A trough is 2 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x10 from x=−1 to x=1. The trough is full of water. Find the amount of work required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot. Your answer must include the correct units. (You may enter lbf or lb*ft for ft-lb.)
I got 5194.10 and 81.158 and both were wrong, I am so stuck I have been working on this question for over an hour!!!! and the assignment is due at midnight, please help!!!!
I do not know what
y = x10 means
Whatever the function is you must find the distance from the top (y =1) down to the centroid, which is the center of mass of the water. That is the distance you have to lift the water, call it h.
Then find the volume V , the area times 2 feet.
work = 62 lbs/ft^2 * volume in ft^3 * h
typo, 62 lbs/ft^3
To find the amount of work required to empty the trough, we need to calculate the total weight of the water and then convert it into work.
First, let's determine the volume of water in the trough. Since the trough has a cross-section shaped like the graph of y = x^2, we can use integration to find the volume.
The equation of the cross-section is y = x^10. However, this seems to be a typo in your question as x raised to the power of 10 would result in a very steep cross-section. Let's assume it was meant to be y = x^2 instead.
Using integration, the volume of a trough from x = a to x = b can be calculated as:
V = ∫[a,b] πy^2 dx
Given that the trough's vertical cross-section is shaped like y = x^2, and the trough length is 2 feet, we integrate from -1 to 1:
V = ∫[-1,1] π(x^2)^2 dx
= ∫[-1,1] πx^4 dx
Integrating:
V = π/5 [x^5] from -1 to 1
= π/5 (1^5 - (-1)^5)
= 2π/5
Since the trough is filled with water, the volume of water is the same as the volume of the trough.
Next, we need to calculate the weight of the water. The weight of water is 62 pounds per cubic foot. Therefore, the weight W is given by:
W = V * 62
W = (2π/5) * 62
W = 124π/5 pounds
Finally, we can convert the weight of the water into work. The work W is given by the equation:
Work = Weight * Height
The height of the trough is 1 foot, so the work required to empty the trough is:
Work = (124π/5) * 1
Work = 124π/5 foot-pounds
Therefore, the amount of work required to empty the trough by pumping the water over the top is (124π/5) foot-pounds.
To find the amount of work required to empty the trough, we need to calculate the weight of the water in the trough and then convert it to work.
Step 1: Calculate the volume of the trough
The vertical cross-section of the trough is shaped like the graph of y = x^10 from x = -1 to x = 1. This means that the width of the trough at any given point x is given by the function y = x^10.
To find the width of the trough, we need to find the difference between the y values at x = -1 and x = 1.
Width = y(1) - y(-1) = (1^10) - (-1^10) = 2
The length of the trough is given as 2 feet, and the height is given as 1 foot.
The volume of the trough can be calculated as:
Volume = Length * Width * Height = 2 * 2 * 1 = 4 cubic feet
Step 2: Calculate the weight of the water in the trough
The weight of water is given as 62 pounds per cubic foot.
Weight = Volume * Weight per cubic foot = 4 * 62 = 248 pounds.
Step 3: Convert weight to work
To convert the weight to work, we need to multiply it by the height from which the water is being lifted. In this case, the water is being pumped over the top of the trough, so the height is 1 foot.
Work = Weight * Height = 248 * 1 = 248 ft-lb (pounds multiplied by feet)
Therefore, the amount of work required to empty the trough by pumping the water over the top is 248 ft-lb.