What Is the Tension in the Horizontal Segment Above the Attached Object?

[ML1]

Homework Statement

A bridge, constructed of 11 beams of equal length and negligible mass, supports an object of mass as shown. Real bridges of this sort have steel rockers at the ends, these assure that the support forces on the bridge are vertical even when it expands or contracts thermally.

Assuming that the bridge segments are free to pivot at each intersection point, what is the tension in the horizontal segment directly above the point where the object is attached? If you find that the horizontal segment directly above the point where the object is attached is being stretched, indicate this with a positive value for . If the segment is being compressed, indicate this with a negative value for .

Homework Equations

\Sigma \tau=FrsinTheta

The Attempt at a Solution

I know that Theta equals 60Degrees but I'm not sure how to go about the problem. And from a previous problem: Fp=2mg/3 and Fq=mg/3

I assume I find the components of something but I don't know what.

 

[tiny-tim]

Welcome to PF!

Hi ML1! Welcome to PF!

(have a theta: θ and a tau: τ and a sigma: ∑ and try using the X₂ tag icon above the Reply box )

There are only three forces at P, and you know the magnitude of one and the direction of all three …

so do a force diagram (a https://www.physicsforums.com/library.php?do=view_item&itemid=99" ) to find the other two magnitudes …

then do the same at the top-left joint

 

[ML1]

I don't know if I fully understand the question. I know I'm looking for the tension in the top joint. But why does torque apply here? Why use torque equations if it isn't moving. Physics frustrates me...lol

 

[tiny-tim]

Hi ML1!

(just got up :zzz: …)

But why does torque apply here?

I don't think it does …

you say you've already found F_p and F_q (for which I assume you did use torque), but I don't think you need anything more than linear equations now.

Why use torque equations if it isn't moving.

Same reason as you use force equations when it isn't moving!

Does that make sense, or do you want me to be more specific? ​

 

[ML1]

Does that make sense, or do you want me to be more specific? ​

[/QUOTE]


Could you be more specific? The question says that the segments are free to pivot. Why is this relevant to finding tension? Hahaha, i think I'm just overthinking it and that's why I'm not fully understanding it.

I appreciate the help you're offering!

 

[tiny-tim]

The question says that the segments are free to pivot. Why is this relevant to finding tension?

oh, that's just one of those phrases like "frictionless pulley" …

it tells you that the question is simple, and you can leave things out …

in this case, if they weren't free to pivot, there would be frictional torques (or stress torques) at the joints …

because they're free to pivot, that means the only forces at the joints are the tensions or compressions along the line of each segment

 

[BjarkeTN]

The trick here is to find an expression for the torque around the pivot from which the mass is dangling. The rods directly connected to this point exert no torque around the point because the "lever arm" is zero. The tension in the upper rod, however, does contribute. We see that since the triangle defined by three rods is equilateral, the angles must all be 60deg. Thus the lever arm from the pivot point of interest to the point where the tension 'attacks' must be
L \cos(\pi/3) corresponding to 30deg (since we're looking for the half-angle, drawing a diagram should make this clear). Thus the magnitude of the torque due to the tension in the upper rod is
TL \cos(\pi/3) = \frac{\sqrt{3} TL}{2}
Taking into account the torque due to F_P and setting the total torque equal to zero (one of the conditions for mechanical equilibrium) should yield the correct result.