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Finger [1]
3 years ago
15

A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

A "door closer" is attached to the door and the top of the door frame so that the door can close on its own after being opened. When the door is open and at rest, the door closer exerts a torque of 5.2 Nm. What is the least force that you need to apply to the door to hold it open (in Newtons)?
Physics
1 answer:
olasank [31]3 years ago
5 0

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

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% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

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