Static Equilibrium of mass and wire

AI Thread Summary
A 345 kg mass is supported by a wire attached to a 15 m steel bar, which is pivoted at a wall and supported by a cable. The tension in the cable for part (a) was calculated to be 10003.6976 N, with the wall exerting a force of 5001.8488 N in the i direction and -4347.056 N in the j direction. In part (b), a longer cable is used, and the user is attempting to find the new tension and wall force but is struggling with the calculations. The user initially calculated the tension incorrectly and is seeking assistance to solve part (b) effectively. The discussion highlights the importance of accurately applying torque and force equations in static equilibrium problems.
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Homework Statement



A 345 kg mass is supported on a wire attached to a 15 m long steel bar that is pivoted at a vertical wall and supported by a cable. The mass of the bar is 95 kg. (Take right and up to be positive.)

http://img522.imageshack.us/img522/8374/1265alt.gif"

(a) With the cable attached to the bar 5.0 m from the lower end as shown, find the tension in the cable.
10003.6976 N (Done)

Find the force exerted by the wall on the steel bar.
5001.8488 N i + -4347.056 N j (Done)

(b) A somewhat longer cable replaces the old and is attached to the steel bar 5.0 m from its upper end, connecting to the same place on the wall as before and maintaining the same angle between the bar and the wall. Find the tension in the cable.
? N

Find the force exerted by the wall on the steel bar.
? N i + ? N j

Homework Equations



Sum of Torques = 0
Sum of Fy = 0
Sum of Fx = 0

The Attempt at a Solution



(b)
[(95kg)(9.81)(7.5m)(cos60) + (345kg)(9.81)(15m)(cos60)] / 10m
= 2887 N This is wrong

I haven't attepmted the second part of B because I need the tension to do so.
 
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unable to portrait the picture of problm..
can you please put up the figure..
 
Added link to picture. Sorry about that
 
it shud be:
[(95kg)(9.81)(7.5m)(sin60) + (345kg)(9.81)(15m)(sin60)] / (10cos30)
 
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