Recent content by Al_Pa_Cone

I was checking my initial calculations regarding the weight of the cat impacting the roof from 1 metre and I got 442Nm of force in my method. I have just recalculated and got 44.2Nm. Does this look correct in the method?

Yeah If you drink your pints by the litre I always have a couple, so in that case I owe you 3.51951 Pints. Cheers!

Yes, I was trying to visually represent the Torque as it helps me to try and understand the logic of the situation. I think my fault in understanding your method is because I have not used the units gram-meters anywhere in my course so far and I am unfamiliar with it. I have just googled a...

I am trying to understand how the mass at twice the distance to the right of the pivot can be equal to the mass balancing it at half the distance to the left of the pivot? From your information I worked out the torque about the pivot to be 0.2 x 1000g x 10 which gives 2000g at 0.2m and the...

So the torque about the pivot is equal to the counterweight mass effectively making the -2000g(clockwise torque being negative) + 2000g(the counterweight anticlockwise torque) = 0 Torque so no movement.

So to find the mass centre would I be looking to 50% of 833.3g mass to the right of the pivot (416.65g). I thought if the setup is effectively a seesaw then to balance each side equally then it would only take a small change in mass at one side to tip the balance and open the lid?

So the overall weight of the lid would be 1000g but 100mm of it is beyond the left of the tipping point. So would I be correct by starting to divide the weight by 600mm. 1000g/600mm*500mm = 833.3g to the right of the pivot point.166.7g to the left and this ratio would be 1:5. so the weight...

I have redrawn the diagram to show the idea. I have started with working out the net weight of the lid as a UDL and worked out 1000g into Newtons = 9.81N. as a UDL this would be 9.81/0.6m = 16.35N.

Here is a stripped down version with the facts I have set out to find the maximum force required at the big red arrow. I will do some calculations to try and get some results myself.

That sounds like an interesting method which I am unfamiliar with. I am interested to learn a bit more about calculus as I have not used it a great deal throughout my course. I am going to set myself a further challenge to complete which you may be able to assist me with using alternative...

This is what I expected to find! Thank you so much for taking the time to help me again... The world needs more people like yourself!

I wonder why there is a delay from 0 to 0.1 before the natural curve begins to take place? is there still something wrong with my method?

I somehow miscalculated step 2e and step 2f leading me to think I had unusual results. These results look good, and are what I was expecting. Does this look ok to you? I have been sloppy with my calculations all the way through. If I have learned something from this, it is to check my...

I have attempted to rework the method following your advice and in the first instance at step 2d, I get a positive value in the results of 0.3102. I thought there was still something I am missunderstanding so I attempted it again, this time excluding the term if the entire term was negative (not...

Im sorry I don't think I am following, I can still include the 110 as it is not part of the method to eliminate anything in the nominator if one value is a negative? If the ()^4 is raised to a factor and effectively becomes zero e.i (negative value)^4 and zero^1 is still zero, I am only left...