The crane shown in the figure is lifting a 474-kg crate upward with an accelerat

Question

The crane shown in the figure is lifting a 474-kg crate upward with an acceleration of 2.87 m/s^2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 104 kg. The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is 230 kg, and its radius (the same as that of the pulley) is 0.944 m. The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

Explanation / Answer

The torque acting on the drum is

= -T1r1+ = Idd ,

  -T1r1+ = (mdr12)(a/r1

-T1r1+ = mdr1a

   = mdr1a + T1r

Apply Newton secon law to the pulley

= T1r1-T2r2 = Ipp

T1r1-T2r2 = (1/2)(mpr22)(a/r2)

                        T1r1 = (1/2)(mpr2a)+T2r2

the above two equations combined

   = mdr1a + (1/2)(mpr2a)+T2r2

the net force acting on the crate is

F = T2-mcg = mca

T2 = mc(a +g)

= mdr1a + (1/2)(mpr2a)+mc(a +g)r2

since r1= r2 = r

= [mdra + (1/2)(mpra)+mcar+mcgr]

        = (ra)[md + (1/2)(mp)+mc] +mcgr    

        = (0.944 m)(2.87 m/s2)[(230 kg)+(1/2)(104 kg)+(474 kg)]

           +(474 kg)(9.8 m/s2)(0.944 m)

        = 6433.28 N.m

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