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One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of the rope as shown in the drawing. the door has a weight of 145 n and is hinged on the right. what is the maximum magnitude of f at which the door will remain at rest? hint: use the rotational equilibrium condition. include the weight of the door into the equation.

Respuesta :

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

[tex]F\times L = mg \times \frac{L}{2}[/tex]

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

[tex]mg = 145 N[/tex]

now we will have

[tex]F = \frac{mg}{2}[/tex]

[tex]F = \frac{145}{2} = 72.5 N[/tex]

so our applied force is 72.5 N