As shown in the figure, three force vectors act on object. The magnitudes of the

Question

As shown in the figure, three force vectors act on object. The magnitudes of the forces as shown in the figure are F_1 = 80.0 N, F_2 = 60.0 N, and F_3 = 40.0 N, where N is the standard unit of force. The resultant force acting on the object is given by 20.0 N at an angle 34.3 degree with respect to *x-axis. 180 N at an angle 60.0 degree with respect to *x-axis 60.0 N at an angle 90.0 degree with respect to *x-axis 35.5 N at an angle 34.3 degree with respect to *x-axis 40.0 N at an angle 60.0 degree with respect to *x-axis The corresponding of vector A vector are A_x = middot 3 4 and A_y = - 5.5, and the components of vector B vector are given are B_x = -3.1 and B_y = -4. What is the magnitude of the vector B vector - A vector 6.7 1.4 45 9.3 1.3

Explanation / Answer

Total force in the x - direction -

Fx = F1*cos30 - F3 = 80*cos30 - 40 = 29.3 N

and, total force in the y - direction -

Fy = F2 - F1*sin30 = 60 - 80*sin30 = 60 - 40 = 20 N

So, the resultant force, F = sqrt[Fx^2 + Fy^2] = sqrt[29.3^2 + 20^2] = 35.5 N

Angle, theta = tan^-1(Fy/Fx) = tan^-1(20 / 29.3) = 34.3 deg

So, fourth option is the correct answer.

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