Question

A metal bar is in the xy-plane with one end of the bar at the origin. A force F⃗ =( 6.22 N )i^+( -2.81 N )j^ is applied to the bar at the point x= 3.58 m , y= 3.46 m .

Part 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.

Part B: What are the magnitude of the torque with respect to the origin produced by F⃗ ?

Part C: What are direction of the torque with respect to the origin produced by F⃗ ?

Answer 1

Part A: r_x, r_y = (3.58, 3.46) m

Part B: Torque, T = r x F, i.e cross product of r and F.

T = (3.58i + 3.46j + 0k) x (6.22i+(-2.81j) + 0k) = 0i + 0j - 31.58k

Hence, magnitude of torque is sqrt(0+ 0+ (31.58)²) = 31.58N

Part C: Direction of Torque is -z direction.