# 3D Physics problem(Forces and Tension)

1. May 24, 2015

### masterchiefo

1. The problem statement, all variables and given/known data

metal piece = 300N
As this metal piece is in balance, what is the tension in each cable?
Picture of the problem in attachment.
2. Relevant equations

3. The attempt at a solution

W = 300*[0 0 1]

D to A:
vtda = tda [sin(37), sin(37)*sin(90), -cos(37)]

D to B:
vtdb = tdb [sin(37)*sin(90), sin(37), - cos(37)]

D to C
vtdc = tdc [cos(30)*sin(37), -sin(30)*sin(37), -cos(37)]

solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}

I have been trying to find the mistake for 3 hours but I really cant find it :/
When I solve with my TI calculator I get a "false"

thank you very much

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2. May 24, 2015

### PWiz

Let the tensions in the 3 cables be $T_1$, $T_2$ and $T_3$ .
It is clear that you'll need a system of 3 equations to solve this problem. Luckily, the coordinate system is set-up in such a way that the z axis is the perpendicular from the surface (it is an assumption that I'm making after looking at the diagram, and I'm 99% sure that its the actual case).

You already know that the metal piece is in equilibrium. This means that the resultant of the x and y components of the tensions equals 0, whereas the resultant of the z components of the tensions equals to the weight of the piece. Can you set up the required 3 equations?

One tip: always set up the algebraic/vector equation first, and substitute the numbers later. That way, it's always easier to spot any mistakes.

Last edited: May 24, 2015
3. May 24, 2015

### BvU

Hi chief,

1. What is this funny language you are writing in ?
2. What are the units of angle in your TI thingy? degrees or radians ?
3. 37 is approximate, 3/5 for the sine is exact. idem 4/5 for the cos.
4. z is downwards, your coordinate system is left-handed ? Bad habit!
5. In the x-direction vtda and vtdc pull in different directions. I don't see that in your expressions
6. sin(90) = 0. Easier to use the shorter form ? Less insightful, I admit.

 slow typist
@PWiz: solve(...) appears to be solving (almost) exactly the three equations you describe. El Jefe does the right thing, he/she doesn't do the thing completely right, though.

4. May 24, 2015

### masterchiefo

its in degree
I am using directional cosine.

Also is 300*[0 0 1] correct since Z is downward?

5. May 24, 2015

### theodoros.mihos

Yes but ropes are not symmetric around, so z components are not equals.

6. May 24, 2015

### masterchiefo

z is downward? how would they be different? only x and y has to be different for each rope.

7. May 24, 2015

### theodoros.mihos

see again angles and write the correct vectors.

8. May 24, 2015

### masterchiefo

sorry I really dont understand how Z is different to each of the rope, z is same direction for all of them since its downward. and the angle between each rope and the middle line is equal to every rope.

9. May 24, 2015

### BvU

Dear chief,
You are not correct. Z components follow once you found the components in the xy plane.

To get going. do pay attention to my remark #5.

10. May 24, 2015

### theodoros.mihos

to make zero x and y components the tension on C must be larger than other two.

11. May 24, 2015

### masterchiefo

yes I fixed that on my calculator put still get false as an answer.

12. May 24, 2015

### BvU

Theo, chief has allowed the tensions to be different!

13. May 24, 2015

### masterchiefo

W = 300*[0 0 1]

vtda = tda [sin(37), sin(37)*sin(90), cos(37)]

vtdb = tdb [sin(37)*cos(90), sin(37), cos(37)]

vtdc = tdc [-cos(30)*sin(37), -cos(60)*sin(37), cos(37)]

solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}

14. May 24, 2015

### PWiz

Your writing notation is rather difficult to follow (it's almost as if you're posting your calculator input). Can you change this?

15. May 24, 2015

### masterchiefo

thats how we have learned to write notation on paper, I am not sure how to make this more easy to understand :(

16. May 24, 2015

### BvU

Looks good to me. I have no problem solving by hand.
Don't know the proper TI incantation language.

17. May 24, 2015

### PWiz

Well you could perhaps call the origin of the coordinate system O and the bottom of the hook D.
Then you can describe angles much more easily. For example, cos (DAO) = 0.6 , etc.
I don't know if it's just me, but this is how I learned geometric description in middle school

18. May 24, 2015

### BvU

solve([0 0 0]=300*[0 0 1]+vtda+vtdb+vtdb), {tda,tdb,tdc} instead of

solve([0 0 0]=300*[0 0 1])+vtda+vtdb+vtdb, {tda,tdb,tdc}

19. May 24, 2015

### BvU

Oh, and if W is pulling down, it's only fair to have vtda, vtdb and vtdc pull upwards. Since 4/5 is positive, I suggest

solve([0 0 0]=300*[0 0 $\bf -$1]+vtda+vtdb+vtdb), {tda,tdb,tdc}

20. May 24, 2015