Tensions in Hanging Beam with Given Weights

[Karol]

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 T_B=1013.3

 

[NascentOxygen]

Homework help requests typically include a question seeking an answer...

The masses here are partly supported by their ropes, so each does not press down on the rod with its full weight.

 

[haruspex]

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 T_A.
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 T_B there is a slightly quicker way. If you pick the reference axis for torque carefully then T_A doesn't feature.

 

[haruspex]

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.

 

[NascentOxygen]

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 T_A of 1012N and T_B of 1541N. xxxxx (There may be errors.)

 

[Karol]

In order to eliminate T_A i combine the 2 T_A forces into one which acts in the middle between them and then i take moments around that point:
$$1.2\cdot T_B+0.9\cdot T_B=2943\cdot 0.9+2000\cdot 0.5+1962\cdot 0.3=4237.3$$
$$2.1\cdot T_B=4237.3\rightarrow T_B=2017.8$$

 

[haruspex]

In order to eliminate T_A i combine the 2 T_A 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 T_A is of no interest then you can take a shortcut to finding T_B. 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.

 

[Karol]

Sure, but I'm saying that if T_A is of no interest then you can take a shortcut to finding T_B. 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 T_B

 

[NascentOxygen]

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.

 

[Karol]

I agree with all your working except the final step.

$$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=2017.8$$

 

[haruspex]

$$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=2017.8$$

Yes.