Question

A trapeze artist on a rope is momentarily held to one side by a partner on a platform The rope makes an angle of 26.0 with the vertical. Insuch a condition of static equilibrium, what is the magnitude of the horizontal force being applied by the partner? The weight of the artist is 7.61 10 N

Answer 1

Answer:

$F_t=371.16N$

Explanation:

The weight 7.61x10^2N

The angle 26.0

To determine the magnitude of horizontal force applied

$tan(\alpha)=\frac{F_t}{F}$

$F_t=F*tan(\alpha)$

$F_t=7.61x10^2N*tan(26.0)$

$F_t=371.16N$