A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
P is 525 lb, determine the values of the angle a and the required magnitude of the vertical force, F.
1 answer:
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
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Spiky Bob your answerer
The actual weight of the gas = apparent weight + weight.
The actual weight = + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight =
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = + W
Learn more about density here: brainly.com/question/406690