Question

A muscle builder
holds the ends of a massless rope. At the center of the rope, a 15-ko
shown. What is the tension in the rope if the angle teta
one tension in the rope if the angle o in the drawing is 4.5°? (sin 4.50 -
----
in 4.5° = 0.08)
2.
A) 1900 N
B) 940 N
C) 470 N D) 230 N E)150N​

Answer 1

(B )940 N.............

Answer 2

Given:

The center of the rope in the $15Kg$ the tension in the rope if the angle.

To find:

Find the angle $o$ in the drawing.

Explanation:

$2Tsin$∅$=mg$

$T=\frac{mg}{2sin0}$

$=\frac{15*10}{2*0.08}$

$=\frac{15*10^{5}}{2*8} *100$

$T=\frac{7500}{8} N$

$T=940N$

Answer:

Therefore, The angle $o$ in the drawing in the $940N$.