Respuesta :
Answer:
Riemann sum
W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)
Integral = W = ∫⁶⁰₀ 0.5x dx
Workdone in pulling the entire rope to the top of the building = 900 lb.ft
Riemann sum for pulling half the length of the rope to the top of the building
W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)
Integral = W = ∫⁶⁰₃₀ 0.5x dx
Work done in pulling half the rope to the top of the building = 675 lb.ft
Step-by-step explanation:
Using Riemann sum which is an estimation of area under a curve
The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.
The weight of rope in that part would be 0.5Δx.
Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx
So, the Riemann sum for this total work done would be
W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)
The Riemann sum can easily be translated to integral form.
In integral form, with the rope being 60 ft long, we have
W = ∫⁶⁰₀ 0.5x dx
W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft
b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft
Just like in (a)
But the Riemann sum will now be from the start of the curve, to it's middle
Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)
W = ∫⁶⁰₃₀ 0.5x dx
W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft
Hope this Helps!!!
The work done in pulling the rope can be found using the Riemann sum or
by integration.
The correct responses are;
a) The Riemann sum for the work done is; [tex]\displaystyle \underline{W = \limits \sum ^n _{i \to1} 0.5 \cdot (60 - y_i) \cdot \Delta y}[/tex]
- The work done in pulling the rope to the top of the building obtained by integration, is 900 ft.·lb.
b) The Riemann sum for the work done in pulling half the rope to the top of the building are;
- [tex]\displaystyle \underline{W_{top} = \limits \sum ^n _{i \to1} 0.5 \cdot (60 - y_i) \cdot \Delta y - \limits \sum ^n _{i \to1} 0.5 \cdot (30 - y_i) \cdot \Delta y}[/tex]
- [tex]\displaystyle \underline{W_{bottom} = \limits \sum ^n _{i \to1} 0.5 \cdot (30 - y_i) \cdot \Delta y}[/tex]
By evaluating the integral expressions the work done are;
- [tex]W_{top}[/tex] = 675 ft.·lb.
- [tex]W_{bottom}[/tex] = 225 ft.·lb.
Reasons:
a) The length a small segment = Δy
The depth of the segment from the wall's top = 60 - [tex]\mathbf{y_i}[/tex]
The weight of the segment = 0.5·Δy
Work done in pulling the segment to the top, [tex]W_i[/tex] = 0.5·Δy·(60 - [tex]\mathbf{y_i}[/tex])
The Riemann sum of the work done in pulling all small segments is therefore;
[tex]\displaystyle \underline{W = \limits \sum ^n _{i \to1} 0.5 \cdot (60 - y_i) \cdot \Delta y}[/tex]
Expressing the work done as an integral
The work done in pulling the rope to the top of the building is expressed as an integral is; [tex]\displaystyle \int\limits^{60}_0 {0.5 \cdot (60 - y)} \, dy[/tex]
Which gives;
[tex]\displaystyle \int\limits^{60}_0 {0.5 \cdot (60 - y)} \, dy = \mathbf{0.5 \cdot \left[60 \cdot y - \frac{y^2}{2^} \right]^{60}_0} = 0.5 \cdot \left[60 \times 60 - \frac{60^2}{2^} \right] = 900[/tex]
The work done in pulling the rope to the top of the building, W = 900 ft.·lb.
b) Work done in pulling the top half is, [tex]W_{top}[/tex] given as follows;
[tex]\displaystyle W_{top} = \mathbf{W- W_{bottom}}[/tex]
The bottom half is 30 feet long.
The work done in pulling the bottom half is therefore;
[tex]\displaystyle \underline{ W_{bottom} = \limits \sum ^n _{i \to1} 0.5 \cdot (30 - y_i) \cdot \Delta y}[/tex]
Therefore;
[tex]\displaystyle \underline{W_{top} = \limits \sum ^n _{i \to1} 0.5 \cdot (60 - y_i) \cdot \Delta y - \limits \sum ^n _{i \to1} 0.5 \cdot (30 - y_i) \cdot \Delta y}[/tex]
Expressing the work done in pulling the top half as an integral is therefore;
[tex]\displaystyle W_{top} = \mathbf{30 \times 0.5 \times 30 + \int\limits^{60}_{30} {0.5 \cdot (60 - y)} \, dy}[/tex]
[tex]\displaystyle \int\limits^{60}_{30} {0.5 \cdot (60 - y)} \, dy = \mathbf{0.5 \cdot \left[60 \cdot y - \frac{y^2}{2^} \right]^{60}_{30}}[/tex]
[tex]\displaystyle 0.5 \cdot \left[60 \cdot y - \frac{y^2}{2^} \right]^{60}_{30} = .5 \cdot \left( \left(60 \times 60 - \frac{60^2}{2^} \right) - \left(30 \times 30- \frac{30^2}{2^} \right) \right) = \mathbf{225}[/tex]
Therefore;
[tex]\displaystyle W_{top} = 30 \times 0.5 \times 30 + 225 = \mathbf{ 675}[/tex]
The work done in pulling the top half, [tex]W_{top}[/tex] = 675 ft.·lb.
The work done in pulling the bottom half is given as follows;
[tex]\displaystyle W_{bottom} = \int\limits^{30}_{0} {0.5 \cdot (30 - y)} \, dy = \mathbf{ 0.5 \cdot \left[30 \cdot y - \frac{y^2}{2^} \right]^{30}_{0} } = 225[/tex]
The work done in pulling the bottom half, [tex]W_{bottom}[/tex] = 225 ft.·lb.
Learn more about application of integration here:
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