The Cable Has A Tension If The Acrobat Falls

$$ Assuming the system at equilibrium state, then by applying Newton's first law on the vertical axis, we obtain $$ $\begin{aligned} \sum F&=0\\ 2T\sin\theta-mg&=0\\ 2T\frac{75}{\sqrt{75^2+9,000^2}}&=mg\\ T&=\frac{mg}{2}\cdot\frac{\sqrt{75^2+9,000^2}}{75}\\ &=\frac{50\cdot9.8}{2}\cdot\frac{\sqrt{75^2+9,000^2}}{75}\\ &\approx29,401.021\;N\\ &=\boxed{29.401021\;kN} \end{aligned}$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

break the tension into its two components. The cos term cancels each other while 2T sin (theta) = mg.. now since sin (theta ) almost = theta .... 2T theta = 50 * 9.8 . on solving T = 29.4

Assume that the acrobat is standing steadily on the rope(or sth you name it). so we have : T1 + T2 + P = 0 (all are vectors) since T1=T2 then T1 = T2 = 1/2P=P1 (all vectors) now we need to calculate the angle between the T1 and P1 . using cosine you can get the angle is aprrox 89.5225 degree. now we have P1/T1 = cosA <=> 1/2P /T1 = cosA <=> 1/2mg / T1 = cos(89.5225) =>Solve for T1 .(sorry for bad English and not having the diagram)

2T*(75 x 10 to the power -3 / 3) = 50 x 9.8

In equilibrium, the horizontal components cancel out each other. the net upward force is equal to the net downward force hence Tension = (Mg)/(2 cosx)....cosx = 0.075/9 T= 29.4 KN

1 Take 7.5mm as 7.5E-2 m 2. Distribute the force of its weight on both the strings so you would get 245N. 3. Make FBD of it and equate 245 with Tsin 4. You can write sinangle as 7.5E-2/9 and then equate it so it would be app. 29.4.