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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
The workdone is
Explanation:
From the question we are told that
The height of the cylinder is
The face Area is
The density of the cylinder is
Where is the density of freshwater which has a constant value
Now
Let the final height of the device under the water be
Let the initial volume underwater be
Let the initial height under water be
Let the final volume under water be
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
So
=>
Now the work done is mathematically represented as
Substituting values