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18 January, 09:04

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis 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 ffor which the door will remain at rest?

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  1. 18 January, 09:11
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    Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20
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