Recent content by mike41

how about choosing your reference nodes. is there something special to that? choosing the node above the 60v wouldn't be the best choice would it?

ok so i can choose what ever i want for my reference at first, but i must stick to the current flow for the whole circuit and the math will work then?

ok but let's say for Node v1 how do i draw the currents would this be correct. this is the part that is holding me back

i have been trouble with this. iunderstand the concept but i don't know how to set up the current directions onto a circuit when i apply KCL. I am not sure which way the arrows should be going when doing kcl. for example like in this problem.

aha i see. was actually very simple then. thanks a lot.

ok I've been doing some problems but i can't seem to figure out how the top 2k ohm resistor is in parallel with the 1k ohm resistor. can anyone explain this to me?

haha ofund it actually. muFN = Fk so uu only have 2 variables. gah well thanks anyways

Homework Statement Determine the force P required to like the 200kg crate. Coeff. Static friciton is 0.3 on ALL contact of surfaces. Neglect Wedge Mass Homework Equations μS x Fn = Fk Summation of forces eqns The Attempt at a Solution im getting more unknowns than equations see attachment.

thank you it makes sense now, i did other practice problems and got them right too.

Well point A is unloaded then right? and A has two unloaded joints so according to rule 1 they would be zero force members. So AE, AB are zero force members. Then BE is a zero force because its perpendicular to two collinear joints. BC is also a zero force member because it is carrying no load...

Is joint c a support because of the little image under it showing its connected to ground?

this is my reasoning. BC then would NOT be a zero force because if you take the forces at C becue it has 3 forces coming off of it. CE and CB do not equal 0, in any direction but CD = 0 in Y and the x direction

what is the reasoning for joint c not being unloaded? it has no force maybe this is why I am getting stuck on this. Joint C has no force being pushed on it

ok well AB , BC are collinear and BE is the non collinear line so it would be a zero force. BC and CD are two unloaded joints and non collinear. So they would be zero force. This would be the same case with AB and Ae wouldn't it? OR do u have to account for any components ?

well from the rules what i got was AB, BE, BC, CE, DE, AE do you guys see any problems with that. what seems to be really getting me on these is this; do you account for action reaction forces? like at EB you would have P going towards B right. Well at BE then wouldn't you need an x component...