Question

A metal bar is in the xy-plane with one end of the bar at the origin. A force F⃗ =( 6.70 N )i^+( -2.61 N )j^ is applied to the bar at the point x = 2.00 m , y = 3.44 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) The position vector (2.00 m, 3.44 m). B) The magnitude of the torque produced by the force is 6.70 N. C) The direction of the torque with respect to the origin is in the negative z-direction.

Step-by-step explanation:

A) The position vector r for the point where the force is applied is (2.00 m, 3.44 m).

B) To calculate the magnitude of the torque produced by the force, we can use the formula T = rF sin theta, where T is the torque, r is the distance from the pivot point to the point where the force is applied, F is the magnitude of the force, and theta is the angle between the force and the vector from the point of application to the pivot point. Since the force is applied perpendicularly to the lever arm, the angle theta is 90 degrees, and sin 90 degrees = 1. Therefore, the magnitude of the torque is simply the magnitude of the force. In this case, the magnitude of the torque is 6.70 N.

C) The direction of the torque with respect to the origin produced by the force can be determined by the right-hand rule. If you curl your fingers in the direction of the force vector, your thumb points in the direction of the torque. In this case, the direction of the torque with respect to the origin is in the negative z-direction.