Question

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)?

Answer 1

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

Answer 2

The smallest force that you need to use to the door to hold it open is mathematically given as

F= 5.71 N

What is the least force that you need to apply to the door to hold it open?

Question Parameters:

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;

the door closer exerts a torque of 5.2 Nm.

Generally, the equation for the Torque  is mathematically given as

T= r x F

Therefore

T= r F sin\theta

F= T/ (r * sin\theta)

F= 5.2/ (0.91 * 1)

F= 5.71 N

In conclusion, the least force that you need to apply

F= 5.71 N

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