the 3 forces shown act on a object is in equilibrium. F1= 50N (E 60 degrees N), F2= 60N (E 30 degrees S ) Calculate the magnitude of the force F3 and the angle???? How to solve??

Answer 1

F1+F2 = 50N[60o] + 60N[330o]

X = 50*Cos60 + 60*Cos330 = 73.3 N.
Y = 50*sin60 + 60*sin330 = 8.30 N.

(F1 + F2) +F3 = 0
73.3 + 8.3i + F3 = 0
F3 = -73.3 - 8.3i

Tan Ar = Y/X = -8.3/-73.3 = 0.11326
Ar = 6.46o N of E = Reference angle(Q1).
A = 6.46 + 180 = 186.46o(Q4) = 6.46o S of W

F3 = X/Cos A = -73.3/Cos186.46 = 73.8 N.
[186.46o] = 73.8N.[6.46o] S of W

Answer 2

Fnetx = 0 (equilibrium) = T1x + T2x - T3x
Fnety = 0 (equilibrium) = T1y - T2y - T3y
T1
x component = 60cos * 50 = 25
y component = 60sin * 50 = 43.3
T2
x component = 30cos * 60 = 51.96
y component = 30sin * 60 = -30
T3
x component = 0 = T1x + T2x + T3x = 25 + 51.96 - T3x
T3x = 76.96 N
y component = 0 = T1y - T2y - T3y = 43.3 - 30 - T3y
T3y = 13.3 N
using pythagorean theorem: T3= √T3x^2+T3y^2
T3 = 78.1 N
Find angle using Tanθ = T3y/T3x
θ = 9.8°
Therefore T3 = 78.1 N [W 9.8°S]