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Two ladders of uniform density and equal mass m are propped up against each other at angles $\theta$ from the frictionless ground. A rope of tension $T$ connects the two horizontally a distance $\ell$ from the center of each ladder. Each ladder is of length $L$. Diagram of ladders

If the ladder mass $m=1.4$kg, the angle $\theta =21.4 ^{\circ}$ , and the distance $\ell = \frac{1}{6} L$, what is the tension on the rope $T$? Answer in Newtons ($N$).

While I understand the underlying physics of this question, and how to solve the problem, I'm having trouble visualizing the directions the forces are going. At the moment, I am assuming the forces look like the following diagram, however I'm not sure if that is correct. I recall from earlier problems that assuming the directions of unknown forces is risky, and can result in the incorrect answer.Diagram of ladders with force vectors drawn on

First, I assumed that since the diagram was symmetrical, I only needed to enumerate the forces on one side, and consider the forces and torques of one side. I then set the center of mass of the ladder as the torque origin (the purple star on the diagram), so $\vec r_g=0$, $\vec\tau _g=0$.

I won't be writing vector hats on everything from here on out.

Going through the algebra, I ended up with $T=|F_2|$, $|F_n|=mg$ (from breaking the forces into $x$ and $z$ components), and (using torque) $|F_2|=\frac{3mg \cos x-T \sin x}{3 \sin x}$. I note that I may have mistaken a trig identity here, as I am not the best at memorization, and even with Google I am prone to little mistakes. Or I got the angles wrong. For $\ell \times T$ I used $\sin (180- \theta) = \sin \theta$, $\frac12 L \times F_2$ I used $\sin \theta$, and for $\frac12 L \times F_n$ I used $\sin (\theta - 90)=-\cos \theta$.

After finding $|F_2|=\frac{3mg \cos x-T \sin x}{3 \sin x}$, I substituted $\frac{3mg \cos x-T \sin x}{3 \sin x}$ for $|F_2|$ in $T=|F_2|$. Solving for $T$, I found that $T=\frac{3mg}{4\tan \theta}=\frac{3(1.4)(9.8)}{4 \tan 21.4} \approx 26.25699$N$\approx 26.26$N. The correct answer is $T \approx 52.51$N.

As I stated above, I believe my main problem is in getting the directions of the forces, so I would like an explanation of what angles I should be using and where those angles came from.

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  • $\begingroup$ I think there is a sign-error in the $|F_2|$ expression. The $T\sin(x)$ in the numerator should be positive. $\endgroup$ – Steeven Oct 3 '18 at 7:24
  • $\begingroup$ Also, I am not clear on what you mean with the cross-product sentence. For instance, what does "For $l\times T$ I used $\sin(180-\theta)=\sin(\theta)$" mean? Where does the $l\times T$ come from? $\endgroup$ – Steeven Oct 3 '18 at 7:28
  • $\begingroup$ I will have to check for the sign error, however I can clarify the $\ell \times T$. Firstly, I prefer the curly L since it's less ambiguous. The $\ell$ came from the diagram; it is defined in the parameters of the question as well: $\ell$ is the distance between the center of the ladder and where $F_1$ is being applied (I defined the force of the rope to be $F_1$ with $|F_1|=T$). As well, $\ell=1/6 L$ is also defined in the question. $\endgroup$ – Tropingenie Oct 3 '18 at 15:25
  • $\begingroup$ You're just missing a factor of 2 somewhere. Plugging in the equations on wolfram gives me the 52.51 answer. $\endgroup$ – HiddenBabel Oct 3 '18 at 15:57
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This is a homework type of question, so just a couple of insights:

1.Why do you assume $F_2$ is horizontal? At a point contact this is not justifiable.

2.You can produce more than enough equations for additional variables (e.g. $F_{2x}$ and $F_{2y}$) because torque is zero at many points in the system.

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  • $\begingroup$ The tension in the rope is horizontal. This tension propagates to the ladder. It is justified to consider it horizontal. $\endgroup$ – Steeven Oct 3 '18 at 7:30
  • $\begingroup$ @Steeven, I have not solved this homework problem. I am stating that apriori, you do not need to apply any considerations such as tension propagation. You should apriori assume any direction at a single point of contact and solve the problem. You may end up with $F_{2y}=0$. The down vote is not in place. $\endgroup$ – npojo Oct 3 '18 at 7:58
  • $\begingroup$ I might add that for this specific problem, you can apply symmetry consideration along with Newton third law to determine $F_2$ direction. $\endgroup$ – npojo Oct 3 '18 at 8:57
  • $\begingroup$ Ah, I see the error: I assumed that $F_2$ was horizontal because I didn't really understand how the forces were applied. Thinking about an example of a ladder against a wall, a component in the $z$-direction makes sense. To clarify, is your suggestion saying that it is more useful to consider the torques rather than the components of the forces? $\endgroup$ – Tropingenie Oct 3 '18 at 15:30
  • $\begingroup$ You need as many equations as variables. $N$ and $T$ are certainly variables. If you do not a-priori assume anything about $F_2$, than $F_{2x}$ and $F_{2y}$ are two additional variables to a total of four. Now select 4 equations. For example $\Sigma{F_x}=0$, another for $y-axis$ and two torque equations at two rotation-points. $\endgroup$ – npojo Oct 3 '18 at 16:51
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I think this problem will be easier to solve, if you assume the top point of the triangle (the point of contact of the two ladders) as the torque origin or the axis of rotation, since, in this case, you won't have to make any assumptions about the direction of the forces at the top point.

Also, it is easier to find the normal reaction forces of the floor, if you treat the two ladders as one object, in which case, again, you don't need to consider the forces at the top point, because they would be internal to the combined object. So, you can see right away that $F_N=mg$.

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