Asked by Josephine
                Two forces of magnitude 8Nand 5N act at an angle 60°to each other. The magnitude of their resultant is
            
            
        Answers
                    Answered by
            Damon
            
    Well, there are two waays:
1. law of cosines
Draw it as a parallelogram with two vectors 60 degrees apart meeting at point A
AD in x direction length 5
AB in direction 60 deg above x axis length 8
Then finish it with BC parallel to AD and DC parallel to AB
we want length AC
well angle BAD = angle BCD = 60
60 + 60 + 2* angle ABC = 360
so angle ABC = 120 degrees
now law of cosines
AC^2 = 8^2 + 5^2 - 2 * 8 * 5 cos 120
AC^2 = 64 + 25 - 80 (-.5) = 64 + 25 + 40 = 129
AC = sqrt (129) = 11.36
Being a physicist though I do it by components
Fx = 5 + 8 cos 60 = 9
Fy = 8 sin 60 = 6.93
|F| = sqrt (81 + 48) = sqrt 129 = 11.36
remarkable same old thing
    
1. law of cosines
Draw it as a parallelogram with two vectors 60 degrees apart meeting at point A
AD in x direction length 5
AB in direction 60 deg above x axis length 8
Then finish it with BC parallel to AD and DC parallel to AB
we want length AC
well angle BAD = angle BCD = 60
60 + 60 + 2* angle ABC = 360
so angle ABC = 120 degrees
now law of cosines
AC^2 = 8^2 + 5^2 - 2 * 8 * 5 cos 120
AC^2 = 64 + 25 - 80 (-.5) = 64 + 25 + 40 = 129
AC = sqrt (129) = 11.36
Being a physicist though I do it by components
Fx = 5 + 8 cos 60 = 9
Fy = 8 sin 60 = 6.93
|F| = sqrt (81 + 48) = sqrt 129 = 11.36
remarkable same old thing
                    Answered by
             henry2,  
            
    Fr = 8[60] + 5[0o].
Fr = (8*cos60+5*cos0) + (8*sin60+5*sin0)I,
Fr = 9 + 6.93i = 11.4N[37.6o]
    
Fr = (8*cos60+5*cos0) + (8*sin60+5*sin0)I,
Fr = 9 + 6.93i = 11.4N[37.6o]
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