Tensions in Hanging Beam with Given Weights

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Homework Help Overview

The discussion revolves around calculating the tensions in ropes supporting a beam with given weights, specifically focusing on the weight of the beam and two masses. The participants are examining the relationships between forces and moments in a static equilibrium context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to set up equations based on moments and forces to find the tensions in the ropes. There are discussions about the correctness of the calculations and the assumptions regarding the distribution of weight from the masses and the beam itself.

Discussion Status

There is ongoing exploration of different methods to calculate the tensions, with some participants suggesting alternative approaches to simplify the problem. Several participants express uncertainty about the results and question the assumptions made in the calculations.

Contextual Notes

Some participants note discrepancies in the units used for weight and mass, and there is mention of potential errors in earlier calculations. The discussion reflects a collaborative effort to clarify the setup and ensure all factors are considered.

Karol
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Homework Statement


The weight of the beam is 2000N, the masses: m1=200kg, m2=300kg.
What are the tensions in the ropes.

Homework Equations


Moments: ##M=F\cdot L##

The Attempt at a Solution


Moments around A:
$$\mbox{and}\left\{\begin{array}{ll} 1.5\cdot T_B+1.2\cdot T_B+0.6\cdot T_A=1962\cdot 0.6+2000\cdot 0.8+2943\cdot 1.2 \\ 2T_A+2T_B=1962+2000+2943\rightarrow T_A=3452.5-T_B \end{array}\right.$$
$$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=-986.4$$
The sign isn't correct and it should be TB=1013.3
 

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The masses here are partly supported by their ropes, so each does not press down on the rod with its full weight.
 
I agree with all your working except the final step. You should get about 2000. The given answer seems too low, even from a cursory consideration, since it should be greater than TA.
A couple of things look odd. W is off-centre, and W is given in N but the masses in kg, but juggling with those still does not give me around 1000.

Edit: If you are only wanting TB there is a slightly quicker way. If you pick the reference axis for torque carefully then TA doesn't feature.
 
Last edited:
NascentOxygen said:
The masses here are partly supported by their ropes, so each does not press down on the rod with its full weight.
Karol has taken that into account.
 
haruspex said:
Karol has taken that into account.
Ah, yes.

I made a spreadsheet, and with the weight of the beam set to 200N (instead of 2000N), I found a first interval of 0.5m (instead of 0.6m) gave tension TA of 1012N and TB of 1541N. xxxxx (There may be errors.)

:smile:
 
In order to eliminate TA i combine the 2 TA forces into one which acts in the middle between them and then i take moments around that point:
$$1.2\cdot T_B+0.9\cdot T_B=2943\cdot 0.9+2000\cdot 0.5+1962\cdot 0.3=4237.3$$
$$2.1\cdot T_B=4237.3\rightarrow T_B=2017.8$$
 
Last edited:
Karol said:
In order to eliminate TA i combine the 2 TA forces into one which acts in the middle between them and then i take moments around that point:
$$1.2\cdot T_B+0.9\cdot T_B=2943\cdot 0.9+2000\cdot 0.5+1962\cdot 0.3=4237.3$$
$$2.1\cdot T_B=4237.3\rightarrow T_B=2017.8$$
Sure, but I'm saying that if TA is of no interest then you can take a shortcut to finding TB. If you pick your axis for torque carefully TA does not appear in the equation, so the one equation will do it. You don't need the vertical linear balance at all.
 
haruspex said:
Sure, but I'm saying that if TA is of no interest then you can take a shortcut to finding TB. If you pick your axis for torque carefully TA does not appear in the equation, so the one equation will do it. You don't need the vertical linear balance at all.
That's what i did but i got a totally different result for TB
 
Karol said:
That's what i did but i got a totally different result for TB
It's the same answer as you'd have in post #1 had you not made an error in post #1, according to my spreadsheet.
 
  • #10
haruspex said:
I agree with all your working except the final step.
$$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=2017.8$$
 
  • #11
Karol said:
$$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=2017.8$$
Yes.