Answer:
net force is 40.8 Lb in direction 50.8 degree South of West
Explanation:
As we know that the three forces are given as
F1 = 30 lb North East
F2 = 70 lb South
F3 = 50 lb at 20 degree North of West
Now we can write in vector form as
[tex]F_1 = 30 sin45 \hat i + 30 cos45 \hat j[/tex]
[tex]F_1 = 21.2 \hat i + 21.2 \hat j[/tex]
[tex]F_2 = -70 \hat j[/tex]
[tex]F_3 = -50 cos20\hat i + 50 sin20 \hat j[/tex]
[tex]F_3 = -47\hat i + 17.1 \hat j[/tex]
now the net force due to above all
[tex]F = F_1 + F_2 + F_3[/tex]
[tex]F = (21.2 - 47)\hat i + (21.2 - 70 + 17.1)\hat j[/tex]
[tex]F = -25.8 \hat i - 31.7 \hat j[/tex]
magnitude of net resultant force is
[tex]F = \sqrt{25.8^2 + 31.7^2}[/tex]
[tex]F = 40.8 Lb[/tex]
direction of the force is along
[tex]\theta = tan^{-1}(\frac{31.7}{25.8})[/tex]
[tex]\theta = 50.8 degree[/tex]
So net force is 40.8 Lb in direction 50.8 degree South of West
