Dynamics- acceleration of beam on rollers

[deevil_knievel]

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 of circular roller =(mR^2)/2


3. The Attempt at a Solution
i am trying to draw the FBD for the roller and am just confused my it. it is hard to get any equations to solve this problem without a FBD. can the mg of the roller be assumed to be (1100+1200)9.81 from the center of gravity? super confused. our teacher teaches us a little differently, draw FBD. find kinetic equations. then find kinematic equations. i assume once i have the FBD correct and i understand the system i will find the velocity and and angular velocity of the point of the roller that touches the beam from the center of the roller and from the instantaneous center of the roller that touches the ground? then use that data to find the acceleration?

answer given is acceleration of roller is 1.29m/sec^2

this is a study guide so any help would be appreciated.
asked in the advanced thread... but i think it wasn't advanced enough.

 

[Simon Bridge]

An fbd for each object is unlikely to help much.
Have you tried using energy arguments?

When you pull on the beam with force F for a distance d you have expended some energy - where does it go?
You'll also need to look carefully at the geometry - i.e. when the beam moves a distance d, how far do the rollers move?

 

[deevil_knievel]

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.

 

[Simon Bridge]

Well you can do that problem using kinematic equations and Newton's laws - remembering the rotational versions - without fbd's provided you recall the geometry effects - i.e. is the beam moves distance d, how far do the rollers move?

If you insist, the fbd's for the rollers will have a force at the top due to the beam and one at the bottom due to the ground as well as weight, which acts at the center. The beam has a force for each of the rollers as well as the applied force F.

You could try dividing the beam into coupled masses - one for each roller... can you do one mass on top of one roller?

You will have as example of the process - but the common example is rolling down an incline - so the unbalanced force acts at the center. In this case the force acts at the top.

 

[SteamKing]

If you had just a beam of 2200 kg mass and you pulled with a force F = 4000 N, acceleration would be found by applying F = ma.

The beam in this case is supported by a couple of massive rollers. With no slippage between rollers and the beam, then F is not only going to accelerate the beam, but also turn the rollers by overcoming their rotational inertia.

Is the length of the beam given?

 

[deevil_knievel]

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.
http://imgur.com/EE90rHK

 

[deevil_knievel]

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). pulling from the beam creates a moment on the roller but is the point between the roller and the beam an instantaneous center? I'm assuming that the ground/roller is...god I'm confused.