# Tension between two/three ropes using vectors

Say there are three points, $a$, $b$, and $c$, with associated vectors $\vec{r_a}$, $\vec{r_b}$, and $\vec{r_c}$. $a$ and $b$ are both attached to firm surfaces, and each are connected to $c$ by ideal ropes. Force $\vec{F}$ acts on point $c$.

I am trying to find the tension in the rope connecting $a$ and $c$, let's call it $\vec{T_{ac}}$, with $\vec{T_{bc}}$ in the other rope.

First I use Newton's 2nd law to get $\vec{F} = \vec{T_{ac}} + \vec{T_{bc}}$.

Now is it wrong that $\vec{T_{ac}}$ is the projection of $\vec{F}$ onto $\vec{r_{ac}}$? Because that seems too easy. I think it involves balancing the components in each direction.

I can handle that in 2D, but then in the 3D case ($a$, $b$, and $c$ connected to $d$ with $\vec{F}$ acting on $d$), I'm having trouble grasping this.

I can't think of what coordinate system to use to get the components of the $\vec{T_{ij}}$s in each direction.

Any help for me and my 2D brain would be much appreciated.

• Can you add a picture of this? Commented Aug 28, 2018 at 10:06
• This is a 2D problem because any force out of plane (defined by points $a$, $b$ and $c$) cannot possibly be balanced. Unless the location of $c$ is unknown and then you need to specify the individual lengths $ac$ and $bc$. Commented Aug 28, 2018 at 16:57

The same conditions hold for equilibrium. The sum of all forces must be $0$: $$\sum \vec F=0$$

This can be broken into corresponding components: $$\sum F_x=0$$ $$\sum F_y=0$$ $$\sum F_z=0$$

Below is an image of what I think the problem you are working with is. I have drawn the force vectors, and drawn a plane perpendicular to the applied force $\vec F$ at the common point of where all the forces are applied.

Let's say $\vec F$ points in the negative $z$ direction. As you can see, we can use the three triangles drawn to determine the $z$ components of each of the tension vectors. Their sum is determined by the magnitude of $\vec F$.

We can also use the triangles to figure out the projections of the tension vectors onto the specified plane (it can be the $x-y$ plane if you would like). These are the primed vectors specified in the diagram. From there the problem reduces into a 2D equilibrium problem in that plane, which it seems like you are able to grasp.

I will leave the finer details to you.

• Sorry, I've been working at this for a while now and I can't figure out how to isolate $T_{ad}$ without relying on $T_{bd}$ and $T_{cd}$. I've come up with complicated relations between the components of each force/vector using angles, but I can't untangle them. Can you give a bit more help on this? Commented Aug 29, 2018 at 4:27
• @Spuds But $T_{ad}$ will rely on $T_{bd}$ and $T_{cd}$ in equilibrium. For the z-component $T_{ad,z}+T_{bd,z}+T_{cd,z}=F$. For the projections onto the x-y plane $T'_{ad,x}+T'_{bd,x}+T'_{cd,x}=0$ and $T'_{ad,y}+T'_{bd,y}+T'_{cd,y}=0$. You end up with a set of three coupled equations, and if you know $F$ then you should be able to solve for the remaining three magnitudes. Then the problem is less of a physics problem and more of a math problem. Commented Aug 29, 2018 at 10:08
• So I worked with the relations I got, and I ended up with a big messy equation relying on the angles between $\vec{F}$ and $\vec{T_{ad}}$, between $\vec{T_{ad}}$ and $\vec{T_{bd}}$, and between $\vec{T_{ad}}$ and $\vec{T_{cd}}$. Really messy. Is that something you would expect, or should it be simpler? It also doesn't rely on the angles between $\vec{F}$ and the other two tensions. Commented Aug 29, 2018 at 21:12
• @Spuds The easiest angles to use would be the angles the ropes make with the x-y plane and and angles of the projection vectors in the x-y plane Commented Aug 29, 2018 at 21:57
• @Spuds That is a valid way to do it. Define the $0$ angle in the x-y plane to be the angle of $T'_{ad}$. The expressions can get complicated. Commented Aug 29, 2018 at 23:02

The vector sum of the 3 forces must be zero. This means that :

1. The 3 forces must lie in the same plane, and

2. If you choose any 2 orthogonal directions in that plane, the sum of components of the forces in each of those 2 directions will be zero.

$F$ is the vector sum of $-T_a$ and $-T_b$. However $-T_a$ and $-T_b$ are not the orthogonal projections of $F$ onto ac and bc respectively.

• Is this for the 3D case? Aren't there 4 forces in this case? Three for the tension and one for the added force? Commented Aug 28, 2018 at 10:14
• @AaronStevens My interpretation of the problem is that there are only 3 forces acting at c. 2 ropes are attached from c to a and b, and force F acts at c (perhaps via a 3rd rope). Commented Aug 28, 2018 at 11:52
• Ok, so you are talking about the first case. I was talking about the other one with 3 ropes and 4 forces. "I can handle that in 2D, but then in the 3D case (a, b, and c connected to d with $\vec F$ acting on d), I'm having trouble grasping this." I think this case with 4 forces that do not lie in a single plane is what the OP is having trouble with. Commented Aug 28, 2018 at 13:58
• @AaronStevens Ah yes you're right. I have overlooked the increase to 4 forces in equilibrium at a point. Commented Aug 28, 2018 at 14:15