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3 years ago

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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 th

e 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.

1 answer:

Vlada [557]3 years ago

7 0

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

$F\times L = mg \times \frac{L}{2}$

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

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

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