The mechanism shown in the figure is used to raise a crate of supplies from a sh

Question

The mechanism shown in the figure is used to raise a crate of supplies from a ship's hold. The crate has total mass 46.0kg . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.220m and a moment of inertia = 2.70 kg*m^2 about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.120m , the cylinder turns, and the crate is raised.

What magnitude of the force applied tangentially to the rotating crank is required to raise the crate with an acceleration of 0.810m/s^2 ? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

Explanation / Answer

Mass of the crate ( m )= 46.0 kg

Radious   ( R ) = 0.220 m

moment of inertia ( I ) = 2.70 kgm^2

Radious of the vertical circle ( r ) = 0.120 m

acceleration   (a) = 0.810 m/s^2

the tension in the rope is given by

F = m( a+ g) = 46.0 kg( 9.8 m/s^2 + 0.810 m/s^2) = 488.06 N

angular acceleration of the cylinder = a/ R = 0.810 / 0.22 = 3.681 rad/s^2

so, the net torque on the cylinder (T ) = I ( angular acceleration )

                                                          = (2.7 kgm^2 )(3.681 rad/s^2 ) = 9.940 Nm

thus, the torque supplied by the crank = T + F .r

                                                            = 9.40 Nm +( 488.06 N )(0.22 m)

                                                              = 117.314 Nm

hence , required force   =   torque / radious of vertical circle

                                   =   117.314 Nm / 0.12 m

                                   =   977.61 N

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