Question

Three forces act on a body. The magnitudes and directions are as follows : ? Force A = 94 N, 24° south of west. Force B = 70 N, 40° north of east. Force C = 110 N, 36° west of north.

Calculate the magnitude and direction of A - B + C. Include a specific directional angle in your answer.

Answer 1

Arrangement of the forces has been depicted in the diagram,

Now, `vec A-vec B+vec C` can be written as `vec A+vec(-B)+vecC`

So, resultant of `|vec A+vec(-B)|= |vec D|=sqrt(94^2+70^2+2 *94 *70 cos (40-24)) = 162.43 N`

Making an angle of `tan ^-1 ((70 sin 16)/(94+70 cos 16)) = 6.82 degrees` w.r.t `94 N`

So, `vec D` makes an angle of `(40+6.82+90-36)=100.82` degrees w.r.t `vecCc`

So, we can write,

`vec A+vec(-B)+vecC = vec D + vec C`

So, `|vec D + vec C| = sqrt(162.43^2 + 110^2 + 2*162.43*110 cos 100.82)=178.26N`

Making an angle of `tan ^-1 ((162.43 sin 100.82)/(110+ 162.43 cos 100.82)) = 63.51 degrees` w.r.t `110 N` force

So,making an angle of `(36+63.51 ) = 99.51 degrees` west of north or `9.51 degrees` south of west