Thanks guys for you answer. So when looking at this from the ladder diagram... both travel limit switches are N/C and they will activate (open contacts) when the travel distance is reached what will cause the R2 or R3 to de-energise. Apart of activating, the switches P3 and P4 also work us...
Homework Statement
Hi, I'm currently studding a module on PLC and have a question on "what will happen if you do this ...".
Homework Equations
I don't have a problem with explaining operation of each rungs of the ladder diagram, I need some help on explaining the function of the reverse relay...
Hi, I'm currently studding a module on PLC and have a question on "what will happen if you do this ...". As I don't have a problem with explaining operation of each rungs of the ladder diagram I need some help on explaining the function of the reverse relay and travel limit switches. Can anyone...
Good Morning SammyS,
I believe you don't have any further information on my question? Thanks for trying anyway. I think the connection rod is at the max velocity when is perpendicular to the crank. I will use this assumption and try to work out the angle and the velocity.
1) For the mechanism shown in the figure included within the #1 post determine for the angle θ=45°:
a) the velocity of the piston relative to the fixed point O (VBO)
b) the angular velocity of AB about point A (i.e. ωAB)
c) the acceleration of point B relative to A (αBA)
Note: Link AB is...
Homework Statement
A) Determine the value of the angle θ (measured from vertical) when the angular velocity of link AB a maximum.
B) What is the maximum angular velocity of link AB.
Homework EquationsThe Attempt at a Solution
This is a part of a bigger question. I've managed to answer all...
Hi SteamKing, I had another think about it and my above solution is incorrect as I used the moment of inertia from the flywheel where I have to calculate the power input. I believe it should be:
P=T*ω
as α=(ω2-ω1)/t
ω1=0, so
α=ω2/t
ω2=α*t
Substituting this into power equation
P=T(α*t)
as...
SteamKing,
I think I'm to stupid for this, let's take it a step at a time.
You are correct, P=T*ω.
as T=I*α
and α=(ω2-ω1)/t
then T=[I*(ω2-ω1)]/t
as ω1=0
so T=(I*ω2)/t
substituting into power equation
P=(I*ω2*ωi)/t
My attempt:
1. P=T*ω
2. T=I*α
so far easy
Change in angular velocity of the shaft versus time?
Is it α=(ω2-ω1)/t
4. T=I*[(ω2-ω1)/t]
as ω1=0
T=I*(ω2/t)
therefore P=(I*ω2*ωi)/t
as I=14.062 kgm-2
ω2=10π rads-1
ωi=2π rads-1
P=(14.062*10π*2π)/t=2775.7275/t
Just need to now plot a graph of P for t...
Homework Statement
A flywheel is attached on the output of the gearbox. The output shaft rotates in the opposite direction to the input shaft at 5 times its speed. The gearbox has an efficiency of 92%.
If the flywheel is solid, has a mass of 50kg, a diameter of 1.5m and is to accelerate from...