Recent content by new90

thanks but I am tired of this problem

i give up i will try this problem at some oint in the future

the system is in static equilibrium .Use he principle of virtual work to find the weight A and B.Neglect the weight of the strings and the friction in the pulley

ok i will make another tread

ok so what I did was follw what harupex said about the distance = if A moves down 1 units 1kg box will move up 0.5 unit then i used the vitual work method =Σfi times virtual distance = 0 this is =10N(because the force of gravity)times .5unit - 10aN(a = the box of the center)times 1 unit = 0(is...

so what i do was that if A drops 1 units |kg drops 1/2U then i used the the virtual work and i canceled y in both sides this is equal to 10N times 1/2 =10A and that equal to .5 kg wt

sin of the angle times v thets what I used to the left side but i think is not the correct process

i don't understand how you get the square root of 2

i was wrong i don't know to solve it I use the virtual work but the answer is not correct

thanks

THanks so much haruspex for your patience I know I am pretty dumb

sorry I understand now

I dot understand why y/2 and y/square root of 2

the letters are in correct place

I think i can use the pythagorean theorem to find PQ' IM correcT?