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Lengths of vertical cables and maximum tension - Feynman exercises 2.35

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a) Find the proper lengths for the remaining vertical cables A and B.

I focused on the left hand side of the diagram and drew the right triangle BEC. Once here I drew the bisector of the right angle located at C, originating the right triangle CED. Now to get the angle $\theta$ we just need to know that the three angles of a triangle sum 180 degrees. As you can see I applied that doing:

angle E + angle CDE + angle DCE = 180 (where angle E = $\theta$ )

So now it is just about solving for $\theta$, which I got is 45 degrees. This is not correct so I am wondering what I did wrong here. Note after I applied tan $\theta$ in order to get the length BC which, as expected, did not make sense.

I am sure my mistake has to be on calculating $\theta$ but I do not see where is the flaw exactly.
It might be on calculating the angle DCE but it has to be 45 degrees because is the bisection of a right angle.

To find the distance GC (the second largest cable, the one related to letter A in the original exercise) I could not find the proper method.

Given answers A = 5 m; B = 11 m.

b) Find maximum tension in the two longitudinal cables

$$\sum F_y = mg $$

$$T sin 45 - mg = 0$$

$$T = 6.66 x 10^5 N$$

But the given answer is T = 34 x 10^3 kg-wt

asked Sep 23 in Physics Problems by JD_PM (490 points)
edited Oct 20 by JD_PM

1 Answer

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I do not understand your calculations. (You are working partly in kgwt and partly in Newtons, which is confusing. Stick to one unit and use throughout.)

I would start from the outside and work inwards :

There are 3 distinct joints (nodes) to examine. At each of them there are 3 forces in balance.

The tension in the 4 longitudinal cable ends can be found because the vertical components of this tension supports the whole weight of the bridge ($4.80\times 10^4 kg$).

The 12 vertical cables each carry the same weight, which in total is again the full weight of the bridge. Balancing forces at joint B allows you to find the force and angle of cable BA.

Repeat the same process for joint A to find the tension and angle of cable AC where C is the innermost joint/node.

The lengths of the vertical cables can be found from the angles calculated above.

Balancing forces horizontally at each joint shows that the horizontal component of tension is the same in each section of the cable : $$T_1\cos\theta_1=T_2\cos\theta_2=T_3\cos\theta_3=...$$ This shows that the maximum tension occurs where $\cos\theta$ has the smallest value, which is where the angle is maximum - ie at the ends. The maximum tension is in the outermost cable : you already calculated it above.


Calculation

Balance forces vertically and horizontally in turn at the joints A, B, C, D shown in the following diagram, starting from the outer end D :

At the 4 end-points D of the two horizontal cables, the vertical component of the tension $T_1$ in the 1st section of the cable is balanced by a reaction force equal to $\frac14 W$ where $W=4.80\times 10^4 kgwt=4.70\times 10^5 N$ is the full weight of the bridge. $$T_1\sin45^{\circ}=\frac{1}{\sqrt2}T_1=\frac14 W$$ $$T_1=\frac{\sqrt2}{4}W=1.70\times 10^4 kgwt=1.66\times 10^5N$$

At joint B the balance of vertical and horizontal forces gives $$T_1\sin\theta_1=T_2\sin\theta_2+\frac{1}{12}W$$ $$T_1\cos\theta_1=T_2\cos\theta_2$$ from which we get $$\frac{T_2\sin\theta_2}{T_2\cos\theta_2}=\tan\theta_2=1-\frac{W}{12T_1\sin\theta_1}=1-\frac{W}{3W}=\frac23$$ $$\theta_2=0.588 rad=33.7^{\circ}$$ $$\cos\theta_2=\frac{3}{\sqrt{13}}, \sin\theta_2=\frac{2}{\sqrt{13}}$$ $$T_2=\frac{T_1\cos\theta_1}{\cos\theta_2}=\frac{\sqrt{13}}{12}W=5.10\times 10^4 kgwt=5.00\times 10^5 N$$

At joint A we have similar forces so we can immediately write down that $$\tan\theta_3=1-\frac{W}{12T_2\sin\theta_2}=1-\frac{W}{2W}=\frac12$$ $$\theta_3=0.464rad=26.6^{\circ}$$ $$\cos\theta_3=\frac{2}{\sqrt5}, \sin\theta_3=\frac{1}{\sqrt5}$$ $$T_3=\frac{T_1\cos\theta_1}{\cos\theta_3}=\frac{\frac14}{\frac{2}{\sqrt5}}W=\frac{\sqrt5}{8}W=1.34\times 10^4 kgwt=1.31\times 10^5 N$$

We do not need to go any further. I leave it to you to calculate the lengths of the vertical cables.

Note : There may be something wrong with my calculation because the largest decrease is from $\theta_3=26.6^{\circ}$ to $\theta_4=0^{\circ}$. The largest decrease should be at the outside and the smallest at the inside. I am too tired to check this right now.

answered Sep 23 by sammy gerbil (26,096 points)
selected Sep 28 by JD_PM
Thank you for the advise, now it is fixed. I have updated my previous comments. However, the issue now is that I get a misleading angle (56.31 degrees) ...
Your "misleading angle" of $\alpha=56.31^{\circ}$ is correct. Perhaps your "mistake" is that you have forgotten that you measured $\alpha$ from the vertical instead of the horizontal.

You can avoid such mistakes by using the same convention throughout the calculation : eg measure all angles with respect to the rightward horizontal; use units of either $kgwt$ or $N$ for all forces - but not a mixture of the two; do not use the same symbol (such as $mg$) for more than one variable.
Outstanding answer. I will delve into it and try to find out any mistake in the calculation as soon as possible
OK Sr I have been checking and I got different results for $T_2$ and $\theta_3$:

$$T_2 = \frac{T_1 cos (45)}{cos (33.64)} = 1.44 x 10^4 Kgwt$$

$$tg \theta_3 = \frac{ T_2 sin (33.64) -\frac{W}{12}}{T_2 cos (33.64)} = 0.33$$

$$\theta_3 = 18.4^{\circ}$$

These results lead me to the right lengths of both cables.
That is good. I thought there was something wrong with my results - both the angles and the tensions.
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