# 3D Physics problem(Forces and Tension)

• masterchiefo
In summary, after 3 hours of trying to find a mistake in a calculator, the issue was found to be a setting error. With the correct setting, the problem could be solved correctly.
masterchiefo

## Homework Statement

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

## 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 can't find it :/
When I solve with my TI calculator I get a "false"

thank you very much

#### Attachments

• savephy1.PNG
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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:
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.

masterchiefo
BvU said:
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.
its in degree
I am using directional cosine.

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

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

theodoros.mihos said:
Yes but ropes are not symmetric around, so z components are not equals.
z is downward? how would they be different? only x and y has to be different for each rope.

see again angles and write the correct vectors.

theodoros.mihos said:
see again angles and write the correct vectors.
sorry I really don't 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.

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.

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

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

Theo, chief has allowed the tensions to be different!

BvU said:
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.
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}

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

PWiz said:
Your writing notation is rather difficult to follow (it's almost as if you're posting your calculator input). Can you change this?
thats how we have learned to write notation on paper, I am not sure how to make this more easy to understand :(

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

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

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}

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}

BvU said:
That is suggestive.

Just curious what the chief has found

Hello everyone,

I appreciate everyone who helped, I thank you.
I found a very stupid mistake in my calculator setting. so everything is good now.

Still curious what the chief has found -- but much more reassured.
For the record, for statistics and for posteriority: what was it ? (And for ego -- or not --) : did it have anything to do with signs, brackets, or any of the items in post #3 ?

BvU said:
Still curious what the chief has found -- but much more reassured.
For the record, for statistics and for posteriority: what was it ? (And for ego -- or not --) : did it have anything to do with signs, brackets, or any of the items in post #3 ?
Its a very weird bug, but when I take the solve function from my calc in the librairy, it gives me FALSE, but when I write solve with the keypad...it works :O

I don't mean the bug, I mean the tensions
And for the solve incantation, which one was the correct one ?
And for the signs: how did they end up ?
Do you also see how to solve this without a calculator (just in case it let's you down again) ?

BvU said:
I don't mean the bug, I mean the tensions
And for the solve incantation, which one was the correct one ?
And for the signs: how did they end up ?
Do you also see how to solve this without a calculator (just in case it let's you down again) ?
tda: -137.26
tdb: -79.2468
tdc: -158.494

yes I figured how to do this by hand.

Well done, thanks!

## 1. What is tension in a 3D physics problem?

Tension is a force that is exerted by a string, rope, or cable on an object when it is pulled tight. It is a reaction force to the weight of the object and is always directed along the length of the string.

## 2. How do you calculate the tension in a 3D physics problem?

The tension in a 3D physics problem can be calculated using the equation T=mg+ma, where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object.

## 3. Can tension be negative in a 3D physics problem?

No, tension cannot be negative in a 3D physics problem. It is always a positive force because it is a reaction force to the weight of the object, which is always positive.

## 4. How does tension affect the motion of an object in a 3D physics problem?

Tension can affect the motion of an object in a 3D physics problem by either accelerating or decelerating the object. If the tension is greater than the weight of the object, it will accelerate the object in the direction of the tension. If the tension is less than the weight of the object, it will decelerate the object in the opposite direction.

## 5. How does the angle of a rope affect the tension in a 3D physics problem?

The angle of a rope affects the tension in a 3D physics problem by changing the direction of the tension force. The tension force will be greater if the angle is closer to 90 degrees and will be smaller if the angle is closer to 0 degrees.

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