# Solving Work Required to Pull Cable to Top of Building

• Enjoicube
In summary, the problem involves finding the work required to pull 10 ft of a 40 ft uniform cable weighing 60 lb to the top of a tall building. The solution involves dividing the problem into two parts: the first part accounting for the constant weight of the remaining 30 ft of cable and the second part accounting for the work required to pull the 10 ft. The teacher's solution involves integrating from 30 to 40 ft, which corresponds to the amount of rope that is hanging.
Enjoicube

## Homework Statement

A uniform cable hanging over the edge of a tall building is 40 ft long and weighs 60 lb. How much work is required to pull 10 ft of the cable to the top?

## Homework Equations

Density=3/2 ft/lb

## The Attempt at a Solution

Ok, the problem is this: I can solve this problem one way, but the way the teacher explains in class (the bounds in particular) is confussing.

Here is my solution: Since you are only going to pull 10 ft of the rope up, the 30 odd feet below this 10 feet will remain a constant, therefore, no integral required for this part.
30ft*3/2 lb/ft*10 ft=450 ft*lbs

So let's now deal with the ten feet you will have to pull up, first divide the 10 ft. into sub intervals dy thick, and let's say that at any given point y on the rope (the top of the building being at height 0), we have y ft to pull up till that segment reaches the top of the building. Therefore, the work to pull one of these slices becomes:
W:3/2 ft/lb * dy * y

so let's integrate 3y/2, from 0 to 10.
10
$$\int(3y/2dy)$$=75 ft*lbs
0
Now, my teachers solution: she says forget the first part of my solution and deal with the problem in one swoop by integrating:

40
$$\int(3y/2dy)$$=525ft*lbs
30

My question: WHERE did she get these bounds? and why?

Last edited:
Enjoicube said:
My question: WHERE did she get these bounds? and why?

I would guess those come from the amount of rope that's hanging. (Just like you're going from 0 to 10, she's going from 30 to 40.)

## What is the equation for calculating the work required to pull cable to the top of a building?

The equation for calculating the work required to pull cable to the top of a building is W = Fd, where W is the work (measured in joules), F is the force applied (measured in newtons), and d is the distance the cable is pulled (measured in meters).

## How do I determine the force needed to pull the cable to the top of a building?

The force needed to pull the cable to the top of a building can be calculated by dividing the weight of the cable by the sine of the angle between the cable and the building, or by using a force meter to measure the force applied.

## Can the work required to pull the cable to the top of a building be reduced?

Yes, the work required to pull the cable to the top of a building can be reduced by either decreasing the distance the cable needs to be pulled or by increasing the force applied. Using smoother pulleys or lubricants can also decrease the amount of work required.

## What factors can affect the work required to pull cable to the top of a building?

The work required to pull cable to the top of a building can be affected by the weight and length of the cable, the angle at which it is pulled, the friction between the cable and the pulleys, and any external forces such as wind or resistance from the building.

## Is there a maximum weight or distance that can be pulled to the top of a building?

There is no specific maximum weight or distance that can be pulled to the top of a building, as it depends on the strength of the cable, the force applied, and other factors mentioned above. However, it is important to ensure that the force applied is within a safe range to prevent any accidents or damage to equipment.

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