Recent content by louza8

nvm - yeah I determined (after posting) that it would be incorrect due to the incorrect integration and the resulting on flow effects. It is quite frustrating when solutions are provided that are incorrect, particularly when I'm just trying to wrap my head around concepts/methods.

Hi nvn. I used the method outlined in the attached file to solve the homework question (with correct integration) but the homework question is for beam length 2L not L as shown in the attached. Hence, the answers differ due to the longer beam under a greater total load deflecting more. To check...

hi nvn, thanks for your reply. both were mistakes. i was trying to be clear but was not, apologies. anyways...i managed to solve the question by simplifying the load using singularity functions then integrating 4 times. i then went and checked the above method (with the correct integrals for...

Hi AlephZero, thanks for the tip. I have added what I hope are more readable versions for the solution :) To be specific I know how to get the moment equation here: EI*y''=wL(3x2 - L)/12 But don't know how they've integrated the above to get here: EI*y'=wx2^2/8 + C2 I thought there would have...

So nobody here can help me along with solving this problem or did you run into similar issues?

Could someone have a look at this one please? I'm still struggling with the understanding. Much appreciated.

Problem and Solution (albeit to a beam of different length L) provided in attachments. Hi, I have been able to follow the provided solution to get the moment equations and understand the x1 and x2 positional references etc. I also understand that integrating the moment equations provides the...

thanks for your time rude man

Homework Statement The tip of the Pitot tube is at the top of the jet. Calculate the flowrate and the angle θ . Homework Equations The Attempt at a Solution Please see attached images. 1 is the question and 1 is my attempted solution. I think I perhaps have done the first part...

is this a reverse square wave? the series looks similar to this?

Ahh I see thanks gneill. Yeah, the diagram illustrates your point nicely.*blushes* silly mistake.

So... I=m(d-a)^2+m(a)^2? Torque=d*F*sin(theta) angular accel=Torque/I m*g*sin(theta)*((d-a)-a)/m(d-a)^2+m(a)^2 angular accel=g*sin(theta)*((d-a)-a)/(d-a)^2+a^2 ------>factorised denominator, got rid of m Is this any better? Is that the expression for Torque required?

Hey Mike thanks for your help! I think I grasp it now but using a different explanation (thank you for your time), the following diagram makes sense to me, is this what is going on in this instance? http://img189.imageshack.us/img189/363/screenshot20110501at114.png

Homework Statement http://img713.imageshack.us/img713/9151/screenshot20110501at113.png Homework Equations Angular acceleration = Torque/Inertia Torque = Fd Mass_moment_Inertia=mr^2 The Attempt at a Solution ->assume masses on the end of seesaw are point particles for calculating inertia. is...

zero? I don't know, I'm obviously missing points of understanding. Is the 0.5m/s the velocity as observed from the 'ground' so it is the vector sum of the center of mass' velocity and the tangential component of velocity such that v_ground=v_center+v_tan=2v_center 0.5=2v v_cm=0.25m/s?