What is the minimum value of for x which the stick will remain in equilibrium?

Question

One end of a uniform meter stick is placed against a vertical wall. 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 0.400.

Let the angle between the cord and the stick be 14.0 degrees. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance from the wall.

http://session.masteringphysics.com/problemAsset/1008117/5/yf_Figure_11_40.jpg

Answer 1

Start by summing the torques where the stick contacts the wall
The tension in the cord I will call T
The mass of the stick and the block I'll call m
The weight will be w=m*g
w*x +.5*w-1*T*sin(14)=0

T=w*(x+.5)/sin(14)

The reactive force of the wall in the horizontal direction is
T*cos(14)
Therefore, the maximum vertical reaction force before friction breaks is
T*cos(14)*0.400

The vertical reaction force is
2*w-T*sin(14)

Therefore
T*cos(14)*0.400=
2*w-T*sin(14)

or

T=2*w/(cos(14)*0.400+sin(14))

Since T=w*(x+.5)/sin(14)

Then
(x+.5)/sin(14)=
2/(cos(14)*0.400+sin(14))

Doing some more algebra

(x+.5)/sin(14)=
2/(cos(14)*0.400+sin(14))

x=
2*sin(14)/
(cos(14)*0.400+sin(14))
-.5

x=0.268 m

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