What force must be applied to push a carton weighing 220 N up a 20° incline, if the coefficient of kinetic friction is 0.39? Assume the force is applied parallel to the incline and the velocity is constant
forces: friction down, pushing up, gravity down the plane.
weight down plane=mgsin20
friction down plane: mg*mu*cos20
pushingforce-mgsin20-mg*mu*cos20=ma=0
To find the force that must be applied to push the carton up the incline, we need to consider the forces acting on the carton.
1. Weight (mg): The weight of the carton creates a force vertically downward due to gravity. Its magnitude is given by the mass of the carton (m) multiplied by the acceleration due to gravity (g). In this case, the weight is 220 N.
2. Normal force (Fn): The normal force is the force exerted by the surface on the carton perpendicular to the incline. It is equal in magnitude and opposite in direction to the weight of the carton when the carton is on a horizontal surface, but it varies when the carton is on an inclined plane. The normal force can be calculated by multiplying the weight of the carton by the cosine of the angle of inclination. In this case, the angle of inclination is 20°, so the normal force is 220 N * cos(20°).
3. Kinetic friction (fk): The kinetic friction force opposes the motion of the carton and acts along the incline. The magnitude of the kinetic friction force can be calculated as the coefficient of kinetic friction (μk) multiplied by the normal force. In this case, the coefficient of kinetic friction is 0.39, so the kinetic friction force is 0.39 * (220 N * cos(20°)).
Since the force being applied is parallel to the incline and the velocity is constant, the applied force (Fapplied) must overcome the kinetic friction force. Therefore, the force that must be applied to push the carton up the incline is equal to the kinetic friction force:
Fapplied = fk = 0.39 * (220 N * cos(20°))
Simplifying the equation and calculating the result will give you the answer.