Question

The three displacement vectors in the drawing have magnitudes of A = 4.95 m, B = 5.93 m, and C = 3.59 m. Find the resultant
((a) magnitude and (b) directional angle) of the three vectors by means of the component method. Express the directional angle as an angle above the positive or negative x axis. A=2nd quadrant 20 degrees B= 1st quadrant 60 degrees C=on Y axis at 270 degrees

Answer 1

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

Rx = R*cos

Ry = R*sin

Now we need to find A + B + C

Using above rule:

A = 4.95 m, direction 20 deg above -ve x-axis = 4.95 m, direction 160 deg above +ve x axis

Ax = 4.95*cos 160 deg = -4.651 m

Ay = 4.95*sin 160 deg = 1.693 m

B = 5.93 m, direction 60 deg above +ve x-axis

Bx = 5.93*cos 60 deg = 2.965 m

By = 5.93*sin 60 deg = 5.136 m

C = 3.59, direction 270 deg with +ve x-axis

Cx = 3.59*cos 270 deg = 0 m

Cy = 3.59*sin 270 deg = -3.59 m

Now

R = A + B + C

R = (Ax + Bx + Cx) i + (Ay + By + Cy) j

R = (-4.651 + 2.965 + 0) i + (1.693 + 5.136 - 3.59) j

R = -1.686 i + 3.239 j

So Magnitude of R will be

|R| = sqrt (Rx^2 + Ry^2)

|R| = sqrt ((-1.686)^2 + (3.239)^2)

|R| = 3.65 m

Direction will be given by:

theta = arctan (Ry/Rx)

theta = arctan (3.239/(-1.686))

theta = -62.5 deg = 62.5 deg angle with above negative x-axis

Since x < 0 and y > 0, then vector will be in 2nd quadrant

Let me know if you've any query.