# Statics:Find the Tension in a rope holding a boom

• fishturtle1
In summary, the problem involves a boom attached to a frictionless pivot, with a weight of 2450 N and an unknown tension in the guy wire. The boom is not uniform and has a center of gravity at 35.5% of its length from the pivot. The sum of torques equation is used to find the tension, with three forces acting on the boom: its weight, the weight of the box, and the tension in the rope. A mistake was made by using sin instead of cos in the calculation of torques, resulting in an incorrect answer. The correct horizontal distance from the axis of rotation to the line of action of the force must be used in the equation. Reorienting the boom horizontally
fishturtle1

## Homework Statement

The boom in the figure below (Figure 1) weighs 2450 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35.5 % of its length.
Find the tension of the guy wire.

Στ=0
∑F=0
τ=F*Dsin(θ)

## The Attempt at a Solution

I made a free body diagram where the boom is the object and i drew three forces acting on it.

The three forces acting on the boom are weight of the boom, weight of the box, and the tension in the rope.

I then tried to use the sum of the torques to find tension:

Στ=0=wboomsin(60°)(.355L) + wboxsin(60°)L - T1sin(30°)L

then rearranged

wboomsin(60°)(.335L) + wboxsin(60°)L = T1sin(30°)L

then i canceled out the L variable, and plugged in and simplified and got this:

T1 = 10,167 N => 10200 N.

This is wrong.
I peaked at the yahoo answer and they used cos(θ) instead of sin(θ) which does not make sense to make because then the force calculated would move the boom linearly rather than creating a torque.

Last edited:
fishturtle1 said:

## Homework Statement

The boom in the figure below (Figure 1) weighs 2450 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35.5 % of its length.

Στ=0
∑F=0
τ=F*Dsin(θ)

## The Attempt at a Solution

I made a free body diagram where the boom is the object and i drew three forces acting on it.

The three forces acting on the boom are weight of the boom, weight of the box, and the tension in the rope.

I then tried to use the sum of the torques to find tension:

Στ=0=wboomsin(60°)(.355L) + wboxsin(60°)L - T1sin(30°)L

then rearranged

wboomsin(60°)(.335L) + wboxsin(60°)L = T1sin(30°)L

then i canceled out the L variable, and plugged in and simplified and got this:

T1 = 10,167 N => 10200 N.

This is wrong.
I peaked at the yahoo answer and they used cos(θ) instead of sin(θ) which does not make sense to make because then the force calculated would move the boom linearly rather than creating a torque.
Post the entire problem statement.

fishturtle1 said:
I peaked at the yahoo answer and they used cos(θ) instead of sin(θ) which does not make sense to make because then the force calculated would move the boom linearly rather than creating a torque.
For the torque of a vertical force around the axis you need the horizontal distance from the axis to the line of action of the force.
If L is the hypotenuse and the angle to the horizontal is θ, what is that horizontal distance?

haruspex said:
For the torque of a vertical force around the axis you need the horizontal distance from the axis to the line of action of the force.
If L is the hypotenuse and the angle to the horizontal is θ, what is that horizontal distance?
the horizontal distance would be Lcos(θ), but why are we treating L as a hypotenuse?

In my diagram I reoriented the boom horizontal and then i did sinθ to find the Fy of each force and then calculated the torques that way.

I am confused when you say "you need a horizontal distance from the axis of the line of action of the force"? So does this mean I need to find the distance from the axis of rotation to the line where the force is being applied?

## 1. What is the definition of statics?

Statics is a branch of mechanics that deals with the study of forces and their effects on stationary objects.

## 2. How do you find the tension in a rope holding a boom?

To find the tension in a rope holding a boom, you need to first determine the forces acting on the boom. This can be done by drawing a free-body diagram and identifying all the external forces acting on the boom. Then, using Newton's second law of motion, you can equate the sum of all the forces in the vertical direction to the weight of the boom. This will give you the tension in the rope.

## 3. What is the importance of finding the tension in a rope holding a boom?

Knowing the tension in a rope holding a boom is important for ensuring the stability and safety of the structure. It can also help in determining the strength and durability of the rope and its ability to withstand the weight of the boom.

## 4. What factors affect the tension in a rope holding a boom?

The tension in a rope holding a boom is affected by several factors such as the weight of the boom, the angle at which the rope is pulling, the properties of the rope (e.g. elasticity), and any external forces acting on the boom (e.g. wind).

## 5. Can the tension in a rope holding a boom be negative?

No, the tension in a rope holding a boom cannot be negative. Tension is a force that always acts away from the object it is pulling and therefore can only have a positive value. If the calculated tension is negative, it means that the direction of the force is opposite to what was assumed, and the magnitude should be taken as positive.

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