How Is the Vertical Force Calculated at the Hinge in a Beam and Sign System?

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Homework Help Overview

The discussion revolves around calculating the vertical force exerted by a hinge on a beam that supports a shop sign. The problem involves a uniform beam and the forces acting on it, including tension in a supporting wire and the weight of the sign and beam.

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  • Mixed

Approaches and Questions Raised

  • Participants share their attempts at calculating the vertical force using equilibrium equations, specifically focusing on vertical force balance. There are questions about the accuracy of the tension value and the impact of the attachment point of the cable on the calculations.

Discussion Status

Several participants are exploring different interpretations of the problem, particularly regarding the tension in the supporting wire and its effect on the vertical force calculation. Some guidance has been offered regarding the importance of considering moments and avoiding rounding errors from earlier calculations.

Contextual Notes

There is an ongoing discussion about the assumptions related to the attachment point of the cable and how it influences the torque calculations. Participants express uncertainty about their results and the implications of their calculations on the final answer.

Anonymous123451234
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A shop sign weighing 215 N hangs from the end of a uniform 155-N beam.

Tension in supporting wire= 642 N
Horizontal force exerted by the hinge= 526N


Find the vertical force exerted by the hinge on the beam at the wall

upload_2017-11-13_22-1-52.png


Fy= Fhy + Ft * sin() - mg - Mg =0

My attempt:
Fhy= -(Ft *sin() - mg - Mg)
=-(642*sin(35) - 155 - 215)
= 1.76 N

My answer is incorrect, but I don't know what I'm doing wrong.
 

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Anonymous123451234 said:
A shop sign weighing 215 N hangs from the end of a uniform 155-N beam.

Tension in supporting wire= 642 N
Horizontal force exerted by the hinge= 526N


Find the vertical force exerted by the hinge on the beam at the wall

View attachment 215006


Fy= Fhy + Ft * sin() - mg - Mg =0

My attempt:
Fhy= -(Ft *sin() - mg - Mg)
=-(642*sin(35) - 155 - 215)
= 1.7 N

My answer is incorrect, but I don't know what I'm doing wrong.
How did you get that tension? I get rather less.
 
haruspex said:
How did you get that tension? I get rather less.
I used the formula Ft= (mg length/2 + Mglength) / (length*sin(35))
 
Anonymous123451234 said:
I used the formula Ft= (mg length/2 + Mglength) / (length*sin(35))
The cable is not attached to the end of the rod.
 
haruspex said:
The cable is not attached to the end of the rod.

It doesn't matter where on the rod it's attached to in this scenario. It goofed me up at first because I thought it did matter when it actually did not. This is the correct tension.
 
Anonymous123451234 said:
It doesn't matter where on the rod it's attached to in this scenario. It goofed me up at first because I thought it did matter when it actually did not. This is the correct tension.
For the torque, it matters.
However, you are right that the tension is 642N, I made a mistake.
 
So, now I wonder whether your 1.7 is not quite accurate enough. I get exactly 5/3N. (There's a lot of cancellation, and angle turns out not to matter.)
 
haruspex said:
So, now I wonder whether your 1.7 is not quite accurate enough. I get exactly 5/3N. (There's a lot of cancellation, and angle turns out not to matter.)
I get exactly 1.7639278626, rounded to 3 sig figs is 1.76 N which is incorrect.
 
Anonymous123451234 said:
I get exactly 1.7639278626, rounded to 3 sig figs is 1.76 N which is incorrect.
To get the most accurate answer, forget the results for earlier parts of the problem. Using those introduces rounding errors.
Take moments about the point of attachment of the wire.
 

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