A Tightrope Mechanic Problem

Let $T_1$ = the tension of the first side of the rope, $T_2$ = the tension of the second side of the rope, and $w$ = the weight.

$T_{1x} = T_1 \cdot cos(3.7^\circ) = 1.0 \cdot T_1$

$T_{1y} = T_1 \cdot sin(3.7^\circ) = 0.065 \cdot T_1$

$T_{2x} = T_2 \cdot cos(176.3^\circ) = -.9979 \cdot T_2$

$T_{2y} = T2 \cdot sin(176.3^\circ) = .06453 \cdot T2$

$w_x = (675 Newtons) \cdot cos(270^\circ) = 0$

$w_y = (675 Newtons) \cdot sin(270^\circ) = -675$ Newtons

Now that we have everything split up into x- and y- components, we can some the forces in each dimension and make sure that they equal zero. (Since the acrobat is staying at the same height the whole time and the rope isn't going up or down, that means we have a Translational/Static Velocity. We are using Newton's second law here ( $\Sigma F = ma$ ) and so know we need to make sure that the sum of the forces in each dimension = 0

In the x- dimension we have:

$1.0 \cdot T+1 + -.9979 \cdot T_2 + 0 = 0$

Although we can't solve anything here, we can rearrange the equation:

$T_1 = 1.0 \cdot T_2$

Now we can go to the y- dimension:

$.065 \cdot T_1 + .06453 \cdot T_2 + -675$ Newtons $= 0$

Since this equation had two unknowns in it, we can use the equation from the y- dimension to replace $T_2$ with $1.0 \cdot T_1$ :

$.065 \cdot T_1 + .06453 \cdot (1.0 \cdot T_1) + -675$ Newtons $= 0$

$T_1 = {{675 Newtons} \over {.130}} = 5.19 \times 10^3$ or 5190