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$\begingroup$ @Xyzar I think the key to understanding this is that density in lb/ft3 is not mass per unit volume, but rather gravitational force per unit volume. So it already has the gravitational constant built in. In other words, the 62.4 lb/ft3 represents rho * g, not just rho in the second equation.
let acceleration = g = 12.176 ft/s^2 (this is the gravitational constant on Mars) let mass = m = 150 lbm. F = m x g = 150 lbm x 12.176 ft/s^2 = 1826.4 (lbm ft)/s^2. Once again, lets convert this quantity from lbm-ft /s2, to something we know (lbf) by using the relationship illustrated above:
This is the weight of the air assumed at sea-level... W = m × g W = m × g. W = 434.3 ⋅ lbm × 32.174 ⋅ ft s2 W = 434.3 ⋅ l b m × 32.174 ⋅ f t s 2. = 13976 ⋅ lbf = 13976 ⋅ l b f. Question: I find it hard to believe that in an average size room the air weighs a whopping 14, 000 ⋅ lbf 14, 000 ⋅ l b f. Did I do something wrong in ...
The 'g' term changes the mass in the density into a force, canceled out in the pressure term. When you use metric units, you don't need an extra extra term to account for the Newtons used in pressure and the kilograms used in density; that's what the gravity term is there for. P=S.G. (mercury) (Y) (h) 13.9 (144)=13.6 (62.4) (h)
Solving this for the maximum value of dry density results in $1919.67 \frac {kg}{m^3}$ at $12.57 \%$ moisture content. Note that you have a measured point that is above your interpolated maximum. This means you could either measure more points (to reduce uncertainty), choose a different fit, etc. But that is the general idea.
For a National Board Exam Review: What is the pressure 7000ft below the water surface of the ocean? Neglect compressibility. Answer is 512000psf $${ P_2 = P_1 + \\gamma h }$$ $${ 14.7 \\frac{l...
Mechanism of Soil Swelling Behavior. The simplified effective stress equation is as follows: σ. Where σ′ is the effective stress, σ is the total stress, and u is the pore water pressure. The above equation assumes a static condition. However, when the simplified effective stress equation is imbalanced, a dynamic condition occurs and the ...
I recently had an interview where the interviewer asked what the young's modulus would be like for an alloy like 5052 versus 6061. was a bit stumped on this one, saying that 5052 might be lower
$\begingroup$ Hi! Thank you so much for the thorough response! I'm however still a bit confused about why in the first image (3) equation forces and reactions on the right side of the hinge contributes to the moment about A; but for the triangular load the force cannot be concentrated at the centroid as a point force?
To get an idea of "correct' sizing, consult a table like the one by Crane for water. It gives minimum pipe sizes to accommodate a given flow at a given pressure drop. You make the runs large enough to accommodate a given combined flow and then size the outlet for the actual flow you want for some estimated pressure.