Final answer:
The moment, or torque, of a force about a pivot point is calculated by the product of the force and the perpendicular distance from the pivot to the line of action of the force. The torque exerted is 375 N·m when a 250 N force is applied perpendicularly to a 1.50 m radius. If friction is considered, the net torque is the applied torque minus the frictional torque.
Step-by-step explanation:
To determine the moment of forces about the axle ab, we must first understand that the moment or torque caused by a force is given by the product of the force and the perpendicular distance from the point of rotation to the line of action of the force. In the provided reference information, the torque is calculated using a force that acts perpendicular to the radius, meaning that the sine of the angle (theta) is 1 since theta is 0 degrees (sin 0 = 1). Therefore, torque (T) is simply the product of the radius (r) and the force (F).
The torque calculation in the question is demonstrated as follows:
T = rF sin ϴ = (1.50 m)(250 N) = 375 N·m.
This indicates that a torque of 375 N·m is exerted about the axle.
To consider angular acceleration (assuming negligible friction), one needs to use Newton's second law for rotation, which connects torque (Τ), moment of inertia (I), and angular acceleration (a): Τ = Ia. If friction is present, then the net torque would be the applied torque minus the torque due to friction. For example, if an opposing frictional force of 20.0 N is at a radius of 1.50 cm from the pivot, the frictional torque would be 0.015 m multiplied by 20.0 N, which needs to be subtracted from the applied torque to determine the net torque. This net torque is then used to determine the angular acceleration.