Recent content by afromanam

thanks a lot, i really appreciate it and i really hope you're Somewhere in the Sunny Caribbean :cool:

ok, i don't know how to start this problem :blushing: the beam in the drawing weighs 3000 kg. It is attached to the wall by means of a simple union and without friction in point A. It is attached to the wall by means of a cable CD. The moment applied in point B is 6000 kg*m. Calculate...

oops, data missing yes, i was asked to find the force exerted by the block on the beam at point B.

i'm really desperate argh! uhmm from what i understood, if P1 = P2 then sliding would ocurr in both surfaces when d1 = 0.4*d2. you know i thought i had it, but nope... MAN, it's frustrating, i really thought i had this one please refer to attachments for diagram and fbd...

no problemo hey, don't worry, even the best make some mistakes what are your thoughts on number 2? thanks for the help

hello, olderdan thanks for helping me out: mhhh, i don't know if what i did is correct, but i added the \sum{F_X} in the cylinder and the \sum{F_X} in the ramp. so we have: N_{A} = F_{C} and doing the same thing with \sum{F_Y} we have F_{A} - W_{cylinder} - W_{ramp} +...

so now I'm undestanding a little bit more, but I'm still confused... 1) a momentum of 96 Nm is applied to the cylinder. The radius of the cylinder is 1 m. \mu_{A} = \mu_{C} = 0.25 The weight of the cylinder is 100 N. What is the minimum vale of \mu_B so the ramp moves to the right...

Thanks, thank you very much once again, you saved me...thank you very much F_B^{'} which is parallel to the plane is 35.55, from that, everything was 'easy'... N_C = 526.05 N \sum F_x = 0 F_N^{'}\cos15° + F_{BN}\sin15° + 0.2N_C - P = 0 P = 283.6 N see you next...

uhmm, i have a couple of questions ok, let's say F &_A=F &_B^' where F &_B^' is the frictional force parallel to the plane if my calculations are right F &_A = R &_A\sin 11.3 F &_A = 274.3 N\sin 11.3 F &_A = 53.74 ok, so if \frac{F &_B^'}{\mu_S} = 268.7 = N &_B^'...

Thanks for the help, i think I'm getting it, but i still have doubts, anyway, here's the problem: A wedge of 15° is pushed under a tube of 50 kg as shown in fig 1. \mu_S = 0.20 in all the surfaces. Determine the required force P to move the wedge. answer is P &= 283 N\leftarrow...

i really do not want this to sound exagerate, but i think God sent you. Thank you. On the other hand: I was so blind :mad: I really need to go over the basics and start making things more simple, thanks again

ok so now i have: U* = Coefficient of static friction FA* = Frictional force in A -W R U* - W R U*^2 + FA* r U* + FA* r U*^2 = M is this equation the one from which i can continue working or I'm missing something? thanks in advance

wow, thanx, could you post a link where i can learn how to use the Tex characters (i take it that they are LaTex characters, I've heard of the software but never used it) oh by the way, could you tell me something that could take me foward, because i don't know what to do with the two last...

ok that thing about clueless :shy: sorry, i was out of line, just gettin frustrated...coz I'm realising that it's true! what I'm doing now is: FA* =Friction force in A FB* =Friction force in B M = -R(FA*+FB*) = -R(U*NA + U*NB) SINCE NA = FB* = U*NB = -R(U*U*NB + U*NB) =...

Uhmm, just one question thanx very much for the info just one question... what about the forces that act in A and B as a result of the Momentum applied (M = Fd, so M=Fr and F=M/r) Would F be equal in A and B (and of course F in A would be opposing to friction vertically, and in B, F would...