Question

A metal bar is in the xy-plane with one end of the bar at the origin. A force F =( 6.39 N )i +( -3.16 N )j is applied to the bar at the point x = 2.29 m, y = 3.19 m.

A) What is the position vector r⃗  for the point where the force is applied?

Enter the x and y components of the radius vector separated by a comma.*

B) What is the magnitude of the torque with respect to the origin produced by F⃗ ?

Express your answer with the appropriate units.

C) What is the direction of the torque with respect to the origin produced by F⃗ ?

What is the direction of the torque with respect to the origin produced by ?

+x-direction

+y-direction

+z-direction

−x-direction

−y-direction

−z-direction

Answer 1

a) r = 2.29, 3.19

b) Use torque formula and take cross product

torque = r x F

torque = (2.29, 3.19) x (6.39, -3.16, 0)

torque = (0, 0, -27.62)

magnitude of torque = 27.62

c) -z-direction b/c the third component of the torque vector is negative