A professor\'s office door is 0.91 m wide, 2.9 m high, and 4.0 cm thick; has a m

Question

A professor's office door is 0.91 m wide, 2.9 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots of frictionless hinges. A "door closer" is attached to the door and the top of the doorframe. When the door is open and at rest, the door closer exerts a torque of 5.21 N*m.

a.) where should you hold the door in order to exert the least amount of force while keeping it open?

b.) what is the value of the least amount of force to hold the door open?

c.) what is the moment of the inertia of the door about its axis through the hinges?

d.) what is the angular acceleration of the door once you let go of it?

Show work please! Thanks!

Explanation / Answer

mass m = 25 kg ,
l = 0.91 m ,
h = 2.9 m
t = 0.04 m

moment of inertia = m*l^2 /3 + m*t^2 /3

 = (25/3)*(0.91^2 + 0.04^2)

= 6.9 kg.m^2

 

b) torque = 5.21 = I*alpha

=> alpha = 5.21/6.9 rad/s^2

            =  0.75 rad/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.