[tex]F=38.57\text{ Newtons}[/tex]
Explanation
Step 1
free body diagram
Step 2
now, let's analyze the forces
a) in y
Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0)
so, as the leg is in rest
[tex]\begin{gathered} \sum ^{\square}_{\text{ y}}=_{}0 \\ T_1\sin 30-T_2sen\text{ 30=0} \end{gathered}[/tex]
let
m= 2.27 Kg , so
[tex]\begin{gathered} \text{weigth}=m\cdot g \\ \text{weigth}=2.27\operatorname{kg}\cdot9.81m/s^2 \\ \text{weigth}=22.2687\text{ N} \end{gathered}[/tex]
hence
[tex]\begin{gathered} T_2=weigth \\ T_2=22.2687 \end{gathered}[/tex]
[tex]\begin{gathered} T_1\sin 30-T_2sen\text{ 30=0} \\ T_1\sin 30-(22.2687)sen\text{ 30=0} \\ T_1\sin 30-(22.2687)sen\text{ 30=0} \\ T_1\sin 30-11.13435=0 \\ \text{add}11.13435\text{ in both sides} \\ T_1\sin 30-11.13435+11.13435=0+11.13435 \\ T_1\sin 30=11.13435 \\ \text{divide both sides by sin 30} \\ \frac{T_1\sin30}{\sin\text{ 30}}=\frac{11.13435}{\sin \text{ 30}} \\ T_1=22.2687\text{ Newtons} \end{gathered}[/tex]
b) now in x ( horizontally)
[tex]\begin{gathered} \sum ^{\square}_{\text{ x}}=_{}0 \\ -F+T_1cos30+T_2cos\text{ 30=0} \\ \text{add F in both sides} \\ -F+T_1cos30+T_2cos\text{ 30+F=0}+F \\ F=T_1cos30+T_2cos\text{ 30} \\ \text{replace} \\ F=(22.2687)cos30+(22.2687)_{}cos\text{ 30} \\ F=38.57\text{ Newtons} \end{gathered}[/tex]
therefore, the answer is
[tex]F=38.57\text{ Newtons to the left}[/tex]
I hope this helps you
