Two Identical Beams leaning against each other in static equilibrium?

I'm really confused on this problem. I drew a FBD and got determined my forces in the x to be N(from plank right on plank left)=frictional force from the floor, and in the y i got the N(from floor onto the planks)=mg. I'm having trouble with the sum of the torque forces and in general figuring out the problem. Any help is greatly appreciated!!!





Two identical uniform beams, each having a mass of 3 kg, are symmetrically set up against each other on a floor with which they have a coefficient of static friction µ = 0.603.





a) What is the minimum angle the beams can make with the floor without falling?


Θ = ? (°)





b) For the angle calculated in part (a), what is the magnitude of the force exerted on one beam by the other beam, at the apex?


|F1 2| = ? (N)





Next a heavy point mass 3 kg is balanced on the apex.


c) What is the normal force exerted on each beam by the floor?


|N| = ? (N)





d) For this case, what is the minimum angle the beams can make with the floor without sliding?


Θ = ? (°)

Two Identical Beams leaning against each other in static equilibrium?
RFx


It is also the frictional force at the floor


summing torque about the floor


m*g*cosθ*L/2=RFX*sinθ*L


RFx=mg/(2*tanθ)


RFx=m*g*µ


tanθ=1/(2*µ)





a) θ=39.7


b)


RFx=m*g/(2*tanθ)


17.7 N





c) The normal force is the sum of the vertical forces 3*g from the beam and 1.5*g from the point mass


Normal force=4.5*9.81


44.1 N





d)


sum torque about the floor again


3*g*cosθ*(L/2+L/2)=RFX*sinθ*L


solve for RFx





RFx=3*g/tanθ


35.5 N








RFx=3*g/tanθ


RFx=4.5*g*µ


combine


4.5*g*µ=3*g/tanθ





solve for θ





tanθ=3/(4.5*µ)





47.87 degrees


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