The mechanism shown in ?Fig. P10.60? is used to raise a crate of supplies from a ship’s hold. The crate has total mass 50 kg. A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.25 m and moment of inertia 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 0.12 m, 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 1.40 m/s2? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

Solution 68P This question involves two torque terms. One is related to raise the crate and the other is to turn the cylinder. Let us now calculate each torque. Given, mass of the crate M = 5c kg Since, the crate moves up with an acceleration of 1.40 m/s , hence total force required to move the crate upward is F = 50 kg × (9.80 + 1.40) m/s = 560 N 1 Torque associated with this force is = F × 0.25 m ( radius of cylinder =0.25 m) 1 1 1 = 560 N × 0.25 m = 140 N.m Torque required to turn the cylinder is 2 = Moment of inertia × angular acceleration 2 = 2.9 kg.m ×2 1.40rad/s 2 0.25 = 16.24 N.m 2 Therefore, net torque involved in the process = + 1 2 = 140 N.m + 16.24 N.m = 156.24 N.m Radius of the vertical circle = 0.12 m Now, = F × 0.12 m F = 156.24 N.m/0.12 m = 1302 N Therefore, the approximate magnitude of the force is 1302 N .