Find cable tension in three dimensions

1. Jun 19, 2009

KEØM

1. The problem statement, all variables and given/known data
A piece of machinery of weight W is temporarily supported by cables AB, AC, and ADE. Cable ADE is attached to the ring at A, passes over the pulley at D and back through the ring, and is attached to the support at E. Knowing that the tension in cable AB is 300 N, determine (a) the tension in AC, (b) the tension in ADE, (c) the weight W. (Hint: The tension is the same portions of cable ADE.)

Here is a link to the picture of the drawing just with a different problem (On page 63 Fig. P2.123):

2. Relevant equations
$$\Sigma$$F = 0
F = FU (u is a unit vector pointing in the direction of F)
U = AB/AB (AB is a distance vector for example describing the length and oreintation of the cables in this problem)

3. The attempt at a solution

I first am finding U for cables AC, AE, AD, and AB and then I multiply that vector by the unknown magnitude of the tension. I then sum all the x, y, and z components of each cable tension and the weight of the machine giving me 3 equations with three unknowns. Solving these equations does not give me the correct answers to this problem. I think I am solving it in the correct way I just can't figure out what I am doing wrong. Is my way of solving this problem correct?

KEØM

Last edited by a moderator: Apr 24, 2017
2. Jun 19, 2009

KEØM

Here is a scanned picture of my work.

KEØM

Attached Files:

• Statics Problem#1.pdf
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3. Jun 20, 2009

djeitnstine

Dont forget that cable AD is not just one cable but has to be treated as 2 because it is looped through the pulley. your expression for summation of forces will have a factor of 2 to account for forces in AE and AD i.e. $$\vec{T_{AC}} + \vec{T_{AB}} + 2 \vec{T_{AD}} + \vec{T_{AE}} = 0$$

(The pdf is pending approval so we cannot view it. Upload it on a filesharing site to show us much quicker)

Since $$T_{AE} = T_{AD}$$ you will have some cancellation of components I believe

Edit: This was just the quick and dirty, don't forget to account for the directions of the components etc...

Last edited: Jun 20, 2009
4. Jun 20, 2009

KEØM

Thank you for the reply. I did not account for the second cable but I don't understand why you would write it as $$2\vec{T_{AD}} 2\vec{T_{AE}}$$.

I have a link to my solution on a file sharing website. I don't have my answers to the problem on this sheet though and I also used $$\lambda$$ instead of U for my unit vector. One other thing is that the dimensions on my problem are different than that of the picture on the google books page.

http://jumalafiles.info/showfile2-91015519004423306881628597981475444/statics_problem_1.pdf" [Broken]

Last edited by a moderator: May 4, 2017
5. Jun 20, 2009

KEØM

Coordinates
A ( 0, -2.4, 0)
B ( -2.7, 0, -3.6)
C ( 0, 0, 1.8)
D ( 1.2, 0, -0.3)
E ( -2.7, 0, 1.2)

$$\Sigma\vec{F} = 0$$
$$\Sigma\vec{F} = \vec{T_{AC}} + \vec{T_{AE}} + \vec{T_{AB}} + \vec{T_{AD}} + \vec{W} = 0$$

$$\vec{T_{AC}} = T_{AC}\vec{U_{AC}}$$

$$\vec{U_{AC}} = \vec{AC}/AC = \frac{(0-0)\vec{i} + (0 - -2.4)\vec{j} + (1.8 - 0)\vec{k}}{\sqrt{(0)^2 + (2.4)^2 + (1.8)^2}}$$

$$\vec{T_{AC}} = 0T_{AC}\vec{i} + 0.8T_{AC}\vec{j} + 0.6T_{AC}\vec{k}$$

I continued this process for each cable in the sum. I then summed all of their components giving me three equations.

$$\Sigma\vec{F_{x}} = 0T_{AC} - 0.5294T_{AB} - 0.7903T_{AE} + 0.4444T_{AD} = 0$$

$$\Sigma\vec{F_{y}} = 0.8T_{AC} + 0.4706T_{AB} + 0.6305T_{AE} - 0.8889T_{AD} - W = 0$$

$$\Sigma\vec{F_{z}} = 0.6T_{AC} - 0.7056T_{AB} + 0.3152T_{AE} - 0.1111T_{AD} = 0$$

Then knowing that $$T_{AE} = T_{AD}$$ and that $$T_{AB} = 300N$$ I simplified the equation and solved for the unknowns.

Last edited: Jun 20, 2009
6. Jun 20, 2009

djeitnstine

Srry late night typo...will fix

7. Jun 20, 2009

djeitnstine

Ok so when you sum each component, you want to write the Forces in the x direction as $$T_{AC}C_1 + T_{AB}C_2 + T_{AE} C_3 + 2T_{AD} C_4$$ where $$C_n$$ is the coefficient of i'th component of that force. So lets consider the equality as mentioned before... Let $$T_{AE} = T_{AD}=T_{ADE}$$

We then have: $$T_{AC}C_1 + T_{AB}C_2 + T_{ADE}( C_3 + 2C_4)$$

8. Jun 20, 2009

KEØM

I understand, so really all I am missing is the 2 in front of the $$\vec{T_{AD}}$$ component right? I just didn't account for the the second cable coming around the pulley.

9. Jun 20, 2009

nvn

KEØM: First, you are dropping far too many significant digits in your first calculation, lamba_ac. Always maintain four significant digits throughout all your intermediate calculations. Secondly, your equations on p. 1 are correct except it appears you did not double-check your calculations. You calculate completely wrong values from correct formulas. Double-check your calculations.

10. Jun 20, 2009

KEØM

Thanks nvn. I will make sure to put in those digits and double check my calculations. Other than those mistakes and not accounting for the second cable am I correct?

11. Jun 20, 2009

djeitnstine

everything else looks fine...your equations are correct.

12. Jun 20, 2009

KEØM

Thanks again for all of your help djeitnstine and nvn.

KEØM