Question

A 19.3 kg door that is 1.19 m wide opens slowly as a pull of constant magnitude 48.5 N is exerted on the doorknob. This applied force, which is always perpendicular to the door, is necessary to balance the frictional torque in the hinges. How large must an applied force of constant magnitude be in order to open the door through an angle of 90.0° in 0.517 s?

Answer #1

= angular displacement = 90 degree = 1.57 rad

t = time taken = 0.517 sec

W_{i} = initial angular velocity = 0 rad/s

= angular acceleration

using the equation

= W_{i} t
+ (0.5)
t^{2}

1.57 = 0 (0.517) + (0.5)
(0.517)^{2}

= 11.75
rad/s^{2}

b = width of door = 1.19 m

I = moment of inertia of door = (1/3) mb^{2} = (1/3)
(19.3) (1.19)^{2} = 9.11 kgm^{2}

Torque is given as

= I

F = applied force

f = frictional force = 48.5

net force = F - f

net torque by the force

= (F - f ) b = I

(F - 48.5) (1.19) = 9.11 x 11.75

F = 138.5 N

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