1. The problem statement, all variables and given/known data Alright, so here's the problem I've got. I'll write it out first verbatim, then say what I've tried and failed with. One end of a uniform meter stick is placed against a vertical wall (Fig. 11.40). The other end is held by a lightweight cord that makes an angle with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.60. http://img171.imageshack.us/img171/665/1141ho5.gif [Broken] (a) What is the maximum value the angle can have if the stick is to remain in equilibrium? (b) Let the angle be 15°. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium? (c) When = 15°, how large must the coefficient of static friction be so that the block can be attached 20 cm from the left end of the stick without causing it to slip? 3. The attempt at a solution (relevant equations included within) OK, so I'm still stuck on a, and I'll be fine with an answer for just that one, cuz I think I can figure out the others if I can get that one... Alright, so I know it's an equilibrium problem. We've got variables Mm (which I'm using for the meter stick), Mb (the box), θ, T (the tension in the wire), and x (the distance from the wall to the box). It's a meter stick, so r=1. So I figure the total force diagram first: Pulling down, we have Mbg + Mθmg Pulling up: Tsinθ + Friction from wall This friction is Fnμ, and Fn comes from the tension's horizontal component, so the total pulling up is: Tsinθ + Tcosθμ So, the force setup: Mbg + Mmg = Tsinθ + Tcosθμ Now, since it's in equilibrium, torque is also gonna be equal, so I made up that equation: Pulling clockwise (down, basically): Mbgx + Mm(1/2) <--r = 1/2 because the center of mass should be halfway down the meterstick. Pulling counterclock: Tsinθ (r=1) So: http://img142.imageshack.us/img142/751/cramsterequation2008111uy8.gif [Broken] I then tried to put those two together using Tsin(theta): http://img89.imageshack.us/img89/8321/cramsterequation2008111ab5.gif [Broken] Shifting everything over to one side, I get Mbg (1-x) + Mmg/2 = Tcos(θ)μ And that's where I'm at. They give μ, but Mb, Mm, θ, x, and T are all variables here, and I don't know how to get rid of them. I think that, for a), I could possibly get rid of x by extending Mb all the way to the end of the meterstick or putting it up against the wall (whichever would give the largest angle), but that still leaves Mb, Mm, T, and θ. I'm solving for θ, but 3 variables is still too many. I feel like there's gotta be some way I can relate Mb, Mm, and T together to get rid of them, but any time I try to fix it, that μ from the wall keeps messing things up. Anybody have any ideas? Thanks.