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.

Solution:

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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.

u_{BA}\,=\,\left(\dfrac{6}{7}i-\dfrac{2}{7}j-\dfrac{3}{7}k\right)

u_{BC}\,=\,\left(\dfrac{6}{14}i+\dfrac{12}{14}j-\dfrac{4}{14}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)}}

 

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}=70\left(\dfrac{6}{7}i-\dfrac{2}{7}j-\dfrac{3}{7}k\right)

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

 

F_{BC}=70\left(\dfrac{6}{14}i+\dfrac{12}{14}j-\dfrac{4}{14}k\right)

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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