A 19.2-kg box rests on a frictionless ramp with a 16.1° slope. The mover pulls on a rope attached to the box to pull it up the incline. If the rope makes an angle of 43.6° with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp?
54
To determine the minimum force required to move the box up the ramp, we need to analyze the forces acting on the box.
1. Resolve the force of gravity into two components: one parallel to the ramp and one perpendicular to the ramp. The component parallel to the ramp is given by F_parallel = m * g * sin(theta), where m is the mass of the box (19.2 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the ramp (16.1°).
F_parallel = 19.2 kg * 9.8 m/s^2 * sin(16.1°)
F_parallel ≈ 51.92 N
2. Determine the force required to counteract the component of gravity parallel to the ramp. This force is equal in magnitude but opposite in direction to the parallel component of gravity. Therefore, the minimum force F required to move the box up the ramp is equal to the parallel component of gravity:
F = F_parallel
F ≈ 51.92 N
Thus, the smallest force the mover will have to exert to move the box up the ramp is approximately 51.92 N.