The man pulls on the rope at C

The man pulls on the rope at C with a force of 70 lb which causes the forces FA and FC at B to have this same magnitude. Express each of these two forces as Cartesian vectors.

The man pulls on the rope at C

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 figure out the locations of points A, B and C with respect to the origin.

The man pulls on the rope at C

Using the diagram, the locations of the points are:

A: (5i-7j+5k) ft

B: (-1i-5j+8k) ft

C: (5i+7j+4k) ft


We can now write position vectors from points B to A and B to C.

r_{BA}\,=\,\left\{(5-(-1))i+(-7-(-5))j+(5-8)k\right\}=\left\{6i-2j-3k\right\} ft

r_{BC}\,=\,\left\{(5-(-1))i+(7-(-5))j+(4-8)k\right\}=\left\{6i+12j-4k\right\} ft

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 figure out the magnitude of each position vector.

magnitude of r_{BA}\,=\,\sqrt{(6)^2+(-2)^2+(-3)^2}=7 ft

magnitude of r_{BC}\,=\,\sqrt{(6)^2+(12)^2+(-4)^2}=14 ft

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.


Next, we will write the unit vectors for each of the position vectors.



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)}}


Let us now express each force in Cartesian vector form by multiplying the unit vector by the magnitude of the force. Remember, the question states that each rope has a force of 70 lb.


F_{BA}=\left\{60i-20j-30k\right\} lb



F_{BC}=\left\{30i+60j-20k\right\} lb


Final Answers:

F_{BA}=\left\{60i-20j-30k\right\} lb

F_{BC}=\left\{30i+60j-20k\right\} lb


This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-96.

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