How to Solve Tension Problems in Physics?

[Wanting to Learn]

Hello,
(I'm not sure if this is the right place to ask.) I'm taking AP Physics 1, having never taken physics before.
In class we are doing forces with Newton's Laws, and a little bit of free body diagrams (if I remember the name correctly).
For homework, we got a worksheet with questions about tension, each similar to the following:

A flowerpot of mass 4.20kg is hung above a window by three ropes, each making an angle of 15° with the vertical. What is the tension in each rope supporting the flowerpot?

We did not really go over this in class, and I couldn't find anything similar in the textbook or anywhere I've searched. Where might I find some resources (such as a website or video?) to learn this concept/how to do this type of problem? I also do not really understand how to draw free body diagrams, so could use a resource to explain that also.

I will probably ask for help on the specific questions later, but for now I'm not even sure how to start and would appreciate being pointed in the right direction to some resources for me to learn.
Thank you in advance.
(It is likely that I will not respond until sometime tomorrow since it is getting late and I have lots of other homework etc. to do.)

 

[CrzCz]

Do you have the answer? Is it 10.6N?

 

[Wanting to Learn]

Do you have the answer? Is it 10.6N?

I have no idea, that's why I'm trying to ask how to get started thinking about it.

 

[CrzCz]

I have no idea, that's why I'm trying to ask how to get started thinking about it.

Well this is what I did... First because an angle was given so you need to use sine. And then the tension is always the weight of the object attached to the end of the rope so... Not sure, hope to get some pros to reply haha...

 

[Wanting to Learn]

(img)
Well this is what I did... First because an angle was given so you need to use sine. And then the tension is always the weight of the object attached to the end of the rope so...

So what exactly did you do?
I understand as far as it looks like you did:
F = m•a
F = 4.2•9.8
4.3 for the mass and 9.8 for gravity? In which case why isn't it negative? (When gravity is supposed to be negative or not in a calculation always confuses me.)
F = 41.16
but why do you multiply that times the angle relative to vertical?
41.16•sin(15°) = 10.65

Also, whatever the answer is, will the tension be the same in each rope? Or different in the vertical rope as opposed to the 2 ropes that are on an angle?
For clarification, the question includes this picture:
(I hope the picture can be seen properly.)
Edit: I cannot figure out how to upload a picture/make a picture (saved on my own computer) to show/be in my post.
In any case, the given picture has the two slanted roped connecting to the same center point at the top with the vertical rope.

 

[CrzCz]

So what exactly did you do?
I understand as far as it looks like you did:
F = m•a
F = 4.2•9.8
4.3 for the mass and 9.8 for gravity? In which case why isn't it negative? (When gravity is supposed to be negative or not in a calculation always confuses me.)
F = 41.16
but why do you multiply that times the angle relative to vertical?
41.16•sin(15°) = 10.65

Also, whatever the answer is, will the tension be the same in each rope? Or different in the vertical rope as opposed to the 2 ropes that are on an angle?
For clarification, the question includes this picture:
(I hope the picture can be seen properly.)
Edit: I am trying to get the picture to show but might not be able to figure it out.
In any case, the given picture has the two slanted roped connecting to the same center point at the top with the vertical rope.

I don't think tension can ever be negative, even though the object has a negative force, tension is pulling upwards.

 

[CrzCz]

So what exactly did you do?
I understand as far as it looks like you did:
F = m•a
F = 4.2•9.8
4.3 for the mass and 9.8 for gravity? In which case why isn't it negative? (When gravity is supposed to be negative or not in a calculation always confuses me.)
F = 41.16
but why do you multiply that times the angle relative to vertical?
41.16•sin(15°) = 10.65

Also, whatever the answer is, will the tension be the same in each rope? Or different in the vertical rope as opposed to the 2 ropes that are on an angle?
For clarification, the question includes this picture:
(I hope the picture can be seen properly.)
Edit: I cannot figure out how to upload a picture/make a picture (saved on my own computer) to show/be in my post.
In any case, the given picture has the two slanted roped connecting to the same center point at the top with the vertical rope.

Actually, I guess always draw a diagram with vectors in it.. I am not sure about if tension is the same in all ropes but for now:

(Sorry I am also only a student too so..)

 

[haruspex]

View attachment 213061
Well this is what I did... First because an angle was given so you need to use sine. And then the tension is always the weight of the object attached to the end of the rope so... Not sure, hope to get some pros to reply haha...

Luckily, that is completely wrong. (Lucky because homework forum rules forbid posing full solutions; just point out errors and offer hints.)

@Wanting to Learn , if the tension in each rope is T, what vertical force does each rope exert on the pot?

 

[Wanting to Learn]

Thanks @haruspex

if the tension in each rope is T, what vertical force does each rope exert on the pot?

Well if I use F=ma, F equaling the Tension, then F = 4.2*9.8 = 41.16N for the total,
and then if each rope is holding an equal amount, 41.16/3 = 13.72N for one rope.
Is this at all correct? How does the fact that two of the ropes are at 15° and one is vertical effect it,
and then how do I get the horizontal tension (assuming I need figure that out?) to include in the final answer?

 

[haruspex]

Thanks @haruspex
Well if I use F=ma, F equaling the Tension, then F = 4.2*9.8 = 41.16N for the total,
and then if each rope is holding an equal amount, 41.16/3 = 13.72N for one rope.
Is this at all correct? How does the fact that two of the ropes are at 15° and one is vertical effect it,
and then how do I get the horizontal tension (assuming I need figure that out?) to include in the final answer?

You did not answer my question. "if the tension in each rope is T, what vertical force does each rope exert on the pot?". To answer that you do not need to think about the pot at all, nor the fact that there are three ropes.

 

[Wanting to Learn]

You did not answer my question. "if the tension in each rope is T, what vertical force does each rope exert on the pot?". To answer that you do not need to think about the pot at all, nor the fact that there are three ropes.

I'm confused/not sure what you mean. Are you referring to that the tension in the rope is the same force that it exerts on the flower pot?
Are there any websites or such resources that you know of that explain this general concept? I don't like asking questions like this when I really don't understand what I'm talking about.

 

[haruspex]

Are you referring to that the tension in the rope is the same force that it exerts on the flower pot?

Yes, if the tension on a rope is T then it is exerting a force T on the pot. But the rope is not vertical, so that force is not vertical. What is the vertical component of the force T?

 

[Wanting to Learn]

What do you mean? Is the vertical component having to do with the mass of the flowerpot and gravity? As in however much that is being pulled down the ropes are pulling up that same amount, which is the tension?

(I apologize for being very frustrated as I am trying and really not understanding.)

 

[haruspex]

What do you mean? Is the vertical component having to do with the mass of the flowerpot and gravity? As in however much that is being pulled down the ropes are pulling up that same amount, which is the tension?

(I apologize for being very frustrated as I am trying and really not understanding.)

Do you understand what is meant by a component of a vector? If a vector has magnitude 2 and points up and to the right of the origin, at an angle of 30 degrees to the X axis, can you say what is its component in the Y direction?

 

[Wanting to Learn]

I'm not exactly sure what is meant by a component of a vector. That confuses me because it sounds like it should be just one side of the triangle (when drawing it), but it is not actually that?

Would the y component be 1?

 

[haruspex]

I'm not exactly sure what is meant by a component of a vector. That confuses me because it sounds like it should be just one side of the triangle (when drawing it), but it is not actually that?
View attachment 213110
Would the y component be 1?

Right.
So back to the rope. The rope exerts a pull T at angle 15 degrees from vertical. What is the vertical component of the pull?

 

[Wanting to Learn]

By the way, here is the flowerpot diagram given.

(now going to attempt the math)

 

[Wanting to Learn]

So back to the rope. The rope exerts a pull T at angle 15 degrees from vertical. What is the vertical component of the pull?

So if the vertical component is how much force from the flowerpot/gravity
F = m•a
y = 4.2•9.8
y = 41.16
then to find the tension, geometry
cos(15°) = y / T
cos(15°) = 41.16 / T
T = 41.16/cos(15°)
T = 42.61
Is this correct?

(My work on paper)

 

[haruspex]

cos(15°) = y / T

Yes! Or to put it the way I asked the question, the vertical component of T is T cos(15°).

y = 4.2•9.8

Yes, except that now you have forgotten to divide by three.

 

[Wanting to Learn]

Okay, so would my final answer be
T = total tension
T/3
42.61/3
tension in each rope = 14.2N

 

[haruspex]

Okay, so would my final answer be
T = total tension
T/3
42.61/3
tension in each rope = 14.2N

Yes.

 

[Wanting to Learn]

@haruspex Thank you so much!

 

[DoctorPhysics]

Assuming its too late to help maybe I can shine the light on some things. Of there is 3 ropes, each is supporting a third of the weight or close to it. Picture you have one rope and then divide it up into the three. mg=T1+T2+T3.
Total mg=mg+mgsin15+mgsin15.
The vertical rope would support more than the other two.

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[DoctorPhysics]

Yes.

No it is not, the two ropes at an angle have different tensions than the vertical one.

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[DoctorPhysics]

Assuming its too late to help maybe I can shine the light on some things. Of there is 3 ropes, each is supporting a third of the weight or close to it. Picture you have one rope and then divide it up into the three. mg=T1+T2+T3.
Total mg=mg+mgsin15+mgsin15.
The vertical rope would support more than the other two.

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I apologize, it would be with cosine, not sine.

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[haruspex]

mg=T1+T2+T3.

Wrong. The mg is vertical, the three tensions are not.

mg=mg+mgsin15+mgsin15.

That cannot possibly be true. It leads to mg=0.

The vertical rope

There is no vertical rope. Reread the question.

 

[Wanting to Learn]

mg=T1+T2+T3.
Total mg=mg+mgsin15+mgsin15

What exactly is mg? It has come up in all of the questions we have done in class (since turning in that homework), but I can't figure out what it means.

 

[Wanting to Learn]

@DoctorPhysics

There is no vertical rope.

is correct. The original question said:

A flowerpot of mass 4.20kg is hung above a window by three ropes, each making an angle of 15° with the vertical. What is the tension in each rope supporting the flowerpot?

 

[DoctorPhysics]

What exactly is mg? It has come up in all of the questions we have done in class (since turning in that homework), but I can't figure out what it means.

That is the mass times the gravitational force (gravity). The product of these two magnitudes produces the weight. Note that mass is kilograms or slugs and weight is Newtons or pounds.

 

[Wanting to Learn]

That is the mass times the gravitational force (gravity). The product of these two magnitudes produces the weight. Note that mass is kilograms or slugs and weight is Newtons or pounds.

Okay, in class we use kg and N. So are Newtons meaning weight and force the same meaning/measurement? I didn't know that Newtons are a unit of weight, just that they are used for force measurements, such as F=ma.
That is another question I have is that in F=ma the units are Newtons = kg * m/s²
How does this work? As in for every other math/chem etc thing/equation units always (in my knowledge/class experience) have to cancel, (ex with PV=nRT in chemistry the units of R always ensure that all the units cancel).

Is a Newton equivalent to a kilogram meter divided by second squared (N=kg*m/s²)?

I guess that is the case, but why? I always wonder how units such as that come to be - is it just for convenience because those happen to be the units that are used, or is there a more specific reason?

(I realize this turned into a really long/different question and it's okay if you're not sure about these questions.)
(Maybe I should make the second half of this a new thread?)​