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17Calculus Integrals - Work - Weight Changing Problems Including Cables, Ropes and Leaking Bags

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These type of problems are interesting in that we need to take into account the weight of the cable or rope itself or the changing weight of the bag of sand or water or whatever is in the bag. In problems you've probably seen before, lifting a weight alone is just force times distance. But if you take into account the changing weight, with each small distance you lift, there is less weight. So the force required is not constant, meaning we need to use an integral to calculate work. So here is how we work these problems.

First of all, let me say that drawing diagrams for these problems so that you understand what is going on is critical. Otherwise it will be nearly impossible to solve these kinds of problems because of the many variations that are possible. We will cover the case where the force required to lift the object is changing. Some problems add static weights to the problem or raise the object beyond a point where the force is changing. These kind of problems are included in the practice problems.

Once you have drawn all the diagrams you need for the problem and you think you understand what is going on, you need to write an equation for the changing force. Some instructors set up an infinite sum then convert to an integral. Several of the practice problems do that too. However, I am going to show you another way to think about it that I think will help you. Let's look at a chain example.

Example

A 10ft chain that weighs 30lbs is initially laying on the ground. Calculate the work to lift one end of the chain 10ft into the air.

Solution

Let's think about what is going on. First, we pick up one end of the chain and start lifting. As we get further up in the air, the weight we are lifting is getting heavier and heavier until we have all 10ft hanging and we are holding 30lbs of chain.
Some instructors have you think about each link in the chain and how much work it takes to raise each link. I prefer to think about only the end link that I am holding and, depending on the height of that one link, it's weight is getting heavier. So at 1ft, I am holding 3lbs (30lbs/10ft * 1ft = 3lbs). As I raise it higher and higher, that one link is getting heavier and heavier, so it takes more work to lift the last foot of the chain than it does to raise it the first foot. Do you have the idea in your head?
So, at each position y, I need to exert a force of \(F = 30lbs/10ft \cdot y = 3y\) lbs. The quantity 30lbs/10ft = 3 lbs/ft is the density of the chain. The more chain there is to raise, the more force required.
So, if we go from \(y=0\) ft to \(y=10\) ft, the work required is \(\int_{0}^{10}{3y~dy} = \left[3y^2/2\right]_{0}^{10} = 3(100)/2=150\) lbs-ft

Final Answer

\(W=150\) lbs-ft

Practice

A cable that weighs 2 kg/meter is lifting a load of 50 kg that is initially at the bottom of a 75 meter shaft. How much work is required to lift the load 40 meters?

Problem Statement

A cable that weighs 2 kg/meter is lifting a load of 50 kg that is initially at the bottom of a 75 meter shaft. How much work is required to lift the load 40 meters?

Final Answer

Total Work = 62720 N-m

Problem Statement

A cable that weighs 2 kg/meter is lifting a load of 50 kg that is initially at the bottom of a 75 meter shaft. How much work is required to lift the load 40 meters?

Solution

For this problem, we separate the work into two parts, [1] the work to lift the 50 kg load and, [2] the work to lift the cable, then we add the two answers to get our final answer.
[1] Work Lifting The 50kg Load
Let's start with the easy one, the 50 kg load. Since the load is constant, the force is also constant, so the work is just force times distance, i.e. \( W = F \cdot d = (mg)d \). Since we are given the mass of 50kg, we need to multiply by gravity to get force. We will use \( g = 9.8 m/s^2 \). So \( W_{load} = (50 kg)(9.8 m/s^2)(40 m) = 19600 \) N-m.
[2] Work Lifting The Cable
Now we need to find the work associated with the cable. It helps to look at each end of the interval to get a feel for how to set up the equations. At the bottom of the shaft, we have 75m of cable, so the force is \( F = 75m(9.8 m/sec^2)(2 kg/m) \). When we have lifted the load 40 meters, then we have \( 75 - 40 = 35m \) of cable. So the force is \( F = 35m(9.8m/sec^2)(2 kg/m) \). So if we choose the range on y as \([0,40]\), our equation for the length of cable is \( l = 75 - y \). The equation for force is \( F = l(9.8)(2) \to F = (75-y)(9.8)(2) \). Now we are ready to set up the integral and evaluate.

\(\displaystyle{ W_{cable} = \int_0^{40}{ (75-y)(9.8)(2) ~dy} }\)

\(\displaystyle{ W_{cable} = 19.6 \int_0^{40}{ (75-y) ~dy} }\)

\(\displaystyle{ W_{cable} = 19.6 [ 75y - y^2/2 ]_0^{40} }\)

\(\displaystyle{ W_{cable} = 19.6 [ 75(40) - (40)^2/2 ] - 0 }\) \( = 43120 \)

The total work is \( W_{total} = W_{load} + W_{cable} = \) \( 43120 + 19600 = 62720\) N-m

Final Answer

Total Work = 62720 N-m

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A cable that weighs 1.5 kg/meter is attached to a bucket that weighs 75 kg. Initially there are 500 kg of grain in the bucket and as the bucket is raised 2 kg of grain leaks out of a hole in the bucket for every meter the bucket is raised. The bucket is 200 meters below a bridge. How much work is required to raise the bucket to the top of the bridge?

Problem Statement

A cable that weighs 1.5 kg/meter is attached to a bucket that weighs 75 kg. Initially there are 500 kg of grain in the bucket and as the bucket is raised 2 kg of grain leaks out of a hole in the bucket for every meter the bucket is raised. The bucket is 200 meters below a bridge. How much work is required to raise the bucket to the top of the bridge?

Final Answer

Total Work \( 882000 \) N-m

Problem Statement

A cable that weighs 1.5 kg/meter is attached to a bucket that weighs 75 kg. Initially there are 500 kg of grain in the bucket and as the bucket is raised 2 kg of grain leaks out of a hole in the bucket for every meter the bucket is raised. The bucket is 200 meters below a bridge. How much work is required to raise the bucket to the top of the bridge?

Solution

There are two things going on here. First, the bucket is losing grain and so the weight of the bucket is changing. Also, the cable is being reeled in and since less cable is being used as time progesses, the force required to raise it is reducing. Let's calculate the work associated with each individually and add the results to get the total work required.
[1] Work To Raise The Bucket
Initially, we have 500kg of grain. So the force is \(F = (500kg)(9.8m/sec^2) = 4900 N\). Notice that in the problem we are given mass. So we need to multiply by gravity to get the force.
As the bucket is raised, we lose 2kg of grain per meter. So the force after raising the bucket y meters is \(F = [500kg - (2kg/m)(y)](9.8m/sec^2) \) \( = 4900 - 19.6y \) Newtons.
The range on y is [0,200]. So now we can set up and evaluate our integral.

\(\displaystyle{ W_{bucket} = \int_0^{200}{ 4900 - 19.6y ~dy} }\)

\(\displaystyle{ W_{bucket} = [4900y - 19.6y^2/2]_0^{200} }\)

\(\displaystyle{ W_{bucket} = [4900(200) - 19.6(200)^2/2] - 0 }\) \( = 588000 \) N-m

[2] Work Associated With The Cable
Initially, we have to hold 200 meters of cable, so the force is \(F = (200m)(1.5kg/m)(9.8m/sec^2) = \) \( 2940 N\). As we raise the bucket, the cable mass is reducing at a rate of 1.5kg/m. When we have raised the bucket y meters, the force required to hold the cable is \(F= (200-y)(1.5)(9.8) \) Newtons. Now we can set up the integral and evaluate it.

\(\displaystyle{ W_{cable} = \int_0^{200}{ (200-y)(1.5)(9.8) ~dy} }\)

\(\displaystyle{ W_{cable} = 14.7 \int_0^{200}{ 200-y ~dy} }\)

\(\displaystyle{ W_{cable} = 14.7 [ 200y - y^2/2 ]_0^{200} }\)

\(\displaystyle{ W_{cable} = 14.7 [ 200(200) - (200)^2/2 ]-0 }\) \( = 294000 \) N-m

Total Work: \( W_{total} = W_{bucket} + W_{cable} =\) \( 588000 + 294000 = 882000 \) N-m

Final Answer

Total Work \( 882000 \) N-m

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Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.

Problem Statement

Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.

Final Answer

total work = 650,000 ft-lbs

Problem Statement

Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.

Solution

352 video

video by Krista King Math

Final Answer

total work = 650,000 ft-lbs

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A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull the entire rope to the top?

Problem Statement

A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull the entire rope to the top?

Solution

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A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull half of the rope to the top?

Problem Statement

A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull half of the rope to the top?

Solution

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What is the total work done to lift a 20ft chain that weighs 60lbs, so the top end is 30ft above the ground?

Problem Statement

What is the total work done to lift a 20ft chain that weighs 60lbs, so the top end is 30ft above the ground?

Solution

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A leaky bucket full of sand has a mass of 25 kg. The bucket is to be lifted to the top of a building 15 meters tall by a rope of negligible weight. Every meter the bucket is lifted results in a loss of 0.2 kg of sand. Find the work done lifting the bucket to the top of the building.

Problem Statement

A leaky bucket full of sand has a mass of 25 kg. The bucket is to be lifted to the top of a building 15 meters tall by a rope of negligible weight. Every meter the bucket is lifted results in a loss of 0.2 kg of sand. Find the work done lifting the bucket to the top of the building.

Final Answer

3454.5 Joules

Problem Statement

A leaky bucket full of sand has a mass of 25 kg. The bucket is to be lifted to the top of a building 15 meters tall by a rope of negligible weight. Every meter the bucket is lifted results in a loss of 0.2 kg of sand. Find the work done lifting the bucket to the top of the building.

Solution

He comes up with a proportion to use as the force that may be difficult to see. We would probably work it differently by saying that the bag starts out at \(144lbs\) but ends the trip at \(72lbs\) (half). We would draw a graph of a straight line (since the leak is a constant rate) between the points \((0,144)\) and \((18,72)\). The equation of that line is the force at any point x, giving us \(F(x)=144-(72)(x/18)\).

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Final Answer

3454.5 Joules

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We are lifting a leaky bag sand that initially weighs 144lbs. The bag is half empty at 18ft. Calculate the work required to lift the bag of sand initially on the ground to 18ft above the ground. Assume the leak is a constant rate.

Problem Statement

We are lifting a leaky bag sand that initially weighs 144lbs. The bag is half empty at 18ft. Calculate the work required to lift the bag of sand initially on the ground to 18ft above the ground. Assume the leak is a constant rate.

Solution

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A bucket with a mass of 30 kg when filled with sand needs to be lifted to the top of a 20 meter tall building with a rope of mass 0.2kg/m using 1 meter to secure to the bucket. Once the bucket reaches the top of the building it has a mass of 19 kg due to the sand leaking out at a constant rate. Find the work done lifting the bucket, sand and rope to the top of the building.

Problem Statement

A bucket with a mass of 30 kg when filled with sand needs to be lifted to the top of a 20 meter tall building with a rope of mass 0.2kg/m using 1 meter to secure to the bucket. Once the bucket reaches the top of the building it has a mass of 19 kg due to the sand leaking out at a constant rate. Find the work done lifting the bucket, sand and rope to the top of the building.

Final Answer

5233.2 Joules

Problem Statement

A bucket with a mass of 30 kg when filled with sand needs to be lifted to the top of a 20 meter tall building with a rope of mass 0.2kg/m using 1 meter to secure to the bucket. Once the bucket reaches the top of the building it has a mass of 19 kg due to the sand leaking out at a constant rate. Find the work done lifting the bucket, sand and rope to the top of the building.

Solution

1914 video

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Final Answer

5233.2 Joules

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A leaky 5 pound bucket is lifted 20 feet into the air at a constant speed. The rope weighs 0.3 lbs/ft. The bucket starts with 10 pounds of water and leaks at a constant rate. There are 5 pounds of water left when it reaches the top. How much work was done to raise the bucket?

Problem Statement

A leaky 5 pound bucket is lifted 20 feet into the air at a constant speed. The rope weighs 0.3 lbs/ft. The bucket starts with 10 pounds of water and leaks at a constant rate. There are 5 pounds of water left when it reaches the top. How much work was done to raise the bucket?

Solution

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Water is to be raised from a well 55 feet deep by means of a bucket attached to a rope that weighs 7 pounds/foot. When the bucket is full of water it weighs 30 pounds. Find the work done in raising the bucket to the top.

Problem Statement

Water is to be raised from a well 55 feet deep by means of a bucket attached to a rope that weighs 7 pounds/foot. When the bucket is full of water it weighs 30 pounds. Find the work done in raising the bucket to the top.

Solution

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A 200 pound weight is connected to a 20 foot rope hanging from a bridge. The rope weighs 2 lb/ft. Find the work done to lift the weight to the bridge.

Problem Statement

A 200 pound weight is connected to a 20 foot rope hanging from a bridge. The rope weighs 2 lb/ft. Find the work done to lift the weight to the bridge.

Solution

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\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

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\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

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\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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