Question

An archer pulls the string of her bow back with her hand with a force of 180 N. If the two halves of the string above and below her hand make an angle of 120 with each other ,what is tension in each half of the string
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Answer 1

The tension component in the direction of the X axis is:

$cos(60^{0})=\frac{T_{x}}{T}\\T_{x}=Tcos(60^{0})$

As they are two components of the tension you have to: $2T_{x}$

Horizontal sums of force.

$\sum F_{x}=F-2T_{x}\\ \sum F_{x}=0\text{ The system is in equilibrium } \\F-2T_{x}=0\\F=2Tcos(60^{0})$

Expression for tension

$T=\frac{F}{2cos(60^{0})}$

Numerically evaluating

$T=\frac{ 180Nw}{2*cos(60^{0})}=180Nw$

The tension in each half of the string is:

$\boxed{T=180Nw}$