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In preparation for the SAT Physics Subject Test, I have been doing extra practice problems. This one states:

A garment bag hangs from a clothesline. The tension in the clothesline is $10 \,\text{N}$ on the right side of the garment bag and $10 \,\text{N}$ on the left side of the garment bag. What is the mass of the garment bag?

The book from which this problem comes has a tendency to state questions unclearly, and some of its solutions have even had typos (which is really frustrating when using the book as a practice resource).

I interpreted the problem as something like this, where $T_1$ and $T_2$ both equal $10 \,\text{N}$.

enter image description here

Then $$F_\text{w} = mg = T_1 \sin\theta_1 + T_2 \sin\theta_2 $$ so

$$m = \frac{T_1 \sin\theta_1 + T_2 \sin\theta_2}{g} = \frac{10 (\sin\theta_1 + \sin\theta_2 )}{g} $$

Since the SAT uses $g = 10 \,\text{m/s}^2$, this simplifies to $$m = \sin\theta_1 + \sin\theta_2$$

Also, if two strings have the same tension, then each string makes the same angle with the horizontal. Thus $\theta_1 = \theta_2,$ and it follows that

$$m = 2\sin\theta_1 .$$

Since the angles were not given, this question cannot be solved, assuming I had interpreted the question correctly.


For reference, this is the solution the book provided (which I can't make any sense of):

Total upward force on the garment bag is equal to the tension in the clothesline. Therefore, the magnitude of $T_{\text{total}}$ equals the garment bag's weight, $mg.$

$$T_{\text{total}} = F_\text{w}$$ $$F_\text{w} = mg$$ $$\implies m = \frac{T_{\text{total}}}{g}$$ $$\implies m = \frac{10 \,\text{N}}{10 \,\text{m/s}^2} = \boxed{1 \,\text{kg}}$$

What confuses me about this solution is how "total upward force on the garment bag is equal to the tension in the clothesline." I don't think the people who wrote this question would be so careless as to omit necessary information. I think if they didn't provide angles, then the question wasn't meant to be interpreted in the above manner. If anybody knows how the problem should be set up, please provide an explanation/diagram :)

Sorry for the lengthy post and thanks for reading.

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    $\begingroup$ @KyleKanos If the rope was taut, can you explain why the upward force is equal to $10 \,\text{N}$? $\endgroup$ – A is for Ambition Nov 12 '15 at 23:06
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    $\begingroup$ I disagree with bruce smitherson ... your solution to the drawing is incorrect. If both angles are the same, then T1 = T2 = T. In addition, the vertical component of each tension is opposite the angle that you have shown, which means that 2Tsin(theta) = mg. Without knowing the angle, you can't get a numeric solution. $\endgroup$ – David White Nov 13 '15 at 1:19
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    $\begingroup$ When I was studying for the PGRE a huge number of the problems I did were missing information, or had errors in the solutions (etc). Don't let it discourage you! It's an even stronger way to test your confidence in your knowledge (which, in my opinion, often contributes a lot to the final score...). $\endgroup$ – DilithiumMatrix Nov 13 '15 at 3:43
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    $\begingroup$ @brucesmitherson, no need to apologize. Note that in this forum, I'm hesitant to be as blunt as I was, but I couldn't think of a way to make the point AND "beat WAY around the bush". $\endgroup$ – David White Nov 14 '15 at 0:37
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    $\begingroup$ @KyleKanos If the rope is taut, there is no lifting force. If the angle is extremely small, then the mass of the bag is extremely small. I'm really not surprised at the solution manual because publishers routinely make mistakes like this. $\endgroup$ – Bill N Jan 28 '16 at 22:37
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enter image description here

$$T_1/\sin \psi=T_2/\sin\psi=W_t/\sin\theta$$ $$10/\sin\psi=10/\sin\psi=W_t/\sin\theta$$ Therefore $$10/\sin\psi=W_t/\sin\theta$$ $$W_t=F $$ $$\theta=360-2\psi$$ $$W_t= \frac{10\sin\theta}{\sin\psi}\tag{using lami's theorem}$$ $$ =\frac{\sin (360-2\psi)*10}{\sin\psi}$$ we know that $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ so $$\sin (360-\psi-\psi)=-sin \psi\cos \psi-\cos\psi\sin \psi$$ $$ =-2\sin \psi\cos \psi$$ Substituting $$W_t=\frac{-2\sin \psi\cos \psi*10}{\sin \psi}$$ $$=-20\cos \psi$$ $$ 180>\psi>90$$ Therefore wt between 20N to 0N So 10N is justifiable

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    $\begingroup$ This answer might be better if you showed how you got the answer, rather than loosely citing a theorem. $\endgroup$ – Kyle Kanos Nov 19 '15 at 16:05
  • $\begingroup$ I am still a rookie. Let me know if you think is ok $\endgroup$ – Vinay5forPrime Nov 22 '15 at 10:56
  • $\begingroup$ A lot of math but no answer $\endgroup$ – Jens Mar 7 '16 at 21:52
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    $\begingroup$ I don't understand why people say "Lami's theorem". It is a silly vector addition rule which has a name. @Vinay5forPrime You should probably do it the simple way, i.e: form a triangle with those three vectors and prove Lami's theorem using the sine rule. $\endgroup$ – Yashas Feb 18 '17 at 5:59
  • $\begingroup$ This doesn't quite answer the question. You go from "the weight of the bag is within this range" to "let's pick an arbitrary value in that range because it agrees with the publisher" rather abruptly. $\endgroup$ – Nuclear Wang Feb 20 '18 at 20:24
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If both the ropes are of identical material, then the fact that they have equal tension in them means, by symmetry, that they make equal angles with horizontal, call it $\theta$. Then the only equation you have is $mg=2Tsin\theta$. So unless $\theta$ is given you cannot solve the problem. If you make, say, $\theta$ small so that the hanging ropes are close to being horizontal, then a large tension $T$ will be required in the ropes so that it has appropriate component along vertical direction to balance the hanging weight.

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