HELPPPPPPPPPPPPP!!!!Torque Questions :(?

Question

One end of a uniform 4.0 m long rod of weight w is supported by a cable. The other end rests against the wall, where it is held by friction (see Fig. P8.24). The coefficient of static friction between the wall and the rod is µs = 0.50. (The wire makes an angle of 37° with the rod.) Determine the minimum distance, x, from point A at which an additional weight w (same as the weight of the rod) can be hung without causing the rod to slip at point A.

The figure can be found at http://www.webassign.net/sf5/p8_24.gif.

Looking at the figure first makes it easier i think. I'm kinda stuck. I have no idea what to do at all. Any help would be great. I feel like I have too many variables and not enough equations or something. help? Please? Thanks to anyone who takes the time to help me out. :)

Answer 1

the diagram is not visible to me when I click the link, but I will try to answer based on the verbal description:

I assume the rod is horizontal

First, sum forces about the point where the rod touches the wall:
T*4*sin(37)-w*2-w*x=0
T is the tension in the cable
T=w*(2+x)/(4*sin(37))

Now, let's compute the maximum force of friction before the rod slips.
The normal force at the wall is the sum of horizontal forces, which is T*cos(37)
therefore, the maximum friction is
T*cos(37)*0.50
this force of friction is a vertical force, so we can sum the forces in the vertical to find:
T*cos(37)*0.50=2*w-T*sin(37)
solving for T:
T=2*w/(cos(37)*0.50+sin(37))

From above
T=w*(2+x)/(4*sin(37))
therefore
2*w/(cos(37)*0.50+sin(37))=w*(2+x)/(4*sin(37))

simplify
8*sin(37)/(cos(37)*0.50+sin(37))-2=x
x=2.81 m

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