142p23s11solution[1]

# 10 17 0 lbs ft ft ft lbs ft distance y 3 dy

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Unformatted text preview: height hi i =1 n ≈ ∑ ( 7 − hi ) ⋅ δ ⋅ π 24 h ⋅ 22 ∆ 123 14 3 i =1 force 7 ( ft ⋅ lbs ft 3 ⋅ ft 3 = ft ⋅ lbs ) distance 5 n ≈ ∑ 200π ( 7 − hi ) ∆h hi i =1 work = 200π ∫ 5 0 ( 7 − h ) dh = 4500π ft ⋅ lbs n 2.9 work ≈ ∑ work to raise the chain, with length yi + 3 ft, vertically ∆y ft i =1 n ≈ ∑ δ ( yi + 3) ⋅ ∆y 1 24 { 43 i =1 force (weight) work = 5∫ 2.10 17 0 ( lbs ⋅ ft ⋅ ft = ft ⋅ lbs ) ft distance ( y + 3) dy = 997.5 ft ⋅ lbs ∆h ∆F ∆F ft lbs bs = 1 sec and = −20 sec . So, = −20 lft . ∆t ∆t ∆h n work ≈ ∑ work to raise the bucket, at height hi ft, vertically ∆h ft i =1 n ≈ ∑ (1000 − 20hi ) ⋅ ∆h 14 244 distance 4 3{ i =1 ( ( lbs − lbs ft ⋅ ft ) ⋅ ft = ft ⋅ lbs ) force (weight) work = ∫ 2.11 (a) (b) 30 0 (1000 − 20h ) dh = 21, 000 ft ⋅ lbs The center of the ellipse is 3 units to the right of the origin. So, we will find the equation of the ellipse with center (0,0) and then add 3 to th...
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