Recent content by Racoon5

Hi, apart from the interesting wording of this question, I'm struggling to understand the concept of what we need to do here. I started by having three equations (summation Fx, summation Fy and Moment about Y). I also calculated the two reaction forces about A and the one rection force about B...

It wasn't a requirement. I went to school in Germany for 13 years which only required a subject for natural sciences (combining chemistry, physics and electrical systems in one). It mostly consisted of doing practical experiments and no theory at all. It seemed fun at the time, but I only...

Okay if we think about it that way, is this approach correct: ##Wf = f*d## Wf = PE = mg(H+h) f = Ff (friction force) d = L so ##mg(H+h) = Ff * L## solved for Ff = ##Ff = \frac {mg(H+h} {L}## ? Im also struggling to use LaTeX engine as you can tell W_f = F_f L

Apologies, I forgot to attach the diagram. I have amended my original post

I approached the problem using the work done by force equation (W=F*d) In my understanding all potential energy would have been converted into kinetic energy (KE) by point P (no friciton) We know d= L ; W= Wf ; f = Ff height = (H+h) So the Energy at point P is entirely kinetic: Which translates...

Thanks! You have a good point calling it point H. Perhaps I should define the points (A = initial, B = end of ramp, C = on the ground).

Hi everyone! I finally decided to go down the path of getting my engineering degree. I never had physics at school so I have to do a uni ready program and catch up on most of the course in my own time. Ive been quietly reading here for a while and it's actually been very interesting to start...

I initially thought about the different forms of energy present at each of the points: Total energy at starting point: PEA+ KEA= mgH at point D: KE_D = 1/2mv2f PED= mgD Energy at point D: PED+ KED D = mgD + 1/2 mv2f because EA= ED mgH = mgD = 1/2 mv2f mg(H-D) = 1/2 mv2f g(H-D) = 1/2...