How to Solve Tension Problems in Physics?

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In summary, the conversation revolves around a student seeking help with a physics problem involving tension, Newton's Laws, and free body diagrams. They are struggling with how to approach the problem and are asking for resources to learn and understand the concepts involved. The conversation also includes a discussion about the correct method to solve the problem, with the student initially using the wrong approach.
  • #1
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.)
 
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  • #2
Do you have the answer? Is it 10.6N?
 
  • #3
CrzCz said:
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.
 
  • #4
Wanting to Learn said:
I have no idea, that's why I'm trying to ask how to get started thinking about it.

Image1508040041.879430.jpg

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...
 
  • #5
CrzCz said:
(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.
 
  • #6
Wanting to Learn said:
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.
 
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  • #7
Wanting to Learn said:
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:
Image1508041843.614970.jpg


(Sorry I am also only a student too so..)
 
  • #8
CrzCz said:
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?
 
  • #9
Thanks @haruspex
haruspex said:
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?
 
  • #10
Wanting to Learn said:
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.
 
  • #11
haruspex said:
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.
 
  • #12
Wanting to Learn said:
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?
 
  • #13
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.)
 
  • #14
Wanting to Learn said:
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?
 
  • #15
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?
img2.JPG

Would the y component be 1?
 
  • #16
Wanting to Learn said:
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?
 
  • #17
By the way, here is the flowerpot diagram given.
flowerpot-diagram.jpg

(now going to attempt the math)
 
  • #18
haruspex said:
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)
work-on-paper.JPG
 
  • #19
Wanting to Learn said:
cos(15°) = y / T
Yes! Or to put it the way I asked the question, the vertical component of T is T cos(15°).
Wanting to Learn said:
y = 4.2•9.8
Yes, except that now you have forgotten to divide by three.
 
  • #20
Okay, so would my final answer be
T = total tension
T/3
42.61/3
tension in each rope = 14.2N
 
Last edited by a moderator:
  • #21
Wanting to Learn said:
Okay, so would my final answer be
T = total tension
T/3
42.61/3
tension in each rope = 14.2N
Yes.
 
  • #22
  • #23
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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  • #24
haruspex said:
Yes.
No it is not, the two ropes at an angle have different tensions than the vertical one.

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  • #25
DoctorPhysics said:
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.

Sent from my SM-G930V using Physics Forums mobile app
I apologize, it would be with cosine, not sine.

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  • #26
DoctorPhysics said:
mg=T1+T2+T3.
Wrong. The mg is vertical, the three tensions are not.
DoctorPhysics said:
mg=mg+mgsin15+mgsin15.
That cannot possibly be true. It leads to mg=0.
DoctorPhysics said:
The vertical rope
There is no vertical rope. Reread the question.
 
  • #27
DoctorPhysics said:
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.
 
  • #28
@DoctorPhysics
haruspex said:
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?
 
  • #29
Wanting to Learn said:
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.
 
  • #30
DoctorPhysics said:
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?)​
 
  • #31
Wanting to Learn said:
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?)​
I would ask what the deep meaning of the origin of the netwon is but yes a Newton is a kilogram meter / second squared. Because weight is a force (measured in Newtons) it is the mass time the gravity (gravity can also be seen as an acceleration downward caused by the earth)
 
  • #32
DoctorPhysics said:
I would ask what the deep meaning of the origin of the netwon is but yes a Newton is a kilogram meter / second squared. Because weight is a force (measured in Newtons) it is the mass time the gravity (gravity can also be seen as an acceleration downward caused by the earth)
Okay, thank you very much.
 
  • #33
Wanting to Learn said:
weight and force the same meaning/measurement?
The weight of an object means the force exerted on it by gravitational attraction (of the planet etc. that it is near). That is why an object weighs less on the moon than on Earth even though its mass is the same. So yes, weight is a force.
The mass of an object is its resistance to being accelerated, i.e. mass is defined as force/acceleration.
That gravitational attraction is proportional to mass is a matter of observation, and is what led Einstein to his General Relativity theory. It was not inevitably true. One could imagine some type of matter which takes force to accelerate it but does not experience gravitational attraction.
Wanting to Learn said:
Is a Newton equivalent to a kilogram meter divided by second squared (N=kg*m/s2)?
I guess that is the case, but why
Because that is how the unit Newton is defined, for convenience. In the SI systems, units are, as far as possible, chosen such that there are no conversion factors needed. The current international standard (there are older ones) is known as MKS, standing for metres, kilograms, seconds. The definitions of Pascal, Newton, Joule and Watt all arise directly from these.
 

Related to How to Solve Tension Problems in Physics?

1. How do I calculate tension in a rope or string?

To calculate tension in a rope or string, you need to know the mass of the object being pulled, the acceleration due to gravity, and the angle of the rope or string. You can use the formula T = mgcosθ, where T is the tension, m is the mass, g is the acceleration due to gravity, and θ is the angle of the rope or string.

2. What is the difference between tension and compression?

Tension and compression are both types of forces that act on objects. Tension is a pulling force that stretches an object, while compression is a pushing force that compresses or squeezes an object. In physics, tension is often represented by a positive value, while compression is represented by a negative value.

3. How do I solve for tension in a system of multiple ropes or strings?

To solve for tension in a system of multiple ropes or strings, you can use the principle of superposition. This means that you can add up the individual tensions in each rope or string to find the total tension in the system. You can also use the equations of equilibrium, which state that the sum of all forces in the x-direction and the sum of all forces in the y-direction must equal zero.

4. What are some common examples of tension problems in physics?

Some common examples of tension problems in physics include calculating the tension in a cable supporting a bridge, determining the tension in a string holding up a hanging object, and finding the tension in a rope used to pull an object up a slope. Tension is also a key factor in understanding the mechanics of pulleys and other simple machines.

5. How does tension affect the motion of an object?

Tension can affect the motion of an object in different ways depending on the situation. In general, tension can either accelerate or decelerate an object, depending on the direction of the tension force. For example, if a rope is pulling an object upwards, the tension force will cause the object to accelerate upwards. However, if the rope is pulling downwards, the tension force will cause the object to decelerate or slow down.

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