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Three Vectors, F1=20 N, F2=40 N, F3=60 N have 10de
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Answer #1

you have asked mulitple questions .as per HomeworkLib guidelines i can only answer one.Please raise different question for others.

For F1:
Fx = 20 cos 10 =19.6N
Fy = 20sin 10 =3.47N

For F2:
Fx = 40 cos (20) =37.5N
Fy = 40 sin (20) =13.6N

For F3:
Fx = 60 cos 30 =25.9N
Fy = 60 sin 30=30N

So sum of:
Fx = 19.6+37.5+25.9=83N
Fy = 47.07N
so answer is E

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Three Vectors, F1=20 N, F2=40 N, F3=60 N have 10degree^5 x 20degree ^6 and 20degree and...
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