The in-situ dry density of a sand is 1.72Mg/m3. The maximum and minimum drydensities, determined by standard laboratory tests, a
1 answer:
Answer:
Relative density = 0.7 or 70%
Explanation:
The following information was provided by this question
Pd = 1.72mg/mg³
Pd max = 1.81 mg/mg³
Pd min = 1.54 mg/mg³
We substitute into the formula. This formula is contained in the attachment.
[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]
= 0.649350 - 0.581395 / 0.649350 - 0.552486
= 0.067955/0.096864
= 0.7015
= 0.7
The relative density is Therefore 0.7 or 70% when converted to percentage
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Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
