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I have the situation in the image where two masses are suspended along a piece of rope (resulting in three segments of rope). I need to find the tension in each of the three segments of rope.

I know how to solve the similar problem with only one mass (and two segments of rope) by finding the horizontal and vertical components for each segment of rope, noting that the system is in equilibrium and then finding two equations in two unknowns and solving.

However, I am a little confused now that there is a second mass involved. If I follow the same method. I have three unknown tensions, but only 2 equations (one for the horizontal and one for the vertical component), and then I can't solve.

What am I missing here? Is this solvable (I suspect so)?

My only other idea is that I need to look at each mass separately somehow, so that I sort of follow the same method as for one mass but do it twice (once for each mass).

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  • $\begingroup$ I just wanted to comment that in my opinion this is a (rare) good example of an "allowed" homework-like question as there is a principle to be explained here, and you are not asking us to "just solve this" for you. $\endgroup$
    – Floris
    Commented Apr 24, 2016 at 11:19

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You can indeed balance the horizontal forces at each point, and the sum of vertical components should equal the weight.

That does seem to leave you with an over constrained problem (four equations with three unknowns) which will only have a solution when the angles are chosen "just so".

If one of the angles was not given you could solve. Pick one and prove that the value for the angle you calculate is the once given... Or prove that there is no solution! Although it is easier to assume the angles as given, and make one of the weights "unknown". Then solve for the value of the weight that gets you the angles.

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  • $\begingroup$ Ok. I'm trying to understand the physical reason why this is over-constrained. Why can't this system be solved? I understand the maths but what does this mean for the real-life picture? $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 3:24
  • $\begingroup$ In a real life system it might find different angles in order to stabilize. Imagine the system was stable, and you move the support point of one of the weights. This puts your system in a position with known angles and tensions - but not in equilibrium. If you stop pushing, the system will find a different position - the one you started from. When they give all the angles, the assumption is that the system is in equilibrium. Otherwise there is a net horizontal force (not necessarily the same force on each of the weights) - your fourth equation... $\endgroup$
    – Floris
    Commented Apr 24, 2016 at 4:03
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Somebody correct me if I'm wrong but I do believe you would have five tensions. There would be the tension in the one rope joining the two ropes holding the masses, the tension in the two ropes from the intersection to the mass, then the tension in the two rope from the top of the diagram to the intersection of all the ropes.

The tension in the rope from intersection to mass would be independent from each other, whereas the other three would be dependent on those masses and each other.

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  • $\begingroup$ Yes, there are 5 segments of rope. We can deal with the two attached to the masses using $F=mg$. I am interested in the two that attach to the ceiling and the one that joins those two together. $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 3:02
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Edit I was wrong. Look at Floris's answer.

If you split it into sections it should be easier than combining it into one.

If you have $F_1$ , $F_2$, $F_3$ as the tensions of the strings

$$ F_{1x} = F_{2x} $$ $$ F_{2x} = F_{3x} $$ $$ F_{1y} + F_{2y} = 12g $$ $$ F_{2y} + 7g = F_{3y} $$

Then you can use trigonometry and then eliminate.

Edit: Eliminate $F_{2y}$

$$ F_{1y} + F_{3y} = 19g $$

Then Trig $$ F_1sin(45) + F_3sin(30) = 19g $$ Along with $$ F_1cos(45) = F_3cos(30) $$

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  • $\begingroup$ For the 4 equations I get (1) $F_1\cos{45}=F_2\cos{10}$, (2) $F_2\cos{10}=F_3\cos{30}$, (3) $F_1\sin{45}+F_2\sin{10}=12g$ and (4) $F_2\sin{10}+7g=F_3\sin{30}$. When I solve this system of equations, I find that there is no solution. Did I do something wrong? Or does this say something about the system? $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 2:59
  • $\begingroup$ Also, is it ok to split it into sections? Are the calculations for one intersection independent of the calculations at the other? Maybe that's why it's not working out... $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 3:08
  • $\begingroup$ Are you sure you get no solution? $\endgroup$
    – Blubber
    Commented Apr 24, 2016 at 3:08
  • $\begingroup$ I will check again, but yes, I am pretty sure. $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 3:09
  • $\begingroup$ There are 4 equations and 3 unknowns. I solved equations (1), (2) and (3) and got $F_1=141.4$, $F_2=101.5$ and $F_3=115.4$. But then for these values, equation (4) does not hold. I used $g=9.8$. $\endgroup$
    – Jack M
    Commented Apr 24, 2016 at 3:11

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