Question

The oars aren't accelerating, and they are rotating at a constant speed, so the net force and net torque on the oars are zero. an oar is 2.80 m long, and the rower pulls with a 290 n force on the handle, which is 0.92m from the pivot.

Assume that the oar touches the water at its very end. What is the drag force from the water on the oar? Assume that the oar is perpendicular to the boat, and that the force of the rower and the drag force are both perpendicular to the oar. Fdrag =

Answer 1

The drag force on the oar is found to be 95.29 N by applying the principle of torque and setting the torque due to the rower's force equal to the torque due to the drag force, with the oar at rotational equilibrium.

Step-by-step explanation:

To find the drag force on the oar, we can apply the principle of torque, which states that the net torque on a body in rotational equilibrium is zero. Since the oars are rotating at a constant speed and are not accelerating, we have:

Torque_due_to_rower's_force = Torque_due_to_drag_force

Let's calculate the torque exerted by the rower:

Torque_due_to_rower = Force_by_rower * Distance_from_pivot

Torque_due_to_rower = 290 N * 0.92 m

Torque_due_to_rower = 266.8 N·m

Now, equating this to the torque due to the drag force, we have:

Torque_due_to_drag_force = Drag_force * Length_of_oar

266.8 N·m = Drag_force * 2.80 m

Drag_force = 266.8 N·m / 2.80 m

Drag_force = 95.29 N

Hence, the drag force from the water on the oar is 95.29 N.