# Tension in Ropes: Solving Forces on Trunk

• sp3sp2sp
In summary: I will happily assist but need to see the diagram ... please scan and attach.In summary, the conversation discusses the problem of two students lifting a heavy trunk using two ropes attached to a small ring at the center of the trunk. The gravitational force acting on the trunk is represented by FG and the students are pulling the trunk at a constant velocity v⃗. The question is about the tension in each rope, which is given by T = (FG)/(2cos(θ)). The conversation also includes a discussion about the x and y components of the tensions, but it is determined that only the y-components need to be considered as the ropes are pulling straight up. The reason for dividing by cos(
sp3sp2sp

## Homework Statement

This is not HW problem specific, but more of a lack of understanding for an equation in answer.
A pair of students are lifting a heavy trunk on move-in day. (Figure 1) Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity v⃗ . Each rope makes an angle θ with respect to the vertical. The gravitational force acting on the trunk has magnitude FG.

(the figure is just 2 people pulling rope on either side of the trunk)
What is the force of tension in each rope? The answer is T = (FG) / (2cos(θ))

## The Attempt at a Solution

I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 and T_1x + T_x2 = 0
But I am still not understanding how the tension in each rope, T = (FG) / (2cos(θ))? Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help

sp3sp2sp said:
mg + T_y = 0
There are two sections of rope, each exerting Ty upwards on the ring.
sp3sp2sp said:
Dont we have to factor in the x-components of the tension also?
That has been done by dividing by cos(θ).

sp3sp2sp said:

## The Attempt at a Solution

I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 ...

That should be mg + 2T_y = 0

Then substitute T_y = TCos(θ).

That's all you need to do.

and T_1x + T_x2 = 0

True but not needed to solve for T.

Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help

The x-components are orthogonal (aka 90 degrees) to mg. So the x-components of the tension doesn't carry any of the force mg.

sp3sp2sp said:
Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?
The x-components must be equal and opposite because the crate is rising straight up as stated by the problem. They do not affect the motion and therefore are not mentioned.

P.S. I do not see a figure. Does everybody else see it?

sp3sp2sp said:

## Homework Statement

This is not HW problem specific, but more of a lack of understanding for an equation in answer.
A pair of students are lifting a heavy trunk on move-in day. (Figure 1) Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity v⃗ . Each rope makes an angle θ with respect to the vertical. The gravitational force acting on the trunk has magnitude FG.

(the figure is just 2 people pulling rope on either side of the trunk)
What is the force of tension in each rope? The answer is T = (FG) / (2cos(θ))

## The Attempt at a Solution

I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 and T_1x + T_x2 = 0
But I am still not understanding how the tension in each rope, T = (FG) / (2cos(θ))? Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help
You know the tensions in the two ropes. So what are the components of these two tensions in the x direction? Are they equal in magnitude and opposite in direction?

OK I now get that they are pulling straight up, so its all y-components, but why are dividing by cos θ? The question asks about force in one rope so we divide by 2, but why are we dividing by cos θ? thanks for any more help

At the moment all the replies above, including mine, assume the ropes are pulling up at an angle (eg the ropes are NOT pulling vertically). On that basis the vertical component is TCos(theta) per rope or 2TCos(theta) in total. Where T is the tension in one rope. When you rearrange that you get 2Cos(theta) on the bottom. Simple geometry.

However if your diagram is different then our replies might all be wrong.

sp3sp2sp said:
OK I now get that they are pulling straight up, so its all y-components, but why are dividing by cos θ? The question asks about force in one rope so we divide by 2, but why are we dividing by cos θ? thanks for any more help
Here are the force balances in the vertical and horizontal directions, respectively:
$$T\cos{\theta}+T\cos{\theta}=mg$$
$$T\sin{\theta}-T\sin{\theta}=0$$
Please tell me if these balances make sense to you. Have you learned how to resolve a vector into horizontal and vertical components?

## What is tension in ropes?

Tension in ropes is the force exerted by a rope when it is pulled tight. It is a reaction force that occurs when an object is being pulled or suspended by a rope.

## How is tension in ropes measured?

Tension in ropes is typically measured in units of force, such as newtons or pounds, using a tension meter or by calculating the forces acting on the rope.

## What factors affect tension in ropes?

The tension in a rope is affected by the magnitude and direction of the applied force, the weight of the object being pulled or suspended, and the properties of the rope itself, such as its length and elasticity.

## Why is it important to solve for forces on the trunk in relation to tension in ropes?

Solving for forces on the trunk allows us to understand the distribution of forces acting on an object that is being supported or pulled by ropes. This information can help us determine the stability and safety of the object and its supporting structure.

## What are some real-world applications of understanding tension in ropes?

Understanding tension in ropes is crucial in a variety of industries, such as construction and engineering, where ropes are used to support heavy objects or structures. It is also important in activities like rock climbing and sailing, where ropes are used for support and navigation.

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