Calculating Forces in a Static Equilibrium Bridge System

AI Thread Summary
The discussion revolves around calculating forces in a static equilibrium bridge system, specifically focusing on a uniform bar supported by two posts. The calculations show that the force on the leftmost support (F_l) is determined to be 1200 N downward, which is confirmed as correct by participants. There is confusion regarding the equilibrium state of the bridge, with some participants questioning the feasibility of the setup. The importance of using moments about the right-hand post for quicker calculations is highlighted, along with a critique of the problem's wording. Overall, the calculations and concepts of static equilibrium are affirmed as valid despite initial doubts about the scenario's plausibility.
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http://img150.imageshack.us/img150/5393/bridgeef8.png
http://g.imageshack.us/img150/bridgeef8.png/1/

Sum(T_z) [pivot at L] = 2 F_r - 4 mg = 0
2 F_r = 4 mg
F_r = 2 mg
F_r = 2 * 120 * 10
F_r = 2400

Sum(F_y) = F_l + F_r - mg = 0
Sum(F_y) = F_l + 2400 - 1200 = 0
F_l = -1200

Answer: 1200 N downward. Is this right? This picture seems to portray an impossible equilibrium, so I'm confused. I think the right end should actually bring the whole thing down by overwheighing...
 
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Hi akan! :smile:

I can't get your picture links to work. :cry:

Can you describe the bridge? :smile:
 
Why can't you get it to work? It's on Image Shack. :S.
Anyway, my description will be weird, but so is the picture - so bear with me.

Problem statement:
A uniform bar of length 8.0 m and mass 120 kg is supported by two vertical posts spaced by 2.0 m, see the figure. Calculate the force on the leftmost support (magnitude and direction!).

Note: please use g = 10 m/s^2 for simplicity. Show all work.

Picture description:
There is a horizontal bridge, whose length is 8 meters. The leftmost end is supported by an upright post support. There is another post support 2 meters to the right from the left one. The force of gravity acts at the center of mass, so I understand it is 4 meters from the leftmost end (or, likewise, the rightmost one). There are no other supports besides these two, so I don't know how there is an equilibrium. But that's the whole problem, as it is stated. Thanks. :)
 
akan said:
Sum(T_z) [pivot at L] = 2 F_r - 4 mg = 0
2 F_r = 4 mg
F_r = 2 mg
F_r = 2 * 120 * 10
F_r = 2400

Sum(F_y) = F_l + F_r - mg = 0
Sum(F_y) = F_l + 2400 - 1200 = 0
F_l = -1200

Answer: 1200 N downward. Is this right? This picture seems to portray an impossible equilibrium, so I'm confused. I think the right end should actually bring the whole thing down by overwheighing...
akan said:
Why can't you get it to work? It's on Image Shack. :S.
Anyway, my description will be weird, but so is the picture - so bear with me.

Hi akan! :smile:

hmm … picture works fine now … it shows up as part of the post … didn't yesterday … mystery :confused:

good description, anyway! :smile:

Yup … 1200N is correct …

though it would have been a lot quicker if you'd just taken moments about the right-hand post, wouldn't it? :wink:

I agree the question is badly worded … "support" begins with "sup", which is the same as "sub", from the Latin meaning "under". :mad:

The equilibrium is as expected … you have equal forces (1200N) at equal distances from the right-hand post, so the whole thing is balanced on that post! :smile:
 
I'm trying to solve a similar problem using this example, but I'm confused as to where the 4 comes from in the first equation:

Sum(T_z) [pivot at L] = 2 F_r - 4 mg = 0
 
Nevermind, it is the downward force due to gravity that is causing a torque force rotating about the axis denoted by L.

Just tired this afternoon... :)
 
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