Maximizing Wall Force on a Dumbell: Solving a Challenging Physics Problem

[DaMastaofFisix]

Got yet another free-response mind bender:

A dumbell is up against a vertical wall. the top end is tipped over. at what agle theita with respect to the floor is the force exetrtedby the wall a maximum?

No freakkin clue how to start, I understand that careful manipulations of the force and torque equations will yield the results and cancel all unwanted terms, but I need a careful nudge if you will. Any takers?

 

[DaMastaofFisix]

Alright, so I started messing around with some variables if you will and came up with about 60 degrees. Problem is that the correct answer is 63 degrees. Something that I must be doing is a bit off. Can someone lay out a method of attack if you will and explain the proper way to solve this problem? Thanks a bunch

 

[andrewchang]

what did you do when you got 60 degrees?

 

[civil_dude]

I think you have some boundary issues such as how does the bottom keep from sliding and is there friction from the wall to the dumbbell.

 

[DaMastaofFisix]

Hey guys sorry for the delay...and a major one at that. Umm, the problem was written by physics mastermind Boris Korsunsky, a crazy smart Russian who wrote a book of EXTREMELY challenging, non-calculus related ohysics problems. SO yeah, the Problem does not specify any mass, any distance L of the dumbell, any friction, coefficients of any sorts, nor any initial forces/torques that could set the thing in motion. The problem reads as follows:

A Dumbell is vertically aligned against a wall. At what angle, with respect to the floor, will the force exerted by the WALL on the dumbell be the greatest?

Okay, so for my work, I have been messing around with the three force/torque equations, in both a regularly aligned and rotated system, both before and during the fall itself. I have ben wondering if maybe conservation of evergy or momentum could lead me in the right direction to either give it a different approach, or help eliminate some variables

SO I got 60 degrees with just the force equations and torque, but the answer is 63. Anybody want to take a shot at thus guy? Remember, no variables were given, not even friction, so any variable introduced must be elimiated somehow or another, and ah yes, it should be done (as it says) without calculus.