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Physics Puzzling Torque Problem

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Ackbach

Indicium Physicus
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Jan 26, 2012
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I seem unable to get the four necessary equations to solve the following problem:

A strut is attached to a wall in such a way that it can pivot. It is currently angled $30^{\circ}$ up from the horizontal, and held in place by a cable. This cable is attached to the wall just above the pivot, and the cable runs out horizontally to the tip of the strut, and then curves over the strut, and down to a weight $w$. The strut is of uniform density, and also has weight $w$. Find the tension in the cable, and the magnitude and direction of the force exerted on the strut by the pivot.

My answer so far: let $\vec{F}_{sp}$ be the force on the strut exerted by the pivot, and let $\vec{F}_{c}$ be the force exerted by the cable on the strut. Let $\theta$ be the angle that $\vec{F}_{sp}$ makes with the horizontal, and let $\varphi$ be the angle that $\vec{F}_{c}$ makes with the negative horizontal. Then the conditions for equilibrium say that
\begin{align*}
F_{sp} \cos( \theta)-F_{c} \cos(\varphi)&=0\\
-F_{c} \sin( \varphi)+F_{sp} \sin( \theta)&=w\\
- \frac{w}{2} \sin(120^{\circ})+F_{c} \sin(150^{\circ}+ \varphi)&=0.
\end{align*}
The first two equations are Newton's Second Law in the $x$ and $y$ directions, respectively, and the third equation is the torque version of Newton's Second Law.

My problem: I cannot find another equation. There are 4 unknowns, and only 3 equations. I tried applying the torque version to another point on the strut (the existing equation was exerted at the pivot), but that does not provide another independent equation.
 

caffeinemachine

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MHB Math Scholar
Mar 10, 2012
834
I seem unable to get the four necessary equations to solve the following problem:

A strut is attached to a wall in such a way that it can pivot. It is currently angled $30^{\circ}$ up from the horizontal, and held in place by a cable. This cable is attached to the wall just above the pivot, and the cable runs out horizontally to the tip of the strut, and then curves over the strut, and down to a weight $w$. The strut is of uniform density, and also has weight $w$. Find the tension in the cable, and the magnitude and direction of the force exerted on the strut by the pivot.

My answer so far: let $\vec{F}_{sp}$ be the force on the strut exerted by the pivot, and let $\vec{F}_{c}$ be the force exerted by the cable on the strut. Let $\theta$ be the angle that $\vec{F}_{sp}$ makes with the horizontal, and let $\varphi$ be the angle that $\vec{F}_{c}$ makes with the negative horizontal. Then the conditions for equilibrium say that
\begin{align*}
F_{sp} \cos( \theta)-F_{c} \cos(\varphi)&=0\\
-F_{c} \sin( \varphi)+F_{sp} \sin( \theta)&=w\\
- \frac{w}{2} \sin(120^{\circ})+F_{c} \sin(150^{\circ}+ \varphi)&=0.
\end{align*}
The first two equations are Newton's Second Law in the $x$ and $y$ directions, respectively, and the third equation is the torque version of Newton's Second Law.

My problem: I cannot find another equation. There are 4 unknowns, and only 3 equations. I tried applying the torque version to another point on the strut (the existing equation was exerted at the pivot), but that does not provide another independent equation.
From what I can picture I'd say the following:

I don't think you have used the FBD of the cable. Let $T$ be the tension in the cable. Then shouldn't $T=w$? (assuming massless cable). Further $F_c=\sqrt{2}T$ in the direction of $-45^\circ$ to the horizontal. Isn't it? So you already now the value of $\varphi$.
 
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Ackbach

Indicium Physicus
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Jan 26, 2012
4,197
From what I can picture I'd say the following:

I don't think you have used the FBD of the cable. Let $T$ be the tension in the cable. Then shouldn't $T=w$? (assuming massless cable). Further $F_c=\sqrt{2}T$ in the direction of $-45^\circ$ to the horizontal. Isn't it? So you already now the value of $\varphi$.
I'm not sure of that. I should have attached a photo of the original problem - let me do that now. I am just working on part (a) for now:

Problem 11-13.png
 

caffeinemachine

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Mar 10, 2012
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MarkFL

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Feb 24, 2012
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Here is another way to work the problem. Consider the following free body diagram:

fbdadrian.jpg

The tension in the vertical cable is obviously $w$. So let's find the tension in the horizontal cable.

To begin, let's sum up the torques about the origin:

\(\displaystyle \tau_O=0\)

\(\displaystyle TL\sin\left(30^{\circ} \right)-w\left(\frac{L}{2}\cos\left(30^{\circ} \right) \right)-w\left(L\cos\left(30^{\circ} \right) \right)=0\)

Combine like terms and divide through by $L$:

\(\displaystyle T\sin\left(30^{\circ} \right)-\frac{3w}{2}\cos\left(30^{\circ} \right)=0\)

Solve for $T$:

\(\displaystyle T=\frac{3w}{2}\cot\left(30^{\circ} \right)=\frac{3\sqrt{3}}{2}w\)

Now, we see by equilibrium, we must have:

\(\displaystyle F_y=2w\)

\(\displaystyle F_x=T=\frac{3\sqrt{3}}{2}w\)

Hence the magnitude $|F|$ and direction $\theta$ of the force exerted on the strut by the pivot can be found as follows:

\(\displaystyle |F|=\sqrt{F_x^2+F_y^2}=\frac{\sqrt{43}}{2}w\)

\(\displaystyle \theta=\tan^{-1}\left(\frac{F_y}{F_x} \right)=\tan^{-1}\left(\frac{4}{3\sqrt{3}} \right)\approx37.589089469^{\circ}\)
 

caffeinemachine

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MHB Math Scholar
Mar 10, 2012
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Here is another way to work the problem. Consider the following free body diagram:

View attachment 1306

The tension in the vertical cable is obviously $w$. So let's find the tention in the horizontal cable.

To begin, let's sum up the torques about the origin:

\(\displaystyle \tau_O=0\)

\(\displaystyle TL\sin\left(30^{\circ} \right)-w\left(\frac{L}{2}\cos\left(30^{\circ} \right) \right)-w\left(L\cos\left(30^{\circ} \right) \right)=0\)

Combine like terms and divide through by $L$:

\(\displaystyle T\sin\left(30^{\circ} \right)-\frac{3w}{2}\cos\left(30^{\circ} \right)=0\)

Solve for $T$:

\(\displaystyle T=\frac{3w}{2}\cot\left(30^{\circ} \right)=\frac{3\sqrt{3}}{2}w\)

Now, we see by equilibrium, we must have:

\(\displaystyle F_y=2w\)

\(\displaystyle F_x=T=\frac{3\sqrt{3}}{2}w\)

Hence the magnitude $|F|$ and direction $\theta$ of the force exerted on the strut by the pivot can be found as follows:

\(\displaystyle |F|=\sqrt{F_x^2+F_y^2}=\frac{\sqrt{43}}{2}w\)

\(\displaystyle \theta=\tan^{-1}\left(\frac{F_y}{F_x} \right)=\tan^{-1}\left(\frac{4}{3\sqrt{3}} \right)\approx37.589089469^{\circ}\)
Hmm.. I think the question asks us to consider the horizontal and vertical strings(or cables) as different and not same.. just as you did. I, in my solution, had assume that its the same string. In that case, no solution exists as the torque about the origin is not coming out to be zero. So its worth mentioning that if a single string is used then the strut cannot maintain an angle of $30^\circ$ wrt the horizontal.
 

Klaas van Aarsen

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Mar 5, 2012
8,878
My problem: I cannot find another equation. There are 4 unknowns, and only 3 equations. I tried applying the torque version to another point on the strut (the existing equation was exerted at the pivot), but that does not provide another independent equation.
Perhaps needless to say by now, but your 4th equation is:
$$F_{c} \sin( \varphi) = w$$
 
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Ackbach

Indicium Physicus
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Jan 26, 2012
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Here is another way to work the problem. Consider the following free body diagram:

View attachment 1306

The tension in the vertical cable is obviously $w$. So let's find the tension in the horizontal cable.

To begin, let's sum up the torques about the origin:

\(\displaystyle \tau_O=0\)

\(\displaystyle TL\sin\left(30^{\circ} \right)-w\left(\frac{L}{2}\cos\left(30^{\circ} \right) \right)-w\left(L\cos\left(30^{\circ} \right) \right)=0\)

Combine like terms and divide through by $L$:

\(\displaystyle T\sin\left(30^{\circ} \right)-\frac{3w}{2}\cos\left(30^{\circ} \right)=0\)

Solve for $T$:

\(\displaystyle T=\frac{3w}{2}\cot\left(30^{\circ} \right)=\frac{3\sqrt{3}}{2}w\)

Now, we see by equilibrium, we must have:

\(\displaystyle F_y=2w\)

\(\displaystyle F_x=T=\frac{3\sqrt{3}}{2}w\)

Hence the magnitude $|F|$ and direction $\theta$ of the force exerted on the strut by the pivot can be found as follows:

\(\displaystyle |F|=\sqrt{F_x^2+F_y^2}=\frac{\sqrt{43}}{2}w\)

\(\displaystyle \theta=\tan^{-1}\left(\frac{F_y}{F_x} \right)=\tan^{-1}\left(\frac{4}{3\sqrt{3}} \right)\approx37.589089469^{\circ}\)
Very nice! I see that you have not assumed the tension is the same everywhere in the cable. Do you happen to know under what conditions a cable of negligible mass would have equal tension throughout its length? I know, for example, that on at Atwood's Machine, if the pulley itself has significant mass, then the tensions on either side of the pulley can be different, but if the pulley has negligible mass, then the tensions are the same.

Hmm.. I think the question asks us to consider the horizontal and vertical strings(or cables) as different and not same.. just as you did. I, in my solution, had assume that its the same string. In that case, no solution exists as the torque about the origin is not coming out to be zero. So its worth mentioning that if a single string is used then the strut cannot maintain an angle of $30^\circ$ wrt the horizontal.
Ah, but does the cable have to have the same tension throughout? That's sort of the question I'm getting at in my response to MarkFL's solution.

Perhaps needless to say by now, but your 4th equation is:
$$F_{c} \sin( \varphi) = w$$
So, Newton's Second Law applied to the hanging mass provides the final equation. Very nice!
 

Klaas van Aarsen

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Mar 5, 2012
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Very nice! I see that you have not assumed the tension is the same everywhere in the cable. Do you happen to know under what conditions a cable of negligible mass would have equal tension throughout its length? I know, for example, that on at Atwood's Machine, if the pulley itself has significant mass, then the tensions on either side of the pulley can be different, but if the pulley has negligible mass, then the tensions are the same.
It appears you already know the answer. :)

What is slightly confusing in your problem is whether the cable is attached to the strut, or whether it can slide or roll freely.
Realistically, from the given set up, it has to be attached, but it would be nice if that were given.
Otherwise, the tension would be the same everywhere in the cable.
If that were the case, either you need an extra torque where the strut is attached to the wall, or the whole things would come collapsing down.
 
Last edited:

MarkFL

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Feb 24, 2012
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I actually assumed the cable was attached to the end of the strut so that we could treat the cable as two separate sections. :D
 

Klaas van Aarsen

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Mar 5, 2012
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I actually assumed the cable was attached to the end of the strut so that we could treat the cable as two separate sections. :D
Ah, assumptions!
If an engineer assumed the end was attached while it was not, I really hope no one would be standing below the contraption. (Angry)
 

caffeinemachine

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MHB Math Scholar
Mar 10, 2012
834
Ah, but does the cable have to have the same tension throughout? That's sort of the question I'm getting at in my response to MarkFL's solution.
The answer is yes if you assume that there is no friction between the cable and the pin. If friction is there then the tension on the two sides are related by factor which is $e$ to the power of something and this something depends on the coefficient of friction and the number of times the cable has been wound on the pin (I don't remember the exact formula but it's easy to work out). Note that the above applies only to the case when there is just ONE cable which is wound on the pin. In MarkFL's solution, which is the correct solution, there are two separate cables. If you see my solution you'll find that just one single cable cannot hold the strut in the position shown in the diagram (although I realized this after seeing MarkFL's solution).
 

Petrus

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Feb 21, 2013
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Klaas van Aarsen

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Mar 5, 2012
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So, Newton's Second Law applied to the hanging mass provides the final equation. Very nice!
A little while ago you noted that I had mixed up Newton's 2nd and 3rd laws.
Let me return the favor. ;)
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,140
My problem: I cannot find another equation. There are 4 unknowns, and only 3 equations. I tried applying the torque version to another point on the strut (the existing equation was exerted at the pivot), but that does not provide another independent equation.
I was just skimming along this one and I have some general advice. If you find yourself needing more equations for a static system usually end points along a bar and center of mass of an object are typically good places to take your torque around. In this case I'd use the point where the bar meets the wall. Note that the reaction force of the wall on the bar is eliminated since the moment arm there is 0 m.

-Dan
 

Ackbach

Indicium Physicus
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Jan 26, 2012
4,197
A little while ago you noted that I had mixed up Newton's 2nd and 3rd laws.
Let me return the favor. ;)
I was just thinking a free-body diagram on the weight reveals that the tension in the vertical cable equals the weight of the hanging mass, since it's not moving. You can get that equation via the Second Law, with a=0. Right?