A force of 80 N Is applied to the end of a 1.20 m door at an angle of 60 degrees

Question

A force of 80 N Is applied to the end of a 1.20 m door at an angle of 60 degrees to the horizontal. Calculate the net torque on the door assuming that the force shown above is the only force acting on it If the door's rotational inertial is 15 kg m^2, what is the angular acceleration of the door. A 45 N force is applied by a doorstop to balance the torque caused by the force shown in the figure. What direction should A force of 80 N Is applied to the end of a 1.20 m door at an angle of 60 degrees to the horizontal. Calculate the net torque on the door assuming that the force shown above is the only force acting on it If the door's rotational inertial is 15 kg m^2, what is the angular acceleration of the door. A 45 N force is applied by a doorstop to balance the torque caused by the force shown in the figure. What direction should the force be applied to maximize the stopping torque and how far from the hinges should it be applied so that it perfectly counters the 80 N force?

Explanation / Answer

Torque = F X r
Torque = 80 * sin(60) * 1.2 Nm
Torque = 83.1 Nm

Torque = Inertia * Angular Acceleration
Angular Acceleration = Torque/Inertia
Angular Acceleration = 83.1/ 15 m/s^2
Angular Acceleration = 5.54 rad/s^2

Direction of the Force should be perpendicular to the door, As that way we can maximise the Force and utilise it completely.
Let the Force be applied x meter from hinges -
45 * x = 80 * sin(60) * 1.2
x = 1.85 m from Hinges.

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