# Solve Tension in Rope Homework | Pi, Area & Density

• Noreturn
In summary: But not for the 2 long ropes!No, once you have the right mass calculate the weight. Then follow what Buffy has been saying about a FBD.Edit: Actually yes, dividing by 4 gets you the tension in the 4 short vertical ropes. But not for the 2 long ropes!So I think I am close. I got the volume which is 2320kg. Then I divided that by 4 and got 580kg for each rope. Then I multiplied that by the gravity which is 9.8 to get 5684N. Then I divided that by 2 to get 2842N for the long ropes. So I got 2842N for the long ropes and 580kg
Noreturn

## Homework Statement

So it's given the pipe has a inside diameter of 60cm and outside diameter of 70cm. the two ropes AC and AB are separated by a spreader bar. Wants us to find tension in the ropes. Also give is the density of concrete which is 2320kg.

pi(r)^2*L=Area

## The Attempt at a Solution

pi(.7-.6m)^2*2.5*2320=23215Newtons

Can we just then divide that answer by sqrt of 2 because the ropes are equal? If no that is where I get lost is how to do the other part of the math.

Thanks!

Hi Noreturn, congrats on the first post !.

Noreturn said:
pi(r)^2*L=Area
##\pi r^2l## =volume

Noreturn said:
Can we just then divide that answer by sqrt of 2 because the ropes are equal? If no that is where I get lost is how to do the other part of the math.
Do:
1) : - Make a FBD.

2) :- Take net force upwards and equate it to net force downwards. For this you have to take the component of all the tension forces.

If I interpreted the diagram correctly then there are four tension forces.

3) :- find the tension. YOU ARE DONE !

So I guess I found the volume which I can use to find the density of the concrete pipe. Then can I dive that by 4 since all four ropes have the same angle and should disperse the force evenly?

So 23215/4=5803.75 or 5804N in each rope

Noreturn said:
I can use to find the density of the concrete pipe.
I think you mean mass of the pipe.
Noreturn said:
hen can I dive that by 4 since all four ropes have the same angle and should disperse the force evenly?
Yes they will disperse evenly but only dividing by 4 does not work as weight and the tension force are not along the same line. Tension makes and angle with the weight. you need to account for it also.

Buffu said:
I think you mean mass of the pipe.

Yes they will disperse evenly but only dividing by 4 does not work as weight and the tension force are not along the same line. Tension makes and angle with the weight. you need to account for it also.

So we don't know the length of AC or BC. So it would be something along the lines of

Sin(45)+sin(45)+mass=0 right? Or what is next? Or teacher isn't very good at explaining these, he didn't even teach us these. Basically a engineering class to show us what we will be doing in engineering but didn't explain how to do any of the problems.

Noreturn said:
So we don't know the length of AC or BC. So it would be something along the lines of

Sin(45)+sin(45)+mass=0 right? Or what is next? Or teacher isn't very good at explaining these, he didn't even teach us these. Basically a engineering class to show us what we will be doing in engineering but didn't explain how to do any of the problems.
Ok answer this question first :-

What is the missing side ?
Hint :- use trignometry.

Buffu said:
Ok answer this question first :-

View attachment 109970
What is the missing side ?
Hint :- use trignometry.
cos(45)=?/T

Using Soh Cah Toa

Noreturn said:
cos(45)=?/T

Using Soh Cah Toa
So, ##? = T cos(45)##.

What is x here in terms of T ? (Sorry for the bad editing, the angle is 45)

Oh, then the T will be what I said earlier. So 5804=Tsin45

So t=8208.1N or 8208N

Perhaps go back a step..

Noreturn said:
So it's given the pipe has a inside diameter of 60cm and outside diameter of 70cm

Noreturn said:
pi(.7-.6m)^2*2.5*2320=23215Newtons

How do you get the area of one end to be π(D-d)2?

Noreturn said:
Oh, then the T will be what I said earlier
Yes correct, assuming your calculations are correct.

CWatters said:
Perhaps go back a step..How do you get the area of one end to be π(D-d)2?

Well it would be outer diameter-inner diameter * length. Or in other words the thickness of the pipe is 10cm or .1m

Is that how you would get the mass of the pipe? the divide that by 4 then plug that in for ?

Noreturn said:
Well it would be outer diameter-inner diameter * length. Or in other words the thickness of the pipe is 10cm or .1m
The area of an annulus is the area of the outer circle minus the area of the inner circle...eg...Pi*352 - Pi*302

Then multiply that by the length to get the volume.

Noreturn said:
Is that how you would get the mass of the pipe? the divide that by 4 then plug that in for ?

No, once you have the right mass calculate the weight. Then follow what Buffy has been saying about a FBD.

Edit: Actually yes, dividing by 4 gets you the tension in the 4 short vertical ropes.

Last edited:

## 1. What is the formula for solving tension in a rope?

The formula for solving tension in a rope is T = (m x g) + (a x m), where T is tension, m is mass, g is gravitational acceleration (9.8 m/s²), and a is acceleration.

## 2. How does the value of pi factor into solving tension in a rope?

The value of pi (π) is used in the formula for calculating the area of a circle, which is often needed in problems involving tension in a rope. This is because the rope is often wrapped around a pulley or other circular object, and the area of this circle is needed to determine the angle of the rope and its effect on tension.

## 3. What role does the area of the rope play in solving tension?

The area of the rope affects the amount of tension it can withstand. A thicker rope will have a larger cross-sectional area, allowing it to handle more tension without breaking. This is why thicker ropes are often used for heavy-duty tasks such as lifting heavy objects.

## 4. How does the density of the rope impact tension calculations?

The density of a rope is directly related to its mass. A rope with a higher density will have a greater mass, and therefore will require more tension to lift or support. This is important to consider when choosing the right rope for a specific task.

## 5. Are there any common mistakes to avoid when solving tension problems involving ropes?

One common mistake is forgetting to consider the direction of the forces acting on the rope. The tension in a rope will be different depending on whether it is being pulled or lifted. It is also important to use consistent units throughout the calculations and to double-check all conversions.

• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
998
• Introductory Physics Homework Help
Replies
7
Views
14K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Classical Physics
Replies
33
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
6K
• Introductory Physics Homework Help
Replies
1
Views
2K