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x° = angle that T2 forms with the ascending vertical

F peso = 7 * 9,8 = 68,6 N;

20 / sin(180° - x°) = F peso / sin(40° + x°);

sin(180° - x°) = sinx°;

20 * sin(40° + x°) = 68,6 * sinx°;

 sin(40° + x°) = sin40° cosx° + cos40° sinx°;

20 * sin40° cosx° + 20 * cos40° sinx° = 68,6 * sinx°;

12,86 * cosx° + 15,32 sinx° = 68,6 * sinx°.

68,6 * sinx° - 15,32 sinx° = - 12,86 * cosx°;

53,28 sinx° = - 12,86 cosx°;

sinx° / cosx° = - 12,86 / 53,28;

tanx° = - 0,241;

x° = 13,6°; angle that T2 forms with the ascending vertical.

40° + 13,6° = 53,6°;

F peso/ sin53,6° = T2 /sin(180° - 40°);

T2 = F peso * sin140°/ sin53,6°;

T2 = 68,6 * 0,643 / 0,805 = 54,8 N.

Ciao.

Presumably the mass of 7 kg is hanging straight down from the junction of the two ropes.  Call the tension in the other rope T.  Lami's Theorem provides that

(7 kg)(9.8 m/s^2)/sin(40 + theta)

= (20 N)/sin(180-theta)

= T/sin(140).

(68.6 N)*sin(theta) = (20N)*sin(40 + theta), so

(68.6)sin(theta) = (20)[0.64279*cos(theta) + 0.76604*sin(theta)] =>

53.279*sin(theta) = 12.856*cos(theta) =>

tan(theta) = about 4 but you can now push some buttons and find theta.

Finally, after you have theta, you can use

(68.6 N)/sin(40 + theta) = T/sin(140)

to find the tension T in the "other" rope.