=\,0.75 m

We can also figure out the l portion as well, by using cosine (adjacent over hypotenuse). This will allow us to figure out the total y-lengths of points A and C.

l\,=\,1.3 m

Let us now write the coordinates of each point.

Using the diagram and the lengths we found, we can now write the location of each point.

B:(0i+0j+0k) m

C:(-2.5i-2.3j+0.75k) m

D:(-0.5i+0j+0k) m

Let us now write the position vectors for points from A to B and C to D.

r_{AB}\,=\,\left\{0i+2.3j-0.75k\right\}

r_{CD}\,=\,\left\{(-0.5-(-2.5))i+(0-(-2.3))j+(0-0.75)k\right\}

r_{CD}\,=\,\left\{2i+2.3j-0.75k\right\}

Let us now find the magnitude of each position vector.

magnitude of r_{AB}\,=\,2.42 m

magnitude of r_{CD}\,=\,\sqrt{(2)^2+(2.3)^2+(-0.75)^2}

magnitude of r_{CD}\,=\,3.14 m

We can now write our unit vectors for r_{AB} and r_{CD}.

u_{CD}\,=\,\left(\dfrac{2}{3.14}i\,+\,\dfrac{2.3}{3.14}j\,-\,\dfrac{.75}{3.14}k\right)

Now we can write each force in Cartesian vector form by multiplying the force by the corresponding unit vector.

F_A\,=\,\left\{0i+285j-93k\right\} N

F_C\,=\,250\left(\dfrac{2}{3.14}i\,+\,\dfrac{2.3}{3.14}j\,-\,\dfrac{.75}{3.14}k\right)

F_C\,=\,\left\{159i+183j-60k\right\} N

#### Final Answers:

F_C\,=\,\left\{159i+183j-60k\right\} N