Recent content by sponsoraw

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...

Thanks Jim for your reply. I was looking for more of a explanation what the travel limit switch and reverse relay does, what's the purpose of it is.

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.

Guys, can anyone help me with question 2b and 3 - see post #4. Much appreciated.

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...

No problem SammyS, I will include the full question and all my workings done so far.

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...

As I=14.062 kgm2 ω2=300 revmin-1=10π rads-1 ωi=2π rads-1 then, P=(14.062*10π*2π)/t=2775.7275/t

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...

Thanks SteamKing for your reply. I will give it a go. Any comment about a) and b)?

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...

Thanks SammyS for you reply. Can you expand a bit more on that?