Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
The power that must be supplied to the motor is 136 hp
<u>Explanation:</u>
Given-
weight of the elevator, m = 1000 lb
Force on the table, F = 500 lb
Distance, s = 27 ft
Efficiency, ε = 0.65
Power = ?
According to the equation of motion:
F = ma
a = 16.1 ft/s²
We know,
To calculate the output power:
Pout = F. v
Pout = 3 (500) * 29.48
Pout = 44220 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = Pout / Pin
0.65 = 44220 / Pin
Pin = 68030.8 lb.ft/s
Pin = 68030.8 / 500 hp
= 136 hp
Therefore, the power that must be supplied to the motor is 136 hp
Answer:
2ib
Explanation:
if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8
there goes your answer.
Answer:
(a) Flow rate of vehicles = No of vehicles per mile * Speed
=No of cars per mile * Speed +No of trucks per mile * Speed
= 0.75*50*60 + 0.25*50*40
=2750 vehicles / hour
(b) Let Density of vehicles on grade = x
Density on flat * Speed =Density on grade * Speed
So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25
So, x= 57.89
So, Density is around 58 Vehicles per Mile.
(c) Percentage of truck by aerial photo = 25%
(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %
