Cable AB exerts a force of 80 N 4

Cable AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom.

Cable AB exerts a force of 80 N

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 write two position vectors, from A to O, and A to B. We will then find the unit vector of the position vector extending from A to B. Once we find the unit vector, we can express the 80 N in Cartesian vector form. We can then calculate the projection using the dot product. To do so, we will first find the locations of points A, O and B in Cartesian vector form.

Cable AB exerts a force of 80 N

Using the diagram, the locations of the points are:

A:(3\cos60^0i+3\sin60^0j+0k)\,=\,(1.5i+2.6j+0k) m

B:(0i+0j+4k) m

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


Let us now write our position vectors.


r_{AO}\,=\,\left\{-1.5i-2.6j+0k\right\} m



r_{AB}\,=\,\left\{-1.5i-2.6j+4k\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 the position vectors.

magnitude of r_{AB}\,=\,\sqrt{(-1.5)^2+(-2.6)^2+(4)^2}\,=\,5

magnitude of r_{AO}\,=\,\sqrt{(-1.5)^2+(-2.6)^2+(0)^2}\,=\,3

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 will now find the unit vectors of 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 express the 80 N force that is along the position vector r_{AB} in Cartesian vector form .


F\,=\,\left\{-24i-42j+64k\right\} N


We can now find the projection of this force along the boom AO. We will use the dot product to do so.

\text{Proj}\,F\,=\,F\cdot U_{AO}


\text{Proj}\,F\,=\,48.4 N

The dot product is the product of the magnitudes of two vectors and the cosine angle \theta between them. For example, if we had two vectors, \mathbf{A} and \mathbf{B}, the dot product of the two is: A\cdot B\,=\,AB\cos \theta. In cartesian vector form, A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z. The angle formed between two vectors can be found by: \theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)


Final Answer:

The magnitude of the projection of the 80 N force along the boom AO is 48.4 N.


This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-146.

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