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.


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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\{-0.75i+0j+3k\right\} m



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



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.




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\,=\,\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


Next, we will find the resultant force.



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.





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





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.