Question

part c=d

A heavy swing door has a mass of m = 7,500 kg, a width W = 1.9 m, and a height H = 3.2 m. The door swings about a vertical axis passing through its center. The moment of inertia of this door about the vertical axis of rotation is given by I = i mw? H (a) You stand on one side and push at the outer edge of the door and perpendicular to the face of the door with a force of F= 12.0 N. Your friend pushes on the other outer edge of the door from the opposite side with the same force. What is the net torque applied to the door? N.m (b) What is the angular acceleration of the door? rad/s2 (c) Now consider a heavy door, with the same mass and dimensions as the one in part (a), that swings about a vertical axis passing through one long edge as shown in the diagram below. The moment of inertia of the door about this new axis of rotation is given by 1 = { mw?, where m is the mass and w is the width of the door.

How much force must you and your friend each apply to the free end of the door on the same side and perpendicular to the plane of the door in order to produce the same torque as that produced by both of you in part (a) above?
N

(d) What is the angular acceleration of the door?
rad/s²

Answer 1

Mass *7500x[1.972 a 12 of the door m=7,500kg width of the door w=i.am Height of the door H=3.2m mwe= MOI of the door I = 1 am I = 2256-25 kg-m? Force applied faian perpendicular distance bloo axis and force applied ý 8= ww = lig Net torque Z= 7, + Lez T= (FX) + (Fxx) T = 2 (Fxx) = 2x12x0.95= 22.8m-m Net torque Z= 22.8 fm - 0.95m 2 r =12 d- I 22.8 2256.25 @ Torque = 0.01 rad/s2 Angular acceleration of the door d=0.01 rad) 57 I'- $mw- x7500x(1.a)!= 9025 kg.mz Wet torque manga (F'xx') + Charge %= 2Fxx' = 2x1.9xF'= 3.8F' Net torque ł= 22.8N-m 3.8 F = 22.8 Force applied is f = 6N 8- W= 1.9m 8=W AllF 7 = I'd a 3200 = 2.596x103 gads Angular acceleration 2-2.526x10? rad/5 = 0.002526 Bad/ I' 9025