Final answer:
The torque applied to the metal bar when a force F=(7.93 N)i^ + (-2.68 N)j^ is applied at the point (3.81 m, 3.50 m) is calculated using the cross product of the position vector and the force, resulting in approximately 17.54 Nm.
Step-by-step explanation:
The student asked: What is the torque applied to a metal bar in the xy-plane with one end at the origin, when a force F=(7.93 N)i^ + (-2.68 N)j^ is applied at the point x=3.81 m, y=3.50 m?
Torque (T) is calculated as the cross product of the position vector (r) where the force is applied and the force (F) itself. In this case, the position vector can be written as r = (3.81 m)i^ + (3.50 m)j^, and since the bar is in the xy-plane and the force is applied parallel to the plane, the angle (θ) between them is 90 degrees, so sin θ will be 1. The torque can then be calculated as:
T = r x F = |r| |F| sin θ
Where |r| and |F| are the magnitudes of the position vector and the force respectively, and θ is the angle between the two.
We can calculate torque explicitly using:
T = (3.81 m * -2.68 N) - (-3.50 m * 7.93 N) = -10.2128 Nm + 27.755 Nm = 17.5422 Nm
Hence, the torque applied to the bar is approximately 17.54 Nm.