A metal bar is in the xy-plane with one end of the bar at the origin. A force F=
ID: 1904682 • Letter: A
Question
A metal bar is in the xy-plane with one end of the bar at the origin. A force F= ( 7.25N ) i + ( -3.21N )j is applied to the bar at the point x= 2.57m , y= 3.41m . Part A What is the position vector ec r for the point where the force is applied? Enter the x and y components of the radius vector separated by a comma. Part B What are the magnitude of the torque with respect to the origin produced by F? Express your answer with the appropriate units. Part C What are direction of the torque with respect to the origin produced by F?Explanation / Answer
a) (2.57,3.41) b) torque = rxF = rx Fy - ry Fx = 2.57*7.25 - 3.41*-3.21=29.58 Nm in the k direction c) since torque is in the k direction the angle is 90 degrees