The door is held opened by means of two chains


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.

The door is held opened by means of two chains

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

The door is held opened by means of two chains

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:

\sin30^0\,=\,\dfrac{h}{1.5}

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.

\cos30^0\,=\,\dfrac{l}{1.5}

l\,=\,1.3 m

 

Let us now write the coordinates of each point.

The door is held opened by means of two chains

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

A:(0i-2.3j+0.75k) m

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\{(0-0)i+(0-(-2.3))j+(0-0.75)k\right\}

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\}

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{(0)^2+(2.3)^2+(-0.75)^2}

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

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 r_{AB} and r_{CD}.

u_{AB}\,=\,\left(\dfrac{0}{2.42}i\,+\,\dfrac{2.3}{2.42}j\,-\,\dfrac{.75}{2.42}k\right)

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

 

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

F_A\,=\,300\left(\dfrac{0}{2.42}i\,+\,\dfrac{2.3}{2.42}j\,-\,\dfrac{.75}{2.42}k\right)

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

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

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-98.

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