Calculating Forces in a Static Equilibrium Bridge System

[akan]

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...

 

[tiny-tim]

Hi akan!

I can't get your picture links to work.

Can you describe the bridge?

 

[akan]

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. :)

 

[tiny-tim]

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...

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!

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

good description, anyway!

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?

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

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!

 

[emoseman]

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

 

[emoseman]

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... :)