Question

Calculate the magnitude and direction of both the equilibrant and the resultant of the following three vectors: 0.4N at 60°, 0.5N at 30°, and 0.2N at 330° .The object of this experiment is to demonstrate the vector property of forces and to gain experience in the addition of vector quantities.

Answer 1

Given that :

magnitude of vec/ A = 0.4 N                              at (theta)_A = 60⁰

magnitude of vec/ B = 0.5 N                            at (theta)_B = 30⁰

magnitude of vec/ C = 0.2 N                             at (theta)_C = 330⁰

components of vec/ A on x & y-axis :

A_X = (0.4 N) Cos (60)⁰                              and                    A_Y = (0.4 N) Sin (60)⁰  

A_X = 0.2 N                                                                          A_Y = 0.34 N

components of vec/ B on x & y-axis :

B_X = (0.5 N) Cos (30)⁰                              and                    B_Y = (0.5 N) Sin (30)⁰  

B_X = 0.17 N                                                                          B_Y = 0..25 N

components of vec/ C on x & y-axis :

C_X = (0.2 N) Cos (330)⁰                              and                    C_Y = (0.2 N) Sin (330)⁰  

C_X = 0.17 N                                                                          C_Y = -0.1 N

Resultant of the vector on x and y-axis which is given as ::

R_X = A_X + B_X + C_X (0.2 N + 0.17 N +0.17 N)

R_x = 0.54 N

And

R_Y = A_Y + B_Y + C_Y (0.34 N +0.25 N - 0.1 N)

R_Y = 0.49 N

magnitude of resultant of the three vectors which will be given as ::

R = (0.54 N)² + (0.49 N)²

R = (0.2916 + 0.2401) N²

R = 0.5317 N²

R = 0.73 N

Direction will be given as :

(theta) = tan^-1 (R_Y / R_x)

(theta) = tan^-1 [(0.49 N) / (0.54 N)]

(theta) = tan^-1 (0.9074)

(theta) = 42.2⁰

The equilibrant vector has the same magnitude as the resultant but opposite direction.

(theta)_equilibrant = (180 + 42.2)⁰

(theta)_equilibrant = 222.2⁰