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LekaFEV
1 month ago
9

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. T?

Physics
2 answers:
ValentinkaMS [3.4K]1 month ago
8 0
-- The torque applied to the door is (force) x (distance from the hinge) = (5 N) x (0.8 m) = 4 N-m. -- Torque = (moment of inertia) x (angular acceleration) 4 N-m = (moment of inertia) x (2 /s²) Moment of inertia = (4 N-m) / (2 /s²) = (4 kg-m²/s²) / (2 /s²) = 2 kg-m². Keep in mind that the mass of the person pushing does not play a role; it is only the force applied perpendicularly to the door. It could even be a mosquito exerting enough force of 5 N (approximately 18 ounces).
Keith_Richards [3.2K]1 month ago
6 0
The door’s moment of inertia around the hinges is defined further. Since the force is applied at a right angle to the door, while it is hinged at one end, this force will create torque within the door. The torque resulting from the applied force is expressed as: Here, is the applied force and is the distance from the rotation axis where the force is applied. Substitute for and for in the above equation. The torque acting on the door gives it angular acceleration, represented in relation to the door's angular acceleration as: Here, is the door's moment of inertia and represents the angular acceleration. By substituting the values of and into the previous equation, the moment of inertia for the door about the hinge is determined.
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