Physics Question (1 Viewer)

theKingPin · May 24, 2015

Could someone please explain to me how to solve this question with working, there are hints on the top however I have not been able to get an answer.

https://isaacphysics.org/questions/towel_washing_line_num?board=af02238b-0d34-49df-82be-e182fb8477d1

Thanks

InteGrand · May 24, 2015

theKingPin said:
Could someone please explain to me how to solve this question with working, there are hints on the top however I have not been able to get an answer.

https://isaacphysics.org/questions/towel_washing_line_num?board=af02238b-0d34-49df-82be-e182fb8477d1

Thanks

$We use trigonometry to find sines and cosines of the acute angle $\theta$ made by the slanted ends of the washing line and the horizontal. By symmetry, the horizontal distance of the peg to the nearest pole on the diagram is $\frac{4b-2a}{2}=2b-a$. Then from trigonometry, we have $\cos \theta = \frac{2b-a}{a}$, which means that $\sin \theta = \sqrt{1-\cos^2 \theta}=\sqrt{1-\left(\frac{2b-a}{a}\right)^2}$.$

$Now we consider the forces on the left peg $P_1$ (by symmetry, the analysis would be the same for the right peg $P_2$). Refer to the force diagrams they gave in one of the hints. There is no acceleration for $P_1$ in the vertical ($y$) direction, so $\sum \vec{F}_y =0\Rightarrow T_1 \sin \theta = mg \Rightarrow T_1 = \frac{mg}{\sin \theta}$ (resolving $T_1$ into its vertical component).$

$Now consider the horizontal forces on $P_1$. The sum of these forces is 0 too, since there is no acceleration in the horizontal ($x$) direction either, so $\sum \vec{F}_x = 0\Rightarrow T_1 \cos \theta = T_2 \Rightarrow T_2 = \frac{mg}{\sin \theta}\cdot \cos \theta = \frac{mg}{\sqrt{1-\left(\frac{2b-a}{a}\right)^2}}}\times \frac{2b-a}{a}$. Now we just substitute in values: $m=0.2\text{ kg},a=0.25\text{ m}, b=0.2\text{ m},g=9.8\text{ m s}^{-2}$, and you should get $T_2 \approx 1.47 \text{ N}$.$

theKingPin · May 24, 2015

Thanks a lot.

Physics Question (1 Viewer)

theKingPin

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InteGrand

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