Final answer:
The moment of force P about point B is calculated as 49.4 N·m, indicating a counterclockwise torque. To determine the moment about point A, trigonometry must be used to find the total perpendicular distance from A to the force, considering the angle of 48° between portions AB and BC of the bar.
Step-by-step explanation:
The question involves calculating the moments of a force about two points on a bent bar, which is a fundamental concept in Physics, specifically in the area of statics and dynamics within mechanics. To find the moment of a force about a point, you can use the equation M = rFsin(θ), where M is the moment, r is the distance from the point to the line of action of the force, F is the magnitude of the force, and θ is the angle between the position vector r and the force vector.
For the moment about point B, since the force P (26 N) is applied perpendicular to BC, the angle (θ) is 90° and sin(θ) = sin(90°) = 1. Thus, the moment about point B is simply the product of the force and the distance, which is MB = 1.9 m * 26 N = 49.4 N·m, and it is positive indicating a counterclockwise direction.
To find the moment about point A, you must consider the additional distance of portion AB, which also forms a 48° angle with portion BC as given in the problem statement. This requires using trigonometry to find the perpendicular distance to the line of action of force P. The total distance from A to the line of action of P is the sum of the length of AB and the perpendicular component of BC, which can be calculated using the sin function for the 48° angle. The total perpendicular distance from A to the force P can be found using the relationship from trigonometry: rA = AB + BC cos(48°). The moment about point A is thus MA = rA * 26 N * sin(90°).