Static Equilibrium of mass and wire

  • Thread starter chara76
  • Start date
  • #1
chara76
10
0

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" [Broken]

(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.
 
Last edited by a moderator:

Answers and Replies

  • #2
vishal007win
79
0
unable to portrait the picture of problm..
can you please put up the figure..
 
  • #3
chara76
10
0
Added link to picture. Sorry about that
 
  • #4
vishal007win
79
0
it shud be:
[(95kg)(9.81)(7.5m)(sin60) + (345kg)(9.81)(15m)(sin60)] / (10cos30)
 
  • #5
sarahmlipinsk
13
0
I have this same problem, but I already got part a. Can someone please help me find part b?
 
  • #6
sarahmlipinsk
13
0
I found the force, and the force in the i direction, but not in the j...
 

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