A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786
kilograms per cubic meter. In each part below, assume that the tank is initially full, and that gravity is 9.8 meters per second squared. Your answers must include the correct units.
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?
1 answer:
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Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
