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.

The cylindrical plate is subjected to the three cable forces

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 cylindrical plate is subjected to the three cable forces

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

 

Final Answers:

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

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-111.

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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.