Final answer:
To move the crate up the 15° incline without tipping, the magnitude of force P must be at least 205.74 N.
Step-by-step explanation:
To determine the corresponding magnitude of the force P needed to move the crate up the incline without tipping, we need to analyze the forces acting on the crate. The force of gravity can be broken down into two components: the force perpendicular to the incline and the force parallel to the incline. The perpendicular component is mg cos(theta), where m is the mass of the crate and g is the acceleration due to gravity. The parallel component is mg sin(theta). The force P should be equal to or greater than the perpendicular component in order to prevent the crate from tipping. Therefore, the magnitude of force P is at least mg cos(theta).
In this case, the mass of the crate is 22 kg and the angle of the incline is 15°. So, the magnitude of force P can be calculated as:
P ≥ 22 kg × 9.8 m/s2 × cos(15°)
P ≥ 205.74 N
Therefore, the magnitude of force P must be at least 205.74 N to move the crate up the 15° incline without tipping.