Consider two vectors where F1 = 53 N, F2 =
81 N, �1 = 320� , and �2 = 13� ,measured from
the positive x axis with the counter-clockwise. What is the magnitude of the resulting vector?
2 answers
1 is angle 1 and 2 is angle 2
α1 =360⁰-320⁰ =40⁰
F(x) =F1(x)+F2(x)
=F1cosα1 + F2cosα2 =
=53•cos40 +81•cos13 =40.6+78.9=119.5 N
F(y) =F1(y)+F2(y)
= - F1sinα1 + F2sinα2 =
= - 53•sin40 +81•sin13 =
= - 34.07 +18.22 = - 15.85 N
F=sqrt{F(x)²+F(y)²} =
=sqrt{119. 5²+15.85²} =120.5 N
F(x) =F1(x)+F2(x)
=F1cosα1 + F2cosα2 =
=53•cos40 +81•cos13 =40.6+78.9=119.5 N
F(y) =F1(y)+F2(y)
= - F1sinα1 + F2sinα2 =
= - 53•sin40 +81•sin13 =
= - 34.07 +18.22 = - 15.85 N
F=sqrt{F(x)²+F(y)²} =
=sqrt{119. 5²+15.85²} =120.5 N
1 answer