# Physics Question (1 Viewer)

#### InteGrand

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

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

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

Thanks a lot.