Question

Tran ties a string to a doorknob 0.70 m from the hinge as shown in the figure. The door is 0.80 m in length (l) and its mass (m) is 15 kg. The moment of inertia for the door is given by: Idoor = 1/3 ml2 . At the instant shown, the Robbie applies a 7.0 N force to the string. i. What is the torque on the door?Determine the angular acceleration of the door just as Tran pulls the string. If a doorstop were placed halfway between the hinge and outer edge of the door, what force would it apply to the door to keep it from moving?

Answer 1

Part A)

Torque = Fr = (7)(.7) = 4.9 Nm

Part B)

Fr = I(alpha)

4.9 = .333(15)(.8)²(alpha)

alpha = 1.53 rad/s²

Part C)

Fr = 4.9

F(.4) = 4.9

F = 12.25 N

Answer 2

Torque = F.r = 7(0.7) = 4.9 Nm

Angular acceleration = 4.9 / I = 4.9 / [ (1/3)ml² ] = 4.9 / [ (1/3)15(0.8)²] = 1.53125 s^-2

Distance of the doorstop from the hinge = (0.7+0.8)/2

F.(0.7+0.8)/2 = 4.9

F = 4.9(2)/1.5 = 6.533