Cables AB and AC can sustain a maximum tension 4


Cables AB and AC can sustain a maximum tension of 500 N, and the pole can support a maximum compression of 300 N. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole.

Cables AB and AC can sustain a maximum tension

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

Show me the final answer↓

If each one of the ropes will break

Using the diagram, the locations of the points A, B, C, and O are:

A:(2i-1.5j+6k)

B:(-4i+1.5j+0k)

C:(0i+1.5j+0k)

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

 

The position vectors for points from A to B, A to C, and A to O are:

r_{AB}\,=\,\left\{(-4-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-6i+3j-6k\right\}

r_{AC}\,=\,\left\{(0-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-2i+3j-6k\right\}

r_{OA}\,=\,\left\{(2-0)i+(-1.5-0)j+(6-0)k\right\}\,=\,\left\{2i-1.5j+6k\right\}
(Why did we write the position vector from O to A instead of A to O? Unlike ropes, which can only be in tension, the pole will actually be in compression. That means a force from the pole is heading upwards, where as in the cables, the force is heading away from the lamp)

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

 

The magnitude of each position vector is:

magnitude of r_{AB}\,=\,\sqrt{(-6)^2+(3)^2+(-6)^2}\,=\,9

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

magnitude of r_{OA}\,=\,\sqrt{(2)^2+(-1.5)^2+(6)^2}\,=\,6.5

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.

 

The unit vector for each position vector is:

u_{AB}\,=\,\left(-\dfrac{6}{9}i+\dfrac{3}{9}j-\dfrac{6}{9}k\right)

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

u_{OA}\,=\,\left(\dfrac{2}{6.5}i-\dfrac{1.5}{6.5}j+\dfrac{6}{6.5}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)}}

 

We can now write each force in Cartesian vector form:

F_{AB}\,=\,F_{AB}\left(-\dfrac{6}{9}i+\dfrac{3}{9}j-\dfrac{6}{9}k\right)

F_{AC}\,=\,F_{AC}\left(-\dfrac{2}{7}i+\dfrac{3}{7}j-\dfrac{6}{7}k\right)

F_{OA}\,=\,F_{OA}\left(\dfrac{2}{6.5}i-\dfrac{1.5}{6.5}j+\dfrac{6}{6.5}k\right)

(Further simplify by expanding the brackets and writing the fractions in decimal form)

F_{AB}\,=\,\left\{-0.667F_{AB}i+0.333F_{AB}j-0.667F_{AB}k\right\}

F_{AC}\,=\,\left\{-0.286F_{AC}i+0.429F_{AC}j-0.857F_{AC}k\right\}

F_{OA}\,=\,\left\{0.308F_{OA}i-0.231F_{OA}j+0.923F_{OA}k\right\}

F\,=\,\left\{0i+0j-W\right\}

(Force F is the weight of the lamp, which only has a z-component)

 

Since the system is in equilibrium, all forces added together must equal zero.

\sum \text{F}\,=\,0

F_{AB}+F_{AC}+F_{OA}+F\,=\,0

 

Furthermore, as each force added together must equal zero, than each individual component (x, y, z-components) added together must also equal zero.

x-components:

-0.667F_{AB}-0.286F_{AC}+0.308F_{OA}\,=\,0

y-components:

0.333F_{AB}+0.429F_{AC}-0.231F_{OA}\,=\,0

z-components:

-0.667F_{AB}-0.857F_{AC}+0.923F_{OA}-W\,=\,0

 

The question states that the pole can only support 300 N in compression. Thus, we will assume force F_{AO} is 300 N and substitute it into our equations. Doing so gives us the following equations:

x-components:

-0.667F_{AB}-0.286F_{AC}+92.4\,=\,0

y-components:

0.333F_{AB}+0.429F_{AC}-69.3\,=\,0

z-components:

-0.667F_{AB}-0.857F_{AC}+276.9-W\,=\,0

 

Solving the three equations simultaneously gives us:

F_{AB}\,=\,103.8 N

F_{AC}\,=\,80.9 N

W\,=\,138.3 N

Notice how our values for the cables are less then 500 N, thus our assumption was right. To double check, you can assume each cable is 500 N and solve the equations again, doing so will show that the forces in either the pole or the cable will exceed the maximum allowable limit.

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-51.

Leave a comment

Your email address will not be published.

4 thoughts on “Cables AB and AC can sustain a maximum tension

  • Jihan

    Hi there, I have been struggling with the part just before you get the final answer, where you have to solve for Fac, Fab, and W. I have gone online looking for a way to solve them, however, they just show me simple numbers like(3, 6, 9, 10, etc..) How do you deal with more advanced numbers for example I am working on one of the questions from your ‘Equilibrium of a particle 3D force Systems’ video on youtube, and here I got to the final part: Equation 1: -Fba(4/14)+Fca(6/14)+Fda(4/14)=0 Equation 2: Fba(6/14)+Fca(6/14)-Fda(6/14)=0 And lastly…
    Equation 3: Fba(12/14)+Fca(12/14)+Fda(12/14)-W=0 We assume Fba has the maximum tension of 15kN
    The answers you get for all of them are: Fba=’assumed’15kN Fda=15kN Fca=0kN W=25.714kN

    Anyways I know this is a lot, it’s the final part I am not able to get, which is just driving me crazy!
    Thank you for all your work in helping people and I just hope you’re getting the credit you deserve
    Kind Regards

    • questionsolutions Post author

      I am not entirely sure I understand your question. Could you email me with a detailed question. contact @ questionsolutions . com

  • khaled

    Hello hope you are doing well. The equestrian states that we need to find the maximum weight that can be used with 500 T in bot AC and AB and compression of 300 N on the pole, i think you found what is the mass and force in each cable not the maximum weight that can be used, and 300 is a compression force it is not a tension force you used vector OA i think we should use AO and do the equations on the Z axis although i am not entirely sure please correct me if i am wrong.
    Kind regards

    • questionsolutions Post author

      Hi. So I think you are confusing the terms mass and weight. The units for mass would be in kg, lb, g, etc. Weight on the other hand is in N. Weight is mass times acceleration due to gravity. So starting off, I didn’t find the mass, the final answer is indeed the weight as required by the question. Also, I think you are misunderstanding the question, the maximum force the cables can take is 500 N and the pole can support a maximum force of 300 N. For compression, you always go from bottom to top, it’s not a cable, so the vector goes from O to A, NOT A to O. Only cables carry forces away from a load, because remember, you can’t push on a rope. With a stiff rod, the force will go from O to A. I don’t know where you looked, the answers are in the green box at the very bottom, so my guess is, either you misunderstood the question, or you didn’t scroll all the way down to see the answer. I hope this cleared it up, if not, let me know, and I will do my best to give more help. Also keep in mind that if you try it with the 500 N force, you will exceed the 300 N force of the rod. Try it our and see, it is a good exercise for you so you get more comfortable with these problems. 👍