Final answer:
Each hin-ge exerts a horizontal force of 165 N on the door.
Step-by-step explanation:
To find the horizontal components of the force exerted on the door by each hin-ge, we need to divide the weight of the door evenly between the two hing-es. The weight of the door is 330 N, so each hin-ge will support 330 N / 2 = 165 N.
Next, we need to determine the horizontal component of the force exerted by each hin-ge. Since the door's center of gravity is at its center and the hin-ge is 0.50 m from the top and 0.50 m from the bottom, the horizontal distance between the hin-ge and the center of gravity is the same for both hin-ges.
Using the formula for torque, τ = force × distance, we can calculate the torque exerted by the weight of the door about each hin-ge. Since the torque is equal to the force times the perpendicular distance from the axis of rotation to the line of action of the force, and the horizontal component of force is perpendicular to the distance, we can set up the following equation: torque = force × distance
Solving for the horizontal component of force, we have: force = torque / distance
By plugging in the torque value for each hin-ge and the distance value, we get:
force = 165 N × 0.50 m / 0.50 m = 165 N
Therefore, the horizontal components of force exerted on the door by each hin-ge are both 165 N.