i'm just confused if i treat it as 2300kg from the CG of the roller? the 2200kg beam is split between the rollers, but technically it sits on top of the roller not from it's CG. however they are both in the y direction meaning the reactive force from the ground should still be 2300(9.81)...
length is not given. it's just slightly longer than the 2m between the rollers. F=ma and M=I(alpha) where I of circular roller =(mR^2)/2.
i guess I'm just trying to figure out how to treat the FBD and get some general equations about the system i have. here is what i am thinking...
solving the problems without a proper FBD or without using kinetic and kinematic equations gets me only 1/10 points on the exam. i am trying to do it his way. t1+sum(u1>u2)=t2 is what the book uses to find the velocity of these problems but it is not applicable to this exam.
1. Homework Statement
2200kg beam is sitting on two rollers spaced 2m apart. rollers weigh 1200kg and have radius of .35m. the block is pulled to the right with 4000N. there is no slippage. find the acceleration of the system (answer given is one of the rollers).
2. Homework Equations
I...