Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound = 
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds = kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Answer:
The circulation around the cylinder is 0.163 
Explanation:
Given :
Velocity of spinning cylinder 
Sea level density 
Sea level span 
Lift per unit circulation is given by,
Where 
circulation around cylinder
Therefore, the circulation around the cylinder is 0.163
Answer:
note:
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
