Question

1) find the vector sum of the three vectors A=0.98N (2cm) at 30˚, B=1.96N (4cm) at 90˚, C=2.94 (6cm) at 225˚. Record the results in graphical (head to tail) and analytical methods.

2) Add the following four displacement vectors using the analytic method. report the x and y components of the resultant, the magnitude of the resultant, and the direction of the resultant. show all work.

A= 5.0 m at 35˚

B= 7.0 m at 90˚

C= 3.0 m at 225˚

D= 2.4 m at 347˚

3)What possible sources of error can you identify for the graphical method of vector addition?

4) What possible sources of error can you identify for the analytical method of vector addition?

5) what possible sources of error can you identify for the experimental method of vector addition?

6) of the graphical and analytical methods, which one do you consider to be more accurate? Why?

7) Why is it not possible to experimently determine the esultant vector directly from the force table?

Answer 1

(1) three vectors are given below as :

= 0.98N (0.02m) = (0.0196 Nm)          at _A = 30⁰

= 1.96N (0.04m) = (0.0784 Nm)                         at _B = 90⁰

=2.94 (0.06m) = (0.1764 Nm)                             at _C = 225⁰

on the x-axis : R_X = (A Cos _A + B Cos _B + C Cos _C)

R_X = [(0.0196 Nm) Cos (30⁰) + (0.0784 Nm) Cos (90⁰) + (0.1764 Nm) Cos (225⁰)]

R_X = -0.107 Nm

on the y-axis : R_Y = (A Sin _A + B Sin _B + C Sin _C)

R_Y = [(0.0196 Nm) Sin (30⁰) + (0.0784 Nm) Sin (90⁰) + (0.1764 Nm) Sin (225⁰)]

R_Y = -0.036 Nm

magnitude of the resultant vector R = (-0.107 Nm)² + (-0.036 m)²

R = 0.0126 Nm

R = 0.112 Nm

(2) Add the following four displacement vectors using the analytic method.

R = A + B + C + D

components of the resultant vector on x & y axis ,

on x-axis :   R_X = A_X + B_X + C_X + D_X   

R_X = (A Cos _A + B Cos _B + C Cos _C + D Cos _D) { eq.1 }

where, A = magnitude of vector A = 5 m       at _A = 35⁰

B = magnitude of vector B = 7 m                   at _B = 90⁰

C = magnitude of vector C = 3 m                  at _C = 225⁰

D = magnitude of vector D = 2.4 m                  at _D = 347⁰

inserting all these values in above eq.

R_X = [(5m) Cos (35⁰) + (7m) Cos (90⁰) + (3m) Cos (225⁰) + (2.4m) Cos (347⁰)]

R_X = 4.31 m

And   R_Y = A_Y + B_Y + C_Y + D_Y   

R_Y = (A Sin _A + B Sin _B + C Sin _C + D Sin _D)                                                                           { eq.2 }

inserting the values in eq.2,

R_Y = [(5m) Sin (35⁰) + (7m) Sin (90⁰) + (3m) Sin (225⁰) + (2.4m) Sin (347⁰)]

R_Y = 7.2 m

magnitude of the resultant vector, R = R_X² + R_Y²                                                                             { eq.3 }

inserting the values in eq.3,

R = (4.31 m)² + (7.2 m)²

R = 70.41 m

R = 8.39 m

And its direction is given as, = tan^-1 (R_Y / R_X)                                                                             { eq.4 }

inserting the values in eq.4,

= tan^-1 [(7.2 m / 4.31 m)]

= tan^-1 (1.67)

= 59.1 degree