Question

Consider two vectors F1 with magnitude 44 N inclined at 61 o and F2 with magnitude 98 N inclined at 139 o , measured from the positive x axis with counterclockwise positive.

Find the direction of this resultant vector (between the limits of 0 and −180◦ from the positive x-axis). Answer in units of o . Your answer must be within ± 5.0%

Answer 1

Suppose given that Force is F and it makes angle with +x-axis, then it's components are given by:

Fx = F*cos

Fy = F*sin

Using above rule:

F1 = 44 N & angle = 61 deg with +x-axis

F1x = F1*cos 1 = 44*cos 61 deg = 21.33 N

F1y = F1*sin 1 = 44*sin 61 deg = 38.48 N

F2 = 98 N & angle = 139 deg with +x-axis

F2x = F2*cos 2 = 98*cos (139 deg) = -73.96 N

F2y = F2*sin 2 = 98*sin (139 deg) = 64.29 N

Fnet = Fnet)x + Fnet)y

Fnet = (F1x + F2x) i + (F1y + F2y) j

Using above values

Fnet = (21.33 - 73.96) i + (38.48 + 64.29) j

Fnet = -52.63 i + 102.77 j

|Fnet| = sqrt (Fnet_x^2 + Fnet_y^2)

|Fnet| = sqrt ((-52.63)^2 + 102.77^2)

|Fnet| = 115.5 N

Direction of Fnet will be

Direction = arctan (Fnet_y/Fnet_x)

Direction = arctan (102.77/(-52.63)) = -62.88 deg = 62.88 deg above -ve x-axis

Direction = 180 - 62.88 = 117.12 deg with positive x-axis

Direction = 117 deg w.r.t. positive x-axis