Cable AB exerts a force of 80 N 2


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.

Solution:

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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\{(0-1.5)i+(0-2.6)j+(0-0)k\right\}

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

 

r_{AB}\,=\,\left\{(0-1.5)i+(0-2.6)j+(4-0)k\right\}

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.

u_{AB}\,=\,\left(-\dfrac{1.5}{5}i-\dfrac{2.6}{5}j+\dfrac{4}{5}k\right)

u_{AO}\,=\,\left(-\dfrac{1.5}{3}i-\dfrac{2.6}{3}j+0k\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 express the 80 N force that is along the position vector r_{AB} in Cartesian vector form .

F\,=\,80\left(-\dfrac{1.5}{5}i-\dfrac{2.6}{5}j+\dfrac{4}{5}k\right)

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\,=\,\left[(-24)(-\dfrac{1.5}{3})+(-42)(-\dfrac{2.6}{3})+(-64)(0)\right]

\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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